[User]
In general relativity, the gravitational potential around a black hole of mass M is

                                                     GM m GM L2
                                         V (r) = −       −        .
                                                      r    mc2 r3
    The second term is a relativistic effect which strengthens the attraction towards the black hole. (It
    has nothing to do with the angular momentum barrier; you still have to add that separately.)

     (a) Explain why this new term allows particles to fall to the center of the black hole, r = 0, and
         why this is impossible in Newtonian gravity.

     (b) For a fixed L, find the values of the circular orbit radii.

      (c) Find the radius of the smallest possible stable circular orbit, for any value of L. What happens
          if you try to orbit the black hole closer than this?

     (d) Find the closest possible approach radius of an unbound object. That is, among the set of all
         trajectories that start and end far away from the black hole (i.e. without falling into it), find
         the smallest possible minimum value of r.

    For all parts, assume the particle is moving nonrelativistically.

       Example 4: Binney 5.1

       For over 150 years, most astronomers believed that Saturn’s rings were rigid bodies, until
       Laplace showed that a solid ring would be unstable. The same instability plagues Larry
       Niven’s Ringworld, a science fiction novel once popular among boomer nerds. Following

  Laplace, consider a rigid, circular ring of radius R and mass m, centered on a planet of
  mass M ≫ m. The ring rotates around the planet with the Keplerian angular velocity
  ω = (GM 2 /R3 )1/3 . Show that this configuration is linearly unstable.

  Solution
  One way to understand the stability of an ordinary planetary orbit is angular momentum
  conservation: if you displace a planet radially inward, then it’ll start moving faster
  tangentially, which will tend to make it go back out, even though the inward gravitational
  force gets stronger too. This tendency is absent for a rigid ring, because the entire ring
  always rotates with the same angular velocity ω = L/mR2 .

  The simplest way to see that this configuration is unstable is to calculate the gravitational
  potential ϕ due to the ring at the planet’s position. If the planet starts at the center of the
  ring, then displacing it along the axis of the ring increases ϕ. But since ∇2 ϕ = 0, displacing
  it towards the ring must decrease ϕ, so the system is unstable. (This is just the gravitational
  analogue of Earnshaw’s theorem from E1.)

  To make this more concrete, fix the planet at the origin, and parametrize the ring by the
  angle θ along it. If the whole ring is shifted by a small distance a in the plane of the ring,
  the elements of the ring are at

                                    r2 = (R cos θ + a)2 + (R sin θ)2 .

  The total gravitational potential energy is
                          Z 2π
                                         GM m 2π dθ
                                              Z
                               dθ 1                           1
            U = −GM m                =−             p                         .
                           0   2π r        R   0 2π 1 + (2a/R) cos θ + a2 /R2

  We have to be a bit careful here, remembering some lessons from P1. The first order term
  in a is going to vanish, because we started at an equilibrium point, which means we need to
  expand everything to second order in a. Using the Taylor series

                                         1     x 3x2
                                    √       =1− +    + O(x3 )
                                        1+x    2  8

  we conclude
                       Z 2π
                                               a2 3 cos2 θ − 1                 a2

             GM m             dθ       a                            GM m
        U ≈−                        1 − cos θ + 2                =−        1+       .
              R         0     2π       R       R       2             R        4R2

  The energy goes down upon a small displacement, so the configuration is unstable. The ring
  will soon crash into the planet.

3    Kepler’s Laws
    Idea 4
    Kepler’s laws for a general orbit are:

      1. The trajectories of planets are conic sections, with a focus at the Sun. Bound orbits are
         ellipses, which contain circles as a special case. Unbound orbits are hyperbolas, which
         contain parabolas as a special case.

      2. The trajectories sweep out equal areas in equal times.

      3. When the orbit is bound, the period T and semimajor axis a obey T 2 ∝ a3 .

    Unlike the other laws, the second is valid for any central force, because the rate of area
    sweeping is rv⊥ /2 ∝ |r × v| ∝ |L|.

    Idea 5
    For a general orbit with semimajor axis a, the total energy is
                                                   GM m
                                             E=−        .
                                                    2a
    This identity also applies to hyperbolas, where a is negative, and the parabola in the limit
    of infinite a, where the total energy vanishes.

    Idea 6
    An ellipse is defined by two foci F1 and F2 separated by a distance 2d. It consists of the
    set of points P so that P F1 + P F2√= 2a is a constant, where a is the semimajor axis. The
    semiminor axis b is related by a = b2 + d2 , as one can show by considering an appropriate
    right triangle, and the area is πab.

    Remark: Virial Theorem
    For bound orbits, the time-averaged values of the kinetic and potential energy are related by

                                             ⟨K⟩ = − ⟨V ⟩.

    In fact, the virial theorem holds for more complicated bound systems of particles as well, as
    long as they interact by a power law potential V (r) ∝ rn . In this case, we have
                                                      n
                                             ⟨K⟩ =      ⟨V ⟩

    where gravity corresponds to the case n = −1.

    You can easily check that the virial theorem works in one dimension for a particle bouncing
    in a uniform gravitational field (n = 1), or a particle on a spring (n = 2). It’s also
    easy to check for a planet in a circular orbit (n = −1). With some more work, you can

       check that it also holds for arbitrary elliptical orbits – to do this most efficiently, convert
       the time integral to an integral over angle θ, and use the form of an ellipse in polar coordinates.

       In astrophysics, the virial theorem is useful because it allows us to estimate V , which can be
       hard to measure, given K. For discussion of the virial theorem along with applications to
       dark matter, see section 1.4.3 of these notes. We will return to these subjects in X3.

Solution:
(a) The effective potential adds a term of L2 /2mr2 to the potential energy term, and
         that makes the effective potential go to +∞ as r → 0. Thus it would take an infinite amount
         of energy to reach r = 0 in Newtonian gravity (unless L = 0). However, the addition of the
         −GM L2 /mc2 r3 term makes the effective potential go to −∞ as r → 0, thus particles can fall
         to the center.

                                  ′ (r) = 0:
 (b) Circular orbits occur when Veff

                                         L2    GM m 3GM L2
                                             =     +
                                         mr3    r2   mc2 r4

                                                  L2    3GM L2
                                     GM mr2 −        r+        =0
                                                  m      mc2
                                                p
                                         L2 ±    L4 − 12(GM mL/c)2
                                    r=
                                                   2GM m2
     Note that there are no extrema
                               √    at all when the discriminant is negative. That is, circular
     orbits only exist when L > 12 GM m/c.

 (c) From part a), we know that limr→0 Veff (r) = −∞ and from part b), we know that there’s two
     1st order extremes. Thus the graph of Veff (r) should look like this, when L is large:

     The particle is trying to minimize its effective potential, so the qualitiative behavior can be
     easily understood p                         ′′ (r) > 0 indicates a stable equilibrium, the larger
                        from the graph. Since Veff
     radius r2 = (L2 + L4 − 12(GM mL/c)2 )/(2GM m2 ) is stable while the smaller one r1 is not.
     As the angular√momentum is decreased, r1 and r2 get closer together. When L reaches the
     critical value 12 GM m/c, the two radii coincide. For smaller L, the curve Veff (r) has no
     extrema, so there are no circular orbits at all.
     Therefore, the radius of the smallest
                                   √       stable circular orbit is the minimum possible value of r2 ,
     which is achieved when L = 12 GM m/c, giving

                                                         6GM
                                                rmin =       .
                                                          c2
     If you orbit in a circular orbit with a smaller radius, it is necessarily unstable, which means
     that under any perturbation, you will either drift into the black hole, or outward away from
     it. If you have rockets, this can be prevented by continual orbital adjustment. (Of course, if
     you try to orbit closer than the Schwarzschild radius 2GM/c2 , you don’t have any choice; you
     must fall into the black hole. But this effect isn’t visible in our simple Newtonian analysis.)

 (d) For this to happen, the effective potential needs√a maximum, so the particle can “bounce” off
     it and get back to r → ∞. Thus we need L > 12 GM m/c. For each value of L, the closest
     we can get while still bouncing off corresponds to the lower critical value of Veff (r), which was
     shown in part (b) to be
                                                 p
                                            L2 − L4 − 12(GM mL/c)2
                                rmin (L) =                               .
                                                      2GM m2
     This function is minimized when L → ∞ (i.e. when the particle is launched from a very large
     impact parameter), so the closest possible approach radius is
                                                    6(GM m/c)2   3GM
                                   lim rmin (L) =           2
                                                               =
                                  L→∞                 2GM m       c2
      where we used the binomial theorem in the first step. (Again, you could get slightly closer,
      but you would necessarily have to fire a rocket to make it back out.)
By the way, the shapes of the orbits in this potential can be quite exciting, featuring “zoom-whirl”
patterns where a particle slowly “zooms” around a black hole, then falls inward and quickly “whirls”
around it several times, then comes back out. Such orbits produce interesting gravitational wave
signatures. You can find a numeric simulation of them here.

   Example 4: Binney 5.1

   For over 150 years, most astronomers believed that Saturn’s rings were rigid bodies, until
   Laplace showed that a solid ring would be unstable. The same instability plagues Larry
   Niven’s Ringworld, a science fiction novel once popular among boomer nerds. Following
   Laplace, consider a rigid, circular ring of radius R and mass m, centered on a planet of
   mass M ≫ m. The ring rotates around the planet with the Keplerian angular velocity
   ω = (GM 2 /R3 )1/3 . Show that this configuration is linearly unstable.

   Solution
   One way to understand the stability of an ordinary planetary orbit is angular momentum
   conservation: if you displace a planet radially inward, then it’ll start moving faster
   tangentially, which will tend to make it go back out, even though the inward gravitational
   force gets stronger too. This tendency is absent for a rigid ring, because the entire ring
   always rotates with the same angular velocity ω = L/mR2 .

   The simplest way to see that this configuration is unstable is to calculate the gravitational
   potential ϕ due to the ring at the planet’s position. If the planet starts at the center of the
   ring, then displacing it along the axis of the ring increases ϕ. But since ∇2 ϕ = 0, displacing
   it towards the ring must decrease ϕ, so the system is unstable. (This is just the gravitational
   analogue of Earnshaw’s theorem from E1.)

   To make this more concrete, fix the planet at the origin, and parametrize the ring by the
   angle θ along it. If the whole ring is shifted by a small distance a in the plane of the ring,
   the elements of the ring are at

                                  r2 = (R cos θ + a)2 + (R sin θ)2 .

    The total gravitational potential energy is
                            Z 2π
                                           GM m 2π dθ
                                                Z
                                 dθ 1                           1
              U = −GM m                =−             p                         .
                             0   2π r        R   0 2π 1 + (2a/R) cos θ + a2 /R2

    We have to be a bit careful here, remembering some lessons from P1. The first order term
    in a is going to vanish, because we started at an equilibrium point, which means we need to
    expand everything to second order in a. Using the Taylor series

                                           1     x 3x2
                                      √       =1− +    + O(x3 )
                                          1+x    2  8

    we conclude
                         Z 2π
                                                 a2 3 cos2 θ − 1                 a2

               GM m             dθ       a                            GM m
          U ≈−                        1 − cos θ + 2                =−        1+       .
                R         0     2π       R       R       2             R        4R2

    The energy goes down upon a small displacement, so the configuration is unstable. The ring
    will soon crash into the planet.

3    Kepler’s Laws
    Idea 4
    Kepler’s laws for a general orbit are:

      1. The trajectories of planets are conic sections, with a focus at the Sun. Bound orbits are
         ellipses, which contain circles as a special case. Unbound orbits are hyperbolas, which
         contain parabolas as a special case.

      2. The trajectories sweep out equal areas in equal times.

      3. When the orbit is bound, the period T and semimajor axis a obey T 2 ∝ a3 .

    Unlike the other laws, the second is valid for any central force, because the rate of area
    sweeping is rv⊥ /2 ∝ |r × v| ∝ |L|.

    Idea 5
    For a general orbit with semimajor axis a, the total energy is
                                                    GM m
                                              E=−        .
                                                     2a
    This identity also applies to hyperbolas, where a is negative, and the parabola in the limit
    of infinite a, where the total energy vanishes.

       Idea 6
       An ellipse is defined by two foci F1 and F2 separated by a distance 2d. It consists of the
       set of points P so that P F1 + P F2√= 2a is a constant, where a is the semimajor axis. The
       semiminor axis b is related by a = b2 + d2 , as one can show by considering an appropriate
       right triangle, and the area is πab.

       Remark: Virial Theorem
       For bound orbits, the time-averaged values of the kinetic and potential energy are related by

                                                 ⟨K⟩ = − ⟨V ⟩.

       In fact, the virial theorem holds for more complicated bound systems of particles as well, as
       long as they interact by a power law potential V (r) ∝ rn . In this case, we have
                                                            n
                                                  ⟨K⟩ =       ⟨V ⟩

       where gravity corresponds to the case n = −1.

       You can easily check that the virial theorem works in one dimension for a particle bouncing
       in a uniform gravitational field (n = 1), or a particle on a spring (n = 2). It’s also
       easy to check for a planet in a circular orbit (n = −1). With some more work, you can
       check that it also holds for arbitrary elliptical orbits – to do this most efficiently, convert
       the time integral to an integral over angle θ, and use the form of an ellipse in polar coordinates.

       In astrophysics, the virial theorem is useful because it allows us to estimate V , which can be
       hard to measure, given K. For discussion of the virial theorem along with applications to
       dark matter, see section 1.4.3 of these notes. We will return to these subjects in X3.


[User]
USAPhO 2024, problem B1. Analyzing the collision rate in a muon collider

                     01h
    using relativistic kinematics and dynamics.

Solution:
USAPhO 2024, problem B1. Analyzing the collision rate in a muon collider
    using relativistic kinematics and dynamics.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
A simple derivation of Kepler’s first law is given in section 7.4 of Morin, and
    centers around solving a differential equation for 1/r(θ). (You can motivate this by noting that the
    polar form of an ellipse is quite simple, 1/r = (1 + e cos θ)/p, where p is the semilatus rectum and
    e is the eccentricity.) However, in this problem, we’ll consider an alternative approach that uses a
    subtle conserved quantity, which is also important in more advanced physics.

      (a) Show that the Laplace–Runge–Lenz vector

                                               A = p × L − GM m2 r̂

          is conserved, where the star is at the origin and r̂ is the radial unit vector at the planet’s
          position r. (Hint: use the fact that L = mr2 ω to evaluate the time derivative.)

     (b) We have A · r = Ar cos θ, where θ is the angle between A and r. Evaluate A · r using the
         definition of A, and the identity a · (b × c) = (a × b) · c, in order to derive an expression for
         r in terms of θ and constants. Then use this to show that the orbit is a conic section.

 (c) As another simple application of the conservation of A, show that the set of velocities during
     an elliptical orbit traces out a circle in velocity space.

The ideas discussed in this problem are almost never required to solve Olympiad problems, but they
can dramatically simplify very tough orbital mechanics problems. For two examples, see Physics
Cup 2021, problem 2 and Physics Cup 2024, problem 4.

Now we’ll consider some really slick problems that can be solved with pure geometry.

   Example 5

   An object is dropped from rest at a distance R above the Earth’s surface, where R is the
   radius of the Earth. How long does it take to hit the Earth’s surface?

   Solution
   The answer doesn’t change much if we give the object a tiny horizontal velocity. In this case,
   the orbit becomes a part of a very thin ellipse, where a ≈ d ≈ ℓ, with one focus at the center
   of the Earth (by the shell theorem) and the other near the starting point.

   If the Earth were replaced by a point mass at its center, then the object could perform a
   full orbit, with total period T . The time until the object actually hits the Earth’s surface is
   determined by the fraction of the orbit’s area swept out. Referring to the diagram, this is

                                       πab/4 + ab/2         1    1
                                 t=T                  =T      +
                                            πab             4 2π

   by summing a quarter of an ellipse and a triangle. All that’s left is to solve for T . Note that
   the semimajor axis is R. Another orbit with the same semimajor axis is simply a circular
   orbit around the Earth, just above its surface. This orbit has

                                             v2  GM
                                                = 2
                                             R   R
            p
   so v =    GM/R. Using T = 2πR/v gives the answer,
                                               r
                                      π         R3
                                  t=      +1         .
                                        2        GM

       Example 6: MPPP 39

       An astronaut jumps out of the international space station directly towards the Earth. What
       happens afterward? In particular, will the astronaut survive?

       Solution
       If you’ve seen certain movies, you might get the impression that the astronaut spirals into
       the Earth, and so will surely die. But that isn’t what Kepler’s laws say! After the jump, the
       astronaut simply performs a Keplerian orbit. Since the change in energy is negligible, so is
       the change in semimajor axis and hence the change in period. The astronaut simply orbits
       in a nearly circular ellipse, with the same period as the space station.

       After one rotation period of the space station, which takes time T = 92 min, the astronaut
       arrives back. They are unharmed as long as their oxygen and cooling supply lasts this long.
       (If you draw some pictures of the orbits, you may think the answer is T /2, because the orbits
       intersect twice. This is incorrect because while the orbits do intersect geometrically halfway
       through, the space station and the astronaut won’t arrive at that point at the same time.)

       Example 7: Wang and Ricardo 8.4

       A particle moves in a circle of radius R, under the influence of a central force. If its minimum
       and maximum speeds are v1 and v2 , what is the period T ?

       Solution
       At first the problem statement might sound confusing, until you realize that the ori-
       gin need not be at the center of the circle; it must be off-center. Now, it would be
       intractable to find the trajectory for a general central force law, but we can infer T by
       thinking about how quickly area is swept out, as in Kepler’s second law. This works because
       conservation of angular momentum holds for all central force laws, not just the inverse square.

       At the furthest and closest points, the distances from the origin must be r1 and r2 , and by
       conservation of angular momentum, the speeds v1 and v2 are achieved at these points, so
                                                                     dA  1
                                r1 v1 = r2 v2 ,   r1 + r2 = 2R,         = r1 v1 .
                                                                     dt  2
       Using the first two equations, we can solve for r1 and plug it into the third, for
                                             2R              dA        R
                                   r1 =              ,          =             .
                                          1 + v1 /v2         dt   1/v1 + 1/v2

       Since dA/dt = πR2 /T , we have

                                                             1   1
                                              T = πR           +         .
                                                             v1 v2

[3] Problem 8 (PPP 88). A rocket is launched from and returns to a spherical planet of radius R so

    that its velocity vector on return is parallel to its velocity vector at launch. The angular separation
    at the center of the planet between the launch and arrival points is θ. How long does the flight take,
    if the period of a satellite flying around the planet just above its surface is T0 ?

[4] Problem 9 (Physics Cup 2012). A cannon at the equator fires a cannonball, which hits the North
    pole. Neglecting air resistance and the Earth’s rotation, at what angle to the horizontal should the
    cannonball be fired to minimize the required speed?

[4] Problem 10 (EFPhO 2015). An asteroid is initially stationary, a distance R from a star of mass
    M . The asteroid suddenly explodes into many pieces, with speed ranging from zero to v0 . What is
    the set of all points that can be hit by a piece of the asteroid? (Hint: this problem requires more
    geometry than the rest. For simplicity, you can begin by treating the problem as two-dimensional,
    but the solution you find will work just as well for three.)

       Idea 7: Reduced Mass
       Consider two objects of mass m1 and m2 with positions r1 and r2 with relative position
       r = r1 − r2 , interacting by a central potential V (r). For the purposes of computing r alone,
       we may replace this system with a single mass µ in the same central potential V (r), where µ
       is the reduced mass, obeying
                                               1     1      1
                                                  =      +     .
                                               µ    m1 m2
       Both systems have the same solutions for r(t).

       Example 8

       Consider two planets of mass m. If one planet is somehow fixed in place, the other can
       perform a circular orbit of radius R with period T . If both planets are allowed to move, they
       can simultaneously perform circular orbits of radius R/2 about their center of mass. What
       is the period of this motion?

       Solution
       First let’s try an explicit solution. In the first case,
                                                                  r
                                          mv 2   Gm2                  Gm
                                               =     ,      v=           .
                                           R      R2                   R
       In the second case, we have
                                                             r
                                       mv 2     Gm2       1    Gm
                                             =      , v=√          .
                                       R/2       R2        2    R
                                                    √
       The velocity in this case is a factor of√1/ 2 smaller, but the arc length of the orbit is a
       factor of 2 smaller, so the period is T / 2.

       We can also handle the problem with reduced mass. Consider the relative position r1 − r2 in
       the second case, which orbits in a circle of radius R. Applying the above idea, we can work
       in the reduced system. In this system, there is a single mass µ = (1/m + 1/m)−1 = m/2 in

       a circular orbit of radius R, experiencing the same force Gm2 /R2 , so
                                                           r
                                        µv 2   Gm2            2Gm
                                             =    2
                                                    , v=           .
                                         R      R               R
                     √
       The speed is 2√bigger than in the first case, but the arc length of the orbit is the same, so
       the period is T / 2.

    Reduced mass is a bit unintuitive, since you need to work in two very different pictures. On the
    other hand, some people like it because it’s mathematically concrete, and can reduce problems to
    one-liners. Whether you use it is up to you.
[2] Problem 11 (MPPP 27). Two permanent magnets are aligned on a horizontal frictionless table,
    separated by a distance d. The magnets are held in such a way so that the net force between them
    is attractive, and there are no torques generated.
        If one of the magnets is held and the other is released, the two collide after time t1 . If instead
    the roles are reversed, the two collide after time t2 . If instead both magnets are released from rest,

                    01W
    how long does it take for them to collide? (For a simpler related problem, see F = ma 2019 B4.)

Solution:
(a) Since the angular momentum is conserved,

                                      dr̂   GM m
               Ȧ = F × L − GM m2         =      (ωmr2 )(−r̂ × ẑ) − GM m2 (ωẑ × r̂) = 0
                                      dt     r2
     as desired.

 (b) We have

                   A · r = r · (p × L) − GM m2 r = (r × p) · L − GM m2 r = L2 − GM m2 r

     which tells us that
                                        Ar cos θ = L2 − GM m2 r.
     But now this can be solved for r to give the trajectory,

                                                    L2
                                          r=                   .
                                               GM m2 + A cos θ
     This is precisely the form of a conic section. Specifically, the general form is
                                                        p
                                               r=
                                                    1 + e cos θ
     and we can identify
                                           L2              A
                                       p=       2
                                                  , e=           .
                                         GM m            GM m2
     As a check, note that A indeed vanishes for circular motion, where

                                                2       2 mv       GM m
                       A = (mv)(mvr) − GM m = mr                −         = 0.
                                                            r       r2

     For an elliptical orbit, A lies in the plane of the orbit and points along the major axis.

 (c) Take the cross product of the vector with L, which is always conserved, for
                              (A + GM m2 r̂) × L = (p × L) × L = −mL2 v
     since p is always perpendicular to L. Now, during an elliptical orbit, the values of A+GM m2 r̂
     trace out a circle because A is conserved and r̂ has constant magnitude. Since A and r̂ are
     perpendicular to L, taking the cross product with L just scales the circle and rotates it by
     90◦ in the orbit plane, so the set of v lies on a circle.
Now we’ll consider some really slick problems that can be solved with pure geometry.

   Example 5

   An object is dropped from rest at a distance R above the Earth’s surface, where R is the
   radius of the Earth. How long does it take to hit the Earth’s surface?

   Solution
   The answer doesn’t change much if we give the object a tiny horizontal velocity. In this case,
   the orbit becomes a part of a very thin ellipse, where a ≈ d ≈ ℓ, with one focus at the center
   of the Earth (by the shell theorem) and the other near the starting point.

   If the Earth were replaced by a point mass at its center, then the object could perform a
   full orbit, with total period T . The time until the object actually hits the Earth’s surface is
   determined by the fraction of the orbit’s area swept out. Referring to the diagram, this is

                                       πab/4 + ab/2         1    1
                                 t=T                  =T      +
                                            πab             4 2π

   by summing a quarter of an ellipse and a triangle. All that’s left is to solve for T . Note that
   the semimajor axis is R. Another orbit with the same semimajor axis is simply a circular
   orbit around the Earth, just above its surface. This orbit has

                                             v2  GM
                                                = 2
                                             R   R
            p
   so v =    GM/R. Using T = 2πR/v gives the answer,
                                               r
                                      π         R3
                                  t=      +1         .
                                        2        GM

       Example 6: MPPP 39

       An astronaut jumps out of the international space station directly towards the Earth. What
       happens afterward? In particular, will the astronaut survive?

       Solution
       If you’ve seen certain movies, you might get the impression that the astronaut spirals into
       the Earth, and so will surely die. But that isn’t what Kepler’s laws say! After the jump, the
       astronaut simply performs a Keplerian orbit. Since the change in energy is negligible, so is
       the change in semimajor axis and hence the change in period. The astronaut simply orbits
       in a nearly circular ellipse, with the same period as the space station.

       After one rotation period of the space station, which takes time T = 92 min, the astronaut
       arrives back. They are unharmed as long as their oxygen and cooling supply lasts this long.
       (If you draw some pictures of the orbits, you may think the answer is T /2, because the orbits
       intersect twice. This is incorrect because while the orbits do intersect geometrically halfway
       through, the space station and the astronaut won’t arrive at that point at the same time.)

       Example 7: Wang and Ricardo 8.4

       A particle moves in a circle of radius R, under the influence of a central force. If its minimum
       and maximum speeds are v1 and v2 , what is the period T ?

       Solution
       At first the problem statement might sound confusing, until you realize that the ori-
       gin need not be at the center of the circle; it must be off-center. Now, it would be
       intractable to find the trajectory for a general central force law, but we can infer T by
       thinking about how quickly area is swept out, as in Kepler’s second law. This works because
       conservation of angular momentum holds for all central force laws, not just the inverse square.

       At the furthest and closest points, the distances from the origin must be r1 and r2 , and by
       conservation of angular momentum, the speeds v1 and v2 are achieved at these points, so
                                                                     dA  1
                                r1 v1 = r2 v2 ,   r1 + r2 = 2R,         = r1 v1 .
                                                                     dt  2
       Using the first two equations, we can solve for r1 and plug it into the third, for
                                             2R              dA        R
                                   r1 =              ,          =             .
                                          1 + v1 /v2         dt   1/v1 + 1/v2

       Since dA/dt = πR2 /T , we have

                                                             1   1
                                              T = πR           +         .
                                                             v1 v2

[3] Problem 8 (PPP 88). A rocket is launched from and returns to a spherical planet of radius R so

    that its velocity vector on return is parallel to its velocity vector at launch. The angular separation
    at the center of the planet between the launch and arrival points is θ. How long does the flight take,
    if the period of a satellite flying around the planet just above its surface is T0 ?

    Solution. Note that the starting and ending points of the trajectory are the two ends of the minor
    axis of that trajectory (which is an ellipse by Kepler’s first law). Therefore, the distance from focus
    to an end of the minor axis is R, so the sum of the distances from the two foci to an end of the
    minor axis is 2R, so 2a = 2R, or a = R. By Kepler’s third law the period depends only on a, so the
    period of this orbit would also be T0 if the Earth were a point mass. However, only part of the orbit
    is actually performed. Hence we need to find the fraction of the ellipse’s area that is swept out.

       We see that the area swept out is A = 12 A0 + 12 a2 sin θ where A0 = πab = πa2 sin 2θ is the area of
    the full ellipse. Thus,
                                                           1 cos 2θ
                                       T /T0 = A/A0 =         +       .
                                                           2      π
[4] Problem 9 (Physics Cup 2012). A cannon at the equator fires a cannonball, which hits the North
    pole. Neglecting air resistance and the Earth’s rotation, at what angle to the horizontal should the
    cannonball be fired to minimize the required speed?

    Solution. The answer is 22.5◦ . See the solutions here.

[4] Problem 10 (EFPhO 2015). An asteroid is initially stationary, a distance R from a star of mass
    M . The asteroid suddenly explodes into many pieces, with speed ranging from zero to v0 . What is
    the set of all points that can be hit by a piece of the asteroid? (Hint: this problem requires more
    geometry than the rest. For simplicity, you can begin by treating the problem as two-dimensional,
    but the solution you find will work just as well for three.)

    Solution. Consider a given piece of the asteroid, with speed v0 . Its trajectory is an ellipse of major
    axis 2a, where a satisfies
                                           1        GM m       GM m
                                             mv02 −         =         .
                                           2           R         2a
    Let S be the position of the sun, and let P be the original point of the asteroid. Let F be the other
    focus of this elliptical trajectory. By the definition of an ellipse, SP + F P = 2a, so F P = 2a − R.

   Let Q be a point that is reached by the piece, and suppose SQ = r. By the triangle inequality,

                         P Q ≤ P F + F Q = 2a − R + 2a − r = 4a − r − R.

Therefore, we see that
                                        QP + QS ≤ 4a − R.
This constraint applies to all points Q that can be hit. Thus, the points that can be hit lie within
an ellipse with foci at the sun and the asteroid, with major axis 4a − R.
   Is it possible to hit every point in this ellipse? It’s intuitive that it’s sufficient to show that
we can hit every point on the boundary, since that’s the hardest thing to do; points inside can
be reached by launching with reduced speed. Let Q be a given point on the boundary. Then the
inequalities above become equalities as long as P Q = P F + F Q, which occurs when F is on P Q.
So the question is reduced to whether we can put the other focus of the orbit at any angle we want,
relative to P . If you play around with a few drawings of orbits, you can see this is always possible
by varying the launch angle, no matter what the launch speed is, so we can get the full ellipse.
   In three dimensions, the answer is the set of points enclosed by rotating the ellipse about the
axis P S. The resulting shape is called a spheroid.

   Idea 7: Reduced Mass
   Consider two objects of mass m1 and m2 with positions r1 and r2 with relative position
   r = r1 − r2 , interacting by a central potential V (r). For the purposes of computing r alone,
   we may replace this system with a single mass µ in the same central potential V (r), where µ
   is the reduced mass, obeying
                                           1     1      1
                                              =      +     .
                                           µ    m1 m2
   Both systems have the same solutions for r(t).

   Example 8

   Consider two planets of mass m. If one planet is somehow fixed in place, the other can
   perform a circular orbit of radius R with period T . If both planets are allowed to move, they
   can simultaneously perform circular orbits of radius R/2 about their center of mass. What
   is the period of this motion?

       Solution
       First let’s try an explicit solution. In the first case,
                                                                     r
                                              mv 2   Gm2                 Gm
                                                   =     ,      v=          .
                                               R      R2                  R
       In the second case, we have
                                                             r
                                       mv 2     Gm2       1    Gm
                                             =     2
                                                     , v=√         .
                                       R/2       R         2    R
                                                     √
       The velocity in this case is a factor of√1/ 2 smaller, but the arc length of the orbit is a
       factor of 2 smaller, so the period is T / 2.

       We can also handle the problem with reduced mass. Consider the relative position r1 − r2 in
       the second case, which orbits in a circle of radius R. Applying the above idea, we can work
       in the reduced system. In this system, there is a single mass µ = (1/m + 1/m)−1 = m/2 in
       a circular orbit of radius R, experiencing the same force Gm2 /R2 , so
                                                            r
                                        µv 2   Gm2            2Gm
                                             =       , v=          .
                                         R      R2              R
                     √
       The speed is 2√bigger than in the first case, but the arc length of the orbit is the same, so
       the period is T / 2.

    Reduced mass is a bit unintuitive, since you need to work in two very different pictures. On the
    other hand, some people like it because it’s mathematically concrete, and can reduce problems to
    one-liners. Whether you use it is up to you.

[2] Problem 11 (MPPP 27). Two permanent magnets are aligned on a horizontal frictionless table,
    separated by a distance d. The magnets are held in such a way so that the net force between them
    is attractive, and there are no torques generated.
        If one of the magnets is held and the other is released, the two collide after time t1 . If instead
    the roles are reversed, the two collide after time t2 . If instead both magnets are released from rest,
    how long does it take for them to collide? (For a simpler related problem, see F = ma 2019 B4.)

    Solution. Let U (r) be the potential energy of the two magnets as a function of r, the distance
    between the two edges of the magnets (r starts off as d, ends at 0). Let the magnet on the right
    have mass m1 , and let the one on the left have mass m2 .
       Consider the case where m2 is fixed. Say the time it takes to collide is t1 , and let v(r) be the
    speed of m1 . By conservation of energy, we have that

                                          m1 v(r)2 = U (d) − U (r) ≡ V (r),

                q
                    2V (r)
    so v(r) =        m1 . Then, v = −dr/dt, so

                                        Z d               Z dr
                                                                  m1        √
                                 t1 =         dr/v(r) =                 dr ≡ m1 τ.
                                         0                0      2V (r)

                     √
   Similarly, if t2 = m2 τ is the time it takes if m1 is fixed.
      Now consider the case where they are both free, and let the velocity of mi be vi . Then,

                                                  m 1 v1 = m 2 v2
                                          m1 v1 + m2 v22 = 2V (r).

   Solving this, we get
                                                   s
                                                        2m2 V (r)
                                        v1 (r) =
                                                       m1 (m1 + m2 )
                                                   s
                                                        2m1 V (r)
                                        v2 (r) =                     .
                                                       m2 (m1 + m2 )

   Now, −dr/dt = v1 (r) + v2 (r), so the time T we are interested in is
                                        Z d
                                                   dr
                                    T =
                                          0 v1 (r) + v2 (r)
                                        Z d
                                                          dr
                                      =      q           q        q
                                          0     2V (r)       m2       m1
                                               m1 +m2        m1  +    m2
                                        Z dr
                                                       m1 m2
                                      =                               dr
                                          0    (m 1 +   m2 ) · 2V (r)
                                        r
                                             m1 m2
                                      =                τ
                                            m1 + m2
                                         t1 t2
                                      = p2        .
                                         t1 + t22


[User]
Consider a uniformly charged ball of total charge Q and radius R. Decompose this
    ball into two parts, A and B, where B is a ball of radius R/2 whose center is a distance R/2 of the
    ball’s center, and A is everything else. Find the potential energy due to the interaction of A and B,
    i.e. the work necessary to bring in B from infinity, against the field of A.
[2] Problem 22 (PPP 149). A distant planet is at a very high electric potential compared with Earth,
    say 106 V higher. A metal space ship is sent from Earth for the purpose of making a landing on the
    planet. Is the mission dangerous? What happens when the astronauts open the door on the space
    ship and step onto the surface of the planet?

       Example 8

       Since Newton’s law of gravity is so similar to Coulomb’s law, the results we’ve seen so far
       should have analogues in Newtonian gravity. What are they? For example, what’s the
       gravitational Gauss’s law?

       Solution
       The fundamental results to compare are
                                               Gm1 m2                q1 q2
                                       F =−           ,    F =
                                                 r2                 4πϵ0 r2
       where the minus sign indicates that the gravitational force is attractive, while the electrostatic
       force between like charges is repulsive. Then we can transform a question involving (only
       positive) electric charges to one involving masses if we map

                                      q → m,          → −G,         E → g.
                                                 4πϵ0
       Thus, while electrostatics is described by
                                                                I
                                                    ρ                           Q
                                 ∇ × E = 0,     ∇·E= ,               E · dS =      ,
                                                    ϵ0                          ϵ0

  the gravitational field is described by
                                                          I
                      ∇ × g = 0,     ∇ · g = −4πGρm ,         g · dS = −4πGM

  where ρm is the mass density. Similarly, the potential energy can be written in two ways,
                               Z                          Z
                             1                3        1
                        U=        ρm (r)ϕ(r) d r = −         |g(r)|2 d3 r
                             2                       8πG

  where ϕ(x) is the gravitational potential. This result was first written down by Maxwell.

  Remark
  Here’s a philosophical question: is potential energy “real”? You likely think the answer
  is obvious, but about half of your friends probably think the opposite answer is obviously
  correct! In fact, in the 1700s, there was a lively debate over whether the ideas of kinetic
  energy and momentum, which at the time were given various other names, were worthwhile.
  Which one of the two was the true measure of motion? In our modern language, proponents
  of energy pointed out that the momentum always vanished in the center of mass frame,
  which made it “trivial”, while supporters of momentum replied that kinetic energy was
  clearly not conserved in even the simplest of cases, like inelastic collisions.

  In the 1800s, thermodynamics was developed, allowing the energy seemingly lost in inelastic
  collisions to be accounted for as internal energy. But there still remained the problem that
  kinetic energy was lost in simple situations, such as when balls are thrown upward. By the
  mid-1800s, the modern language that “kinetic energy is converted to potential energy” was
  finally standardized, but it was still common to read in textbooks that potential energy was
  fake, a mathematical trick used to patch up energy conservation. After all, potential energy
  has some suspicious qualities. If a ball has lots of potential energy, you can’t see or feel it, or
  even know it’s there by considering the ball alone. It doesn’t seem to be located anywhere
  in space, and its amount is arbitrary, as a constant can always be added.

  In the late 1800s, a revolution on physics answered some of these questions. Maxwell and his
  successors recast electromagnetism as a theory of fields, and showed that the dynamics of
  charges and currents were best understood by allowing the fields themselves to carry energy
  and momentum. We’ll cover this in detail in E7, but for now, it implies that electrostatic
  potential energy is fundamentally stored in the field, with a density of ϵ0 E 2 /2. This implies
  that its location and total amount are directly measurable.

  Maxwell believed that the dynamics of fields emerged from the microscopic motions and
  elastic deformations of an all-pervading ether, in the same way that, say, a fluid’s velocity
  field emerges from the average motion of fluid molecules. This makes it manifestly positive,
  so he was disturbed to find that the energy density of a gravitational field is negative! He
  therefore concluded that gravity could not be described as a vector field.

  A few decades later, the arrival of special relativity answered some questions and reopened

  others. On one hand, it demolished Maxwell’s vision of the ether. On the other hand, it finally
  answered the question of whether all kinds of potential energy are “real”, and it got rid of
  the freedom to add arbitrary constants. That’s because in special relativity, the total energy
  of a system at rest is related to its mass by E = mc2 , and the mass is directly measurable.
  This finally puts thermal energy, elastic potential energy, and field energy on an equal footing.

  Here’s the most modern view of energy conservation. All particles and their interactions are
  fundamentally described by relativistic quantum fields. A famous result called Noether’s
  theorem implies that whenever such a theory is time-translationally symmetric, there
  is a conserved quantity which we call the energy. (The distinction between kinetic and
  potential energy becomes irrelevant; it’s all just energy.) The density of energy in space
  can be computed from the state of the fields, but it doesn’t need to be explained, as
  Maxwell imagined, by the internal motion of whatever the fields are made of. The fields are
  fundamental: they aren’t made of anything; instead, they make up everything!

  What happens when we throw gravity into the mix? As we’ll discuss further in R3, it turns
  out that at nonrelativistic velocities, the dynamics of gravitating particles can be described
  by “gravitoelectromagnetism”, a theory closely analogous to electromagnetism, where moving
  masses also source “gravitomagnetic” fields Bg , which result in mv × Bg forces. But the
  situation gets much more subtle when we upgrade to full general relativity. Here, the notion
  of a gravitational field disappears completely, and is replaced by the curvature of spacetime,
  making it hard to define an energy density for it at all. For an accessible overview of the
  debate, see this paper. Ultimately, though, it doesn’t matter that much, since it doesn’t
  impair our ability to use either Newtonian gravity or general relativity.

  Example 9

  For an infinite line of linear charge density λ, find the potential V (r) by dimensional analysis.

  Solution
  This example illustrates a famous subtlety of dimensional analysis. The only quantities in
  the problem with dimensions are λ, ϵ0 , and r. To get the electrical units to balance, we have
                                                     λ
                                           V (r) =      f (r)
                                                     ϵ0
  where f (r) is a dimensionless function. But there are no nontrivial dimensionless functions
  of a single dimensionful quantity r. The only possibilities are that f (r) is a dimensionless
  constant, or that f (r) is infinite. In the first case, the electric field would vanish, which can’t
  be right. In the second case, it is unclear how to calculate the electric field at all.

  In fact, the electric potential is infinite, if you insist on the usual convention of setting
  V (∞) = 0. In that case, we have
                                    Z ∞              Z ∞
                                                            λ dr
                            V (r) =       E(r) dr =               =∞
                                      r               r   2πϵ0 r

    for any r. But this is useless; to get a finite result we can actually work with, we need to
    subtract off an infinite constant from the potential. Equivalently, we need to set the potential
    to be zero at some finite distance r = r0 . This process is known as renormalization, and it is
    extremely important in modern physics. After renormalization, we have
                                           Z r0
                                                            λ       r0
                                   V (r) =      E(r) dr =       log
                                            r             2πϵ0      r

    which is perfectly consistent with dimensional analysis.

    Notice that in the process of renormalization, a new dimensionful quantity r0 appeared out
    of nowhere. This phenomenon is known as dimensional transmutation. Nothing measurable
    depends on this new scale (e.g. the electric field is independent of r0 ), but you can’t write
    down quantities like the potential without it.

3    Conductors
    Idea 9
    In electrostatic conditions, E = 0 inside a conductor, which implies the conductor has constant
    electric potential V . This further implies that E is always perpendicular to a conductor’s
    surface. By Gauss’s law, the conductor has ρ = 0 everywhere inside, so all charge resides on
    the surface.

    Example 10

    Consider a point on the surface of a conductor with surface charge density σ. Show that the
    outward pressure on the charges at this point is σ 2 /2ϵ0 .

    Solution
    Gauss’s law tells us that the difference of the electric fields right inside and outside the
    conductor at this point is
                                                        σ
                                          Eout − Ein =
                                                       ϵ0
    by drawing a pillbox-shaped Gaussian surface. But we also know that Ein = 0 since we’re
    dealing with a conductor, so Eout = σ/ϵ0 .

    Let’s think about how this electric field is made. If there were no charges around except for
    the ones at this surface, then the interior and exterior fields would have been ±σ/2ϵ0 . This
    means that all of the other charges, that lie elsewhere on the surface of the conductor, must
    provide a field σ/2ϵ0 here, so that Ein cancels out.

    The pressure on the charges at this point on the surface is equal to the product of the surface
    charge density with the field due to the rest of the charges, since the charges at this point

       can’t exert an overall force on themselves, so

                                                            σ2

                                                      σ
                                           P =σ           =
                                                    2ϵ0     2ϵ0
                                                               2 /2.
       as required. Equivalently, we can conclude that P = ϵ0 Eout

       Example 11

       Is the charge density at the surface of a charged conductor usually greater at regions of higher
       or lower curvature?

       Solution
       We can’t answer this question in general, because it is usually impossible to solve for
       the charge distribution of an irregularly shaped conductor. Charges at any point in the
       conductor will influence the charges everywhere else.

       However, we can get some insight by considering the limiting case of a conductor made of
       two spheres of radii R1 and R2 , connected by a very long rod. For the potential to be the
       same at both spheres, we must have Q1 /R1 = Q2 /R2 , so the charge is proportional to the
       radius, and the charge density is inversely proportional to the radius. Thus, there’s generally
       higher charge density at sharper points of the conductor, provided that those points are sharp
       enough, or far enough away from the rest of conductor for the charges on the rest of the
       conductor not to matter much. That’s basically all we can say for sure.

Solution:
If we tried to compute the potential energy directly, by integrating over A and B, we
    would get messy integrals. Instead, let’s consider bringing in B in three steps:

       1. At infinity, compress B into a point charge Q/8.

       2. Move this point charge to the center of the B-shaped hole in A.

       3. Expand the point charge back into the original shape of B.

    Our first claim is that the total work needed to do steps (1) and (3) is zero. These two steps are very
    close to being opposites; the only difference in that in step (3), the expansion takes place within
    the field of A. By the same reasoning as in problem 1, the field of A within the B-shaped hole is
    constant, with magnitude
                                                  k(Q/8)      kQ
                                             E=          2
                                                           =
                                                   (R/2)      2R2
    This constant field does no net work when B is expanded, because the positive work done on one
    half of B is cancelled by the negative work on the other half.
       Therefore, we only have to calculate the work done for step (2), which is easy. Applying
    superposition and the shell theorem, the work needed to bring the point charge to the point where
    the surface of A meets the surface of the B-shaped hole is

                                              Q kQ k(Q/8)
                                        W1 =          −            .
                                               8   R      R/2

    Next, moving the point charge from this point to the center of the B-shaped hole takes work

                                                   RQ     kQ2
                                            W2 =       E=     .
                                                   2 8    32R
    The total work is
                                                           kQ2
                                              W1 + W2 =        .
                                                           8R
[2] Problem 22 (PPP 149). A distant planet is at a very high electric potential compared with Earth,
    say 106 V higher. A metal space ship is sent from Earth for the purpose of making a landing on the
    planet. Is the mission dangerous? What happens when the astronauts open the door on the space
    ship and step onto the surface of the planet?

    Solution. As the space ship approaches the planet, its potential gradually increases from that of
    the Earth, to that of the distant planet. Meanwhile, all the astronauts inside are doing just fine
    since the ship acts like a Faraday cage. Once the ship lands, it’s already at the same potential as
    the planet, and when the astronauts step out, nothing happens. In other words, it’s electric field
    that’s dangerous, not potential, and the electric fields in this problem are always small.
       Another way to see that there’s no danger is to replace electric fields with gravitational fields,
    and thus electric potential with gravitational potential. An elevator in a skyscraper takes you from
    a low to a very high gravitational potential. But nothing violent happens when you get off!

       Example 8

       Since Newton’s law of gravity is so similar to Coulomb’s law, the results we’ve seen so far
       should have analogues in Newtonian gravity. What are they? For example, what’s the
       gravitational Gauss’s law?

  Solution
  The fundamental results to compare are
                                            Gm1 m2             q1 q2
                                   F =−            ,   F =
                                              r2              4πϵ0 r2
  where the minus sign indicates that the gravitational force is attractive, while the electrostatic
  force between like charges is repulsive. Then we can transform a question involving (only
  positive) electric charges to one involving masses if we map

                                 q → m,           → −G,       E → g.
                                             4πϵ0
  Thus, while electrostatics is described by
                                                          I
                                                ρ                         Q
                            ∇ × E = 0,      ∇·E= ,             E · dS =      ,
                                                ϵ0                        ϵ0
  the gravitational field is described by
                                                          I
                      ∇ × g = 0,     ∇ · g = −4πGρm ,          g · dS = −4πGM

  where ρm is the mass density. Similarly, the potential energy can be written in two ways,
                               Z                          Z
                             1                3        1
                        U=        ρm (r)ϕ(r) d r = −         |g(r)|2 d3 r
                             2                       8πG

  where ϕ(x) is the gravitational potential. This result was first written down by Maxwell.

  Remark
  Here’s a philosophical question: is potential energy “real”? You likely think the answer
  is obvious, but about half of your friends probably think the opposite answer is obviously
  correct! In fact, in the 1700s, there was a lively debate over whether the ideas of kinetic
  energy and momentum, which at the time were given various other names, were worthwhile.
  Which one of the two was the true measure of motion? In our modern language, proponents
  of energy pointed out that the momentum always vanished in the center of mass frame,
  which made it “trivial”, while supporters of momentum replied that kinetic energy was
  clearly not conserved in even the simplest of cases, like inelastic collisions.

  In the 1800s, thermodynamics was developed, allowing the energy seemingly lost in inelastic
  collisions to be accounted for as internal energy. But there still remained the problem that
  kinetic energy was lost in simple situations, such as when balls are thrown upward. By the
  mid-1800s, the modern language that “kinetic energy is converted to potential energy” was
  finally standardized, but it was still common to read in textbooks that potential energy was
  fake, a mathematical trick used to patch up energy conservation. After all, potential energy
  has some suspicious qualities. If a ball has lots of potential energy, you can’t see or feel it, or
  even know it’s there by considering the ball alone. It doesn’t seem to be located anywhere

  in space, and its amount is arbitrary, as a constant can always be added.

  In the late 1800s, a revolution on physics answered some of these questions. Maxwell and his
  successors recast electromagnetism as a theory of fields, and showed that the dynamics of
  charges and currents were best understood by allowing the fields themselves to carry energy
  and momentum. We’ll cover this in detail in E7, but for now, it implies that electrostatic
  potential energy is fundamentally stored in the field, with a density of ϵ0 E 2 /2. This implies
  that its location and total amount are directly measurable.

  Maxwell believed that the dynamics of fields emerged from the microscopic motions and
  elastic deformations of an all-pervading ether, in the same way that, say, a fluid’s velocity
  field emerges from the average motion of fluid molecules. This makes it manifestly positive,
  so he was disturbed to find that the energy density of a gravitational field is negative! He
  therefore concluded that gravity could not be described as a vector field.

  A few decades later, the arrival of special relativity answered some questions and reopened
  others. On one hand, it demolished Maxwell’s vision of the ether. On the other hand, it finally
  answered the question of whether all kinds of potential energy are “real”, and it got rid of
  the freedom to add arbitrary constants. That’s because in special relativity, the total energy
  of a system at rest is related to its mass by E = mc2 , and the mass is directly measurable.
  This finally puts thermal energy, elastic potential energy, and field energy on an equal footing.

  Here’s the most modern view of energy conservation. All particles and their interactions are
  fundamentally described by relativistic quantum fields. A famous result called Noether’s
  theorem implies that whenever such a theory is time-translationally symmetric, there
  is a conserved quantity which we call the energy. (The distinction between kinetic and
  potential energy becomes irrelevant; it’s all just energy.) The density of energy in space
  can be computed from the state of the fields, but it doesn’t need to be explained, as
  Maxwell imagined, by the internal motion of whatever the fields are made of. The fields are
  fundamental: they aren’t made of anything; instead, they make up everything!

  What happens when we throw gravity into the mix? As we’ll discuss further in R3, it turns
  out that at nonrelativistic velocities, the dynamics of gravitating particles can be described
  by “gravitoelectromagnetism”, a theory closely analogous to electromagnetism, where moving
  masses also source “gravitomagnetic” fields Bg , which result in mv × Bg forces. But the
  situation gets much more subtle when we upgrade to full general relativity. Here, the notion
  of a gravitational field disappears completely, and is replaced by the curvature of spacetime,
  making it hard to define an energy density for it at all. For an accessible overview of the
  debate, see this paper. Ultimately, though, it doesn’t matter that much, since it doesn’t
  impair our ability to use either Newtonian gravity or general relativity.

  Example 9

  For an infinite line of linear charge density λ, find the potential V (r) by dimensional analysis.

    Solution
    This example illustrates a famous subtlety of dimensional analysis. The only quantities in
    the problem with dimensions are λ, ϵ0 , and r. To get the electrical units to balance, we have
                                                       λ
                                             V (r) =      f (r)
                                                       ϵ0
    where f (r) is a dimensionless function. But there are no nontrivial dimensionless functions
    of a single dimensionful quantity r. The only possibilities are that f (r) is a dimensionless
    constant, or that f (r) is infinite. In the first case, the electric field would vanish, which can’t
    be right. In the second case, it is unclear how to calculate the electric field at all.

    In fact, the electric potential is infinite, if you insist on the usual convention of setting
    V (∞) = 0. In that case, we have
                                      Z ∞              Z ∞
                                                              λ dr
                              V (r) =       E(r) dr =               =∞
                                        r               r   2πϵ0 r

    for any r. But this is useless; to get a finite result we can actually work with, we need to
    subtract off an infinite constant from the potential. Equivalently, we need to set the potential
    to be zero at some finite distance r = r0 . This process is known as renormalization, and it is
    extremely important in modern physics. After renormalization, we have
                                           Z r0
                                                            λ       r0
                                   V (r) =      E(r) dr =       log
                                            r             2πϵ 0     r

    which is perfectly consistent with dimensional analysis.

    Notice that in the process of renormalization, a new dimensionful quantity r0 appeared out
    of nowhere. This phenomenon is known as dimensional transmutation. Nothing measurable
    depends on this new scale (e.g. the electric field is independent of r0 ), but you can’t write
    down quantities like the potential without it.

3    Conductors
    Idea 9
    In electrostatic conditions, E = 0 inside a conductor, which implies the conductor has constant
    electric potential V . This further implies that E is always perpendicular to a conductor’s
    surface. By Gauss’s law, the conductor has ρ = 0 everywhere inside, so all charge resides on
    the surface.

    Example 10

    Consider a point on the surface of a conductor with surface charge density σ. Show that the
    outward pressure on the charges at this point is σ 2 /2ϵ0 .

       Solution
       Gauss’s law tells us that the difference of the electric fields right inside and outside the
       conductor at this point is
                                                           σ
                                             Eout − Ein =
                                                          ϵ0
       by drawing a pillbox-shaped Gaussian surface. But we also know that Ein = 0 since we’re
       dealing with a conductor, so Eout = σ/ϵ0 .

       Let’s think about how this electric field is made. If there were no charges around except for
       the ones at this surface, then the interior and exterior fields would have been ±σ/2ϵ0 . This
       means that all of the other charges, that lie elsewhere on the surface of the conductor, must
       provide a field σ/2ϵ0 here, so that Ein cancels out.

       The pressure on the charges at this point on the surface is equal to the product of the surface
       charge density with the field due to the rest of the charges, since the charges at this point
       can’t exert an overall force on themselves, so

                                                             σ2

                                                      σ
                                           P =σ           =
                                                    2ϵ0      2ϵ0
                                                               2 /2.
       as required. Equivalently, we can conclude that P = ϵ0 Eout

       Example 11

       Is the charge density at the surface of a charged conductor usually greater at regions of higher
       or lower curvature?

       Solution
       We can’t answer this question in general, because it is usually impossible to solve for
       the charge distribution of an irregularly shaped conductor. Charges at any point in the
       conductor will influence the charges everywhere else.

       However, we can get some insight by considering the limiting case of a conductor made of
       two spheres of radii R1 and R2 , connected by a very long rod. For the potential to be the
       same at both spheres, we must have Q1 /R1 = Q2 /R2 , so the charge is proportional to the
       radius, and the charge density is inversely proportional to the radius. Thus, there’s generally
       higher charge density at sharper points of the conductor, provided that those points are sharp
       enough, or far enough away from the rest of conductor for the charges on the rest of the
       conductor not to matter much. That’s basically all we can say for sure.


[User]
9. THERMAL ACCELERATION (9 points)

Consider a cube of side length a=1cm, made
of aluminium (density ρ = 2.7g/cm3
, molar
mass MA = 23g/mol). The heat capacitance
of one mole of aluminium is given as a func-
tion of temperature in the graph below. The
speed of light c = 3×108
m/s, universal gas
constant R = 8.31J/(kg·K). The initial tem-
perature of the cube is T0 =300K.
Cv
R
2R
3R
T(K)
100 200 300 0
i) (1 point) What is the total heat energy of
such a cube at the initial temperature T0?
ii) (3 points) Now, the cube has 5
faces painted in white (reflects all relevant
wavelengths) and one face painted in black
(absorbs all these waves). The cube is sur-
rounded by vacuum at a very low temperat-
ure (near absolute zero); there is no gravity
field. Initially, the cube is at rest; as it cools
down due to heat radiation, it starts slowly
moving. Estimate its terminal speed v1.
iii) (2 points) At very low temperatures,
the heat capacitance of aluminium is pro-
portional to T3
, where T is its temperat-
ure. Which functional dependance f(t) de-
scribes the temperature as a function of time
[T = A· f(Bt), where A and B are constants]
for such very low temperatures under the
assumptions of the previous question?
iv) (3 points) Now, the cube has 5 faces
covered with a thermal insulation layer (you
may neglect heat transfer through these
faces). One face is left uncovered. The cube
is surrounded by hydrogen atmosphere at
a very low temperature (molar mass of hy-
drogen molecules MH = 2g/mol). The cube
starts cooling down due to heat transfer to
the surrounding gas; you may neglect the
heat radiation. Initially, the cube is at rest;
as it cools down, it starts slowly moving. Es-
timate the order of magnitude of its terminal
speed v2. Assume that the surrounding gas
is sparse, so that the mean free path of the
molecules is much larger than a. Assume
that v2 ¿ cs where cs is the speed sound in
the atmosphere surrounding the cube.

Solution:
9. Thermal acceleration

1) For the heat energy of one mole of material
dq = CvdT. There is no heat energy by T = 0,
hence q =
RT
0
CvdT. Using the graph we find
this as the area under the curve, q ≈ R·560J/K.
The number of moles ν = a3
ρ/MA ≈ 0.117mol,
hence the total heat energy Q = qν ≈ 546J.
2) Each photon of frequency ν radiated by the
cube carries away heat energy equal to E = hν,
and carries momentum p = h/λ = hν/c = E/c.
If the photon departs at the angle α with re-
spect to the surface normal then the compon-
ent parallel to the surface normal pk = E
c
cosα.
The total momentum given by the photons to
the cube equals by modulus to the total mo-
mentum carried by the photons; when averaged
over all the photons, the perpendicular to the
surface normal components cancel out (photons
go to all the directions). The average value of
the parallel component can be estimated just as
pk ∼ E
c
.
If we want to obtain an exact result, we need
to integrate over all the directions while keeping
in mind that the light intensity is proportional to
cosα. So, the momentum averaged over all the
directions p̄k = E
c
1
2π
R
cos2
αdΩ, where the solid
angle dierential dΩ = 2π sinαdα. Therefore, p̄k =
E
c
Rπ/2
0
sinαcos2
αdα = E
c
Rπ/2
0
cos2
αdcosα = E
3c .
Since the momentum-energy ratio is the
same for all the photons, equal to 1/c, the overall
momentum equals to Q/c. Thus, a3
ρv ∼ Q/c,
hence
v ∼
Q
ρa3c
≈ 0.67mm/s.
If we apply the exact factor
1
3
(obtained above
via integration), we end up with v ≈ 0.22mm/s.
3) The heat balance at very low temperatures
can be written as AT3
dT = −σST4
dt, where
A is a constant, σ is the Stefan-Boltzmann con-
stant, and S  the radiating area. This simpli-
fies to
dT
T
= −Bt ⇒ T = A · e−Bt
.
4) In the case of a hydrogen atmosphere, the
momentum is given to the cube due to the fact
that the molecules colliding with the coated
faces bounce back with the same speed as the
they came, but uncoated face gives away heat
energy, and the molecules leave at higher tem-
perature. If we assume that the departing
molecules have the same temperature as the
cube (which serves us only as an estimate
when particles of dierent masses collide, only
a part of the energy is transferred), then the
momentum-to-heat ratio is estimated as 1/vT ,
where vT =
p
RT/MH is the thermal speed of
the molecules after the collision with the cube
for the motion along the surface normal. So we
estimate a3
ρv ∼ Q/vT , hence
v ∼
Q
ρa3
r
MH
RT
≈ 180m/s.
It should be noted that in fact, one should have
been careful with such an estimate, because the
thermal speed is at the denominator. This will
increase the relative contribution of the heat ra-
diated at low temperatures. However, the re-
maining heat at low temperatures is propor-
tional to T4
, and therefore the contribution of
those molecules which collide with the cube at
low temperatures to the overall momentum re-
mains still small.

THERMAL ACCELERATION


[User]
1. Rock Climber (6 points)

Arockclimberofmassm = 80kgascendsalong
a vertical rock. For self-protection, the climber
uses the following method. One end of an elastic
rope is anchored to the ground. The rope goes
through smooth protection loops (carabiners),
whichareanchoredtotherock. Theheightofthe
last carabineris H = 20m. The other end of the
rope goes through a special braking clip which
is tied to the harness of the climber. During the
climb, this clip keeps rope tight, but enables the
climber to lengthen protective part of the rope.
(Assume that the rope between the clip and ca-
rabiners is always tight) When falling, the max-
imum acceleration must not exceed amax = 5g
(to protect from injuries). You may assume that
the rope is always vertical, the distance between
the clip and the centre of mass of the climber is
verysmall, andfrictionbetweentheropeandca-
rabiners is negligible. Relationship between the
strain and stress of the rope is sketched on the
graph below.
1)Assumethatthedistancebetweentheclimber
and the last carabiner is L (see Figure). If the
climberhappenstofall,thedistancebetweenthe
highest carabiner and the climber will reach a
maximal value l (afterwards, the elasticity of the
ropestarts lifting the climber). Which inequality
should be satisfied for l ? (1.5 pts)
2) Find the maximal safe length L between the
climber and the last carabiner (upon reaching of
which hehasto anchora nextcarabiner;4.5pts).
σ(kN)
ε(%) 10 20 30
2
4
6
H L

Solution:
1. Rock Climber

1) In the case of falling, the acceleration should not exceed 5g, which
means that σ(ε)
m
− g < 5g. Maximum strain is the solution of the fol-
lowingequationσ(ε) = 6gm = 6×9.8m
s2 ×80kg = 4.7kN. According
to the graph, ε = 0.315; hence, l < 0.315(L + H) + L
2) In the case of falling, the climber reaches the lowest point, when its
velocity become zero. This means that the energy absorbed by the rope
becomesequal to the change of the potential energy:
E = mg(2L + x),
where x = l − L. Energy absorbed by the rope is given by
E =

σ(ε)dx =

σ(ε)(L + H)dε = (L + H)

σ(ε)dε.
We know that the maximal value is ε = 0.315, which makes it possible
to calculate the integral numerically, as the area under the graph.
S(ε) =
 0.31
0
σ(ε)dε ≈ 564.8N
Thus,
(L + H)S(ε) = mg(2L + x) = mg(2L + ε(L + H)),
hence
L =
H(mgε − S(ε))
S(ε) − mg(ε + 2)
≈ 5.08m.
So, the new carabiner mustbe anchored within next L = 5.08m.

Rock Climber


[User]
Kärbes otsustas
lennates püsida ainult elektrivälja ekvipotentsiaalpindade peal. Ta lendab sisse ruumi,
mis on täidetud homogeense elektriväljaga $\vec{E}$, välja jõujoontega risti.
Elektriväljas hoitakse paigal ka laengut $-Q$ nii, et kärbse
trajektoori esialgse puutuja ja laengu vahemaa on $d$ (vt. joonist; $-Q < 0$).
Kui lähedale kärbes laengule jõuab? Eeldage, et $Q \le \pi\epsilon_0Ed^2$.
\begin{center}
	\includegraphics[width=0.4\textwidth]{2010-lahg-07-muha_tekst}
\end{center}

Solution:
Valime potentsiaali nullnivooks kärbse asümptootilise asukoha (lõpmatuses).
Olgu laeng $Q$ koordinaatide alguspunktiks ning olgu $x$-telg horisontaalne ja $y$-telg vertikaalne.
Kärbes peab püsima laengu lähedal kõverduval null-potentsiaalil.
Potentsiaal avaldub homogeense välja $E$ potentsiaali ja punktlaengu potentsiaali superpositsioonina, seega
\[
E(d - y) - \frac{1}{4\pi\epsilon_0}\frac{Q}{\sqrt{y^2 + x^2}} = 0.
\]
Ülesande sümmeetriast on selge, kärbse vahemaa on minimaalne $x=0$ korral ehk teisisõnu kehtib $E(d-y) = kQ/y$. Antud võrrandist saame ruutvõrrandi lahenditega
\[
y_{1,2} = {d \over 2} \pm \sqrt{\frac {d^2}4 - {kQ \over E}} = {d \over 2}\left( 1 \pm \sqrt{1 - 4kQ / Ed^2}\right).
\]
Juhul kui laeng on väike, liigub kärbes ilmselgelt sirgjooneliselt. Ometigi leidub kaks lahendit. Lahendi kahesus tuleneb sellest, et laengu läheduses leidub samuti
null-potentsiaaliga suletud kõver. Laengu kasvades need kaks null-potentsiaaliga joont lähenevad üksteisele, kuni $Q=\pi\epsilon_0Ed^2$ juures
puutuvad kokku ja edaspidi moodustub null-potentsiaalist juba üksainus kõver, mis kulgeb ümber laengu.
Seega vastab kärbse trajektoor ruutvõrrandi suuremale lahendile:
$$y= {d \over 2}\left(1 + \sqrt{1 - {4kQ \over Ed^2}}\right).$$

Kärbes

Kärbes otsustas
lennates püsida ainult elektrivälja ekvipotentsiaalpindade peal. Ta lendab sisse ruumi,
mis on täidetud homogeense elektriväljaga $\vec{E}$, välja jõujoontega risti.
Elektriväljas hoitakse paigal ka laengut $-Q$ nii, et kärbse
trajektoori esialgse puutuja ja laengu vahemaa on $d$ (vt. joonist; $-Q < 0$).
Kui lähedale kärbes laengule jõuab? Eeldage, et $Q \le \pi\epsilon_0Ed^2$.
\begin{center}
	\includegraphics[width=0.4\textwidth]{2010-lahg-07-muha_tekst}
\end{center}

Kärbse potentsiaali on mugav avaldada $x-y$ koordinaadistikus. Ülesande sümmeetriast on suhteliselt lihtne näha, et kärbse kaugus laengust on minimaalne siis, kui ta asub otse laengu kohal.


[User]
1. Spool (12 points)

Aspoolwith inner radius r andouterradius
R lies on a horizontal table; the axis of the spoolis horizontal. A
weightless rope is wound around the inner part as shown in the
picture. Thelooseendoftheropemakesanangleαwiththehori-
zont(theangle αcan bealsonegative). Themomentofinertia of
the spool is J and mass — M. In what follows you may assume
that the spoolrollson the table without slipping.
i)(2 pts ) We pull the looseend of the ropewith velocity u(par-
allel to the loose part of the rope; that loose part can be thought
to be very long). Whatis the velocity ofthe spool?
ii)(3pts)Supposenowthatthespoolisatrest,andweapplyforce
F to the loose end of the rope (parallel to the loose part of the
rope). Whatisthe acceleration ofaxisofthe spool?
iii)(2pts)Howlargedoesthecoefficient offrictionμneedtobe
(asafunctionofα)toensurethatthereisnoslippingbetweenthe
spooland the table?
iv)(1.5 pts) Now the spoolrolls,again, with velocity u; however
there is no rope. The spool hits a threshold of height H (see Fig-
ure); the impact is perfectly inelastic. What is the speed v of the
spoolimmediately after the impact?
v)(1.5pts)Whatisthespeedw ofthespoolafterrollingoverthe
threshold? Assume that u is such that the spoolwill roll over the
thresholdwithout losing contact with its edge.
vi)(2pts)Ifthespeeduistoolarge,u > u0,thespoolwilljump
upandlosecontactwiththeedgeofthethreshold. Determineu0.

Solution:
1. Spool

i) Firstsolution. The momentary rotation
centre of the spool is the contact point P with the floor (since
this point is at rest). So, the velocity of the spool is u′
= Rω,
whereωistheangularvelocity. ConsidertrianglePOA, whereA
is defined as the point where the loose end of the rope meets the
innerpartofthespoolatthecurrentmomentoftime,butwhichis
actually a point of the spool, i.e. it rollstogether with the spool);
O isthecentreofthespool. Letusdenote6 PAO = β; itiseasy
to see that 6 AOP = π − α. The velocity ~ vA of the point A
is perpendicular to PA and, hence, formsangle β with the loose
end of the rope. Its projection to the rope equals to u, therefore
vA = u/cosβ. Further, ω = vA/l, where l = |AP| can be
found from the cosine theorem: l =
√
R2 + r2 + 2Rrcosα.
The angle β can be found using the sine theorem for the triangle
AOP: sinβ = R sinα
l
. Combining everything together we end
up with
u′
=
uR
√
R2 cos2 α + r2 + 2Rrcosα
=
uR
|Rcosα + r|
.
Second solution. Let us decompose the velocity ~ vA into two
components: the tangential component (parallel to the rope)
equals (by modulus) to u; let us denote the radial component as
ur. Since the distance between O and A is constant, the projec-
tion ofthe velocitiesofO and Atothe line OAareequal:
ur = v sinα ⇒ v = ur/sinα ⇒ ω = ur/Rsinα.
TheverticalcomponentofthevelocityofthepointAremainsun-
changed if we switch the laboratory system of reference with the
systemassociated with point O; hence,
usinα − ur cosα = ωrsinα = urr/R ⇒
v =
ur
sinα
=
uR
Rcosα + r
.
ii) (2 pts) The easiest way to solve this part is to use the energy
balance for infinitesimal displacement of the cylinder and apply
the answerto the previousquestion:
Fu·dt = d

M
2
v2

1 +
J
MR2

= Mvdv

1 +
J
MR2

⇒
a =
dv
dt
=
Fu
Mv 1 + J
MR2
 =
F
M
·
cosα + r
R
1 + J
MR2
.
iii)Letuswritetheforcebalanceprojectiontothehorizontalaxis
assuming that the spool is at the edge of slipping, i.e. the friction
force Ff = μminN, where N = mg − F sinα is the normal
force:
Ma = F cosα+μminN = F cosα+μmin(Mg−F sinα).
Using theresultoftheprevioustask,wecanusethisequationdir-
ectly toobtain anexpressionforthe minimal allowed valueofthe
coefficient offriction:
μmin =
r
R
− J
MR2 cosα
1 + J
MR2
 Mg
F
− sinα
iv) The angular moment of the spool with respect to the edge of
thethresholdconservesduringtheimpact(sincetheimpactforce
haszeroarm):
Mu(R − H) + J
u
R
= J + MR2
 v
R
⇒
v = u 1 −
H/R
1 + J
MR2
!
v)Fromthe energy conservation lawwe obtain immediately
(J + MR2
)
v2
R2
= (J + MR2
)
w2
R2
+ 2MgH ⇒
w =
s
v2 −
2gH
1 + J
MR2
.
vi)The spoolis the mostprone to jumping immediately after the
impact; the gravity force needs to be large enough to bind the
centre of mass to the rotational motion around the edge of the
threshold:
Mv2
R
≤ g
R − H
R
⇒ v2
≤
g
M
(R − H) ⇒
u0 =
r
g
M
(R − H)
1 + J
MR2
1 + J
MR2 − H
R
.

Spool


[User] Juku istub puu otsas ja laseb algkiiruseta lahti tema käes oleva palli. All seisab Juhan, kes samal hetkel viskab Juku pihta vertikaalselt üles täpselt samasuguse palli. Pärast pallide põrget jõuab Juku pall täpselt tema kõrgusele tagasi. Kas pall tabab Juhanit enne või pärast seda, kui Juku pall jõuab Jukuni? Lugeda, et pallide põrge on absoluutselt elastne.

Solution:
Ühesuguse massiga pallide elastse kokkupõrke tulemusena vahetavad nad oma
kiirusevektorid (järeldub lihtsalt impulsi ja energia jäävusest massikeskme süsteemis vaadatuna). Seega võime sama hästi öelda, et pallid lähevad üksteisest vabalt läbi, kusjuures ühe palli algkiirus ja teise palli lõppkiirus on võrdsed nulliga. Kui Juku istub kõrgusel $h$, siis saame mõlema vabalt liikuva palli jaoks lennuajaks $t = \sqrt{2h/g}$. Seega tabavad pallid viskajaid üheaegselt.

\vspace{0.5\baselineskip}

\emph{Alternatiivne lahendus}\\
Kui Juku visatud pall jõuab täpselt tagasi oma algkõrgusele, siis peab tema kiiruse absoluutväärtus vahetult kokkupõrke eel ja vahetult kokkupõrke järel olema sama. Elastse kokkupõrke korral kehtib energia jäävus, järelikult võrdub ka Juhani palli kiiruse absoluutväärtus vahetult kokkupõrke eel kiiruse absoluutväärtusega vahetult kokkupõrke järel. Seega liigub kumbki pall tuldud suunas tagasi nii, et liikumise ajagraafik on peegelsümmeetriline põrkehetke suhtes. Seega, kui nad startisid samaaegselt, siis nad ka finišeerivad samaaegselt.

Pallid

Juku istub puu otsas ja laseb algkiiruseta lahti tema käes oleva palli. All seisab Juhan, kes samal hetkel viskab Juku pihta vertikaalselt üles täpselt samasuguse palli. Pärast pallide põrget jõuab Juku pall täpselt tema kõrgusele tagasi. Kas pall tabab Juhanit enne või pärast seda, kui Juku pall jõuab Jukuni? Lugeda, et pallide põrge on absoluutselt elastne.

Kuna pallid on samasuguse massiga ja tegu on elastse kokkupõrkega, vahetavad pallid oma kiirusvektorid. Seega võime sama hästi öelda, et pallid lähevad üksteisest vabalt läbi.


[User]
3. WOBBLE (8 points)

— Taavet Kalda. We
investigate the so‐called astrometric wobble
method for detecting exoplanets. The
method relies on the ability to measure the
change in position in the sky that the host
star experiences due to the gravitational in‐
fluence of its orbiting planets. This method
has become feasible in recent years due to
the availability of more precise instruments.
i) (2.5 points)Consider a system consisting of
a host star, an inner planet A, and an outer
planet B. Below is a measurement of the tra‐
jectory of the centre of the star in the plane
perpendicular to the line of sight, measured
over a period of t = 10yr. In all of the
subsequent parts, you may assume that both
planets orbit in circular orbits in the plane of
thediagram. Whataretheorbitalperiods TA
and TB of the two planets?
ii) (2.5 points) Based on direct imaging of
planet A in the infrared, the orbital ra‐
dius of planet A is measured to be aA =
1.5AU = 2.2 × 108
km. What is the mass
M of the host star, and the mass mA of
planet A? The gravitational constant is G =
6.67 × 10−11
m3
kg−1
s−2
.
iii) (1 point) What is the mass mB of planet
B?
iv) (2 points)Now consider a similar t = 10yr
measurement of a different system shown
below, also consisting of a host star and
two planets A and B (where A is the inner
planet). Similarly to before, find the mass M
of the host star, and the masses mA, mB of
theplanetsinthenewsystem. Asbefore, dir‐
ect imaging yields that the orbital radius of
planet A is aA = 1.3AU = 2.0 × 108
km.

Solution:
3. WOBBLE

— Solution by Taavet
Kalda, grading schemes by ....
i) (2.5 points) The farther away a planet
is from the host star, the bigger its orbital
period. In a fixed period, we then expect
planet A to make more rotations around the
star than planet B. The wobble effect comes
from both the star and the planets orbiting
around their common barycentre (centre of
mass). The effect of one planet makes the
star undergo a circular motion with some ra‐
dius x and frequency ω. The effect of two
planets is additive, so the overall motion is
the sum of two circular motions with radius’
xA andxB withdifferentangularfrequencies
ωA and ωB.
In the figure, we find the centre of cir‐
cular motion of the star from the centre of
the envelope (by for example using a ruler
to find the diameter and then the centre‐
point). From there, we measure that lower
frequencycomponentcoversanangleofα =
290° within the measurement period t =
10yr. The higher frequency component, in
the meantime, undergoes 7.5 full rotations
with respect to the lower frequency compon‐
ent. Hence, ωBt = α and (ωA − ωB)t/(2π) =
7.5. This yields
TB =
2π
ωB
=
2πt
α
= 12.4yr,
TA =
2π
ωA
=
1
1
TB
+ 7.5
t
= 1.20yr.
Grading:
• Realising that the higher frequency com‐
ponent is due to planet A and the lower
frequency component is due to planet B
(0.5pts)
• The higher frequency component under‐
goes 7.5 full rotations (0.4pts)
• The lower frequency component covers
an angle of 290°, or alternatively, 7.5
9 · 2π
(0.4pts)
• Finding TA (0.8pts)
• Finding TB (0.4pts)
ii) (2.5 points) From the figure, we meas‐
ure the extrema of the distance of the star
from the origin as l1 = 1.6 × 105
km and
l2 = 2.8 × 105
km. The extrema corresponds
to when the two circular motions are point‐
ing in the same direction or the opposite dir‐
ection. In other words, the distances are
l1 = xB − xA and l2 = xB + xA. Rearran‐
ging, xA = (l2 − l1)/2 = 6 × 104
km and
xB = 2.2 × 105
km. To infer the masses, we
need to study the dynamics of the system.
First, we can assume that the semi‐major
axis of the planets are much bigger than x.
Indeed, aA/xA = 4000 ≫ 1. Either from
force balance between gravity and centrifu‐
gal force, or through Kepler’s Third Law, we
have 4π2
/(GM) = T2
A/a3
A so the star’s mass
is
M =
4π2
a3
A
GT2
A
= 4.4 × 1030
kg.
The planet’s mass can be inferred from
the property that planet and star rotate
around their common centre of mass:
MxA = mAaA. Hence,
mA = M
xA
aA
= 1.2 × 1027
kg.
Grading:
• Expressing 4π2
/(GM) = T2
A/a3
A (0.6pts)
• Finding M (0.4pts)
• Realising that the position of the centre of
massofAandM isconserved(withrespect
to B) (0.5pts)
• Inferring equation MxA = mAaA (0.5pts)
• Finding xA from the trajectory (0.3pts)
• Finding mA (0.2pts)
iii) (1 point)Once again, from Kepler’s Third
law, we get the semi‐major axis of planet B
to be
aB =
3
√
T2
BGM
4π2
= 1.0 × 109
km.
Similarly to before, we get the mass as
mB = M
xB
aB
= 9.3 × 1026
kg.
Both of the planets are gas giants, similar
to Jupiter (0.63 and 0.50 Jupiter masses re‐
spectively). Thismakessense,asthismethod
is more sensitive to higher mass exoplanets
with bigger orbits.
Grading:
• Finding aB (0.5pts)
• Finding mB (0.5pts)
iv) (2 points) The equations describing the
system still hold, but reading off the fre‐
quencies is trickier. Nevertheless, we can
still locate the origin and measure the ex‐
treme distances l1 = 1.20 × 105
km and
l2 = 1.85 × 105
km. We also note that the
higher frequency component (correspond‐
ing to planet A) has a bigger radius than the
lower frequency one (planet B). Hence, l1 =
xA − xB and l2 = xA + xB, and so xA =
(l1 +l2)/2 = 1.5 × 105
km, xB = (l2 −l1)/2 =
3.3 × 104
km.
The higher frequency component does 3
full turns and, measuring from the figure,
an extra 280° on top of it. This gives TA =
t/(3 + 280°/360°) = 2.65yr. The lower fre‐
quency component, meanwhile, starts and
ends from the farthest away point, and os‐
cillates through the closest point 3 times.
Hence it goes through 3 full rotations. Thus,
(
2π
TA
−
2π
TB
)
t = 6π,
and so
TB =
1
1
TA
− 3
t
= 12.9yr.
By using the equations from the previous
parts, we get
M =
4π2
a3
A
GT2
A
= 6.8 × 1029
kg,
mA = M
xA
aA
= 5.1 × 1026
kg,
mB = MxB
3
√
4π2
T2
BGM
= 3.9 × 1025
kg.
The first planet is once again a gas giant
(0.26 Jupiter masses) while the second one is
a “Super‐Earth” (6.4 Earth masses).
Grading:
• Finding TA (0.4pts)
• Finding TB (0.4pts)
• Finding M (0.3pts)
• Finding mA (0.4pts)
• Finding mB (0.5pts)

WOBBLE


[User]
3. Temporal focusing (10 points)

Suppose that at the point A, there is a source of thermal electrons (of
negligible thermal energy), which are accelerated initially by the vol-
tage U0 = 36 V in horizontal direction (see figure). At the path of
the electrons, there are two voltage gaps B and C of negligible size,
at the distance a from each other. These gaps receive a voltage signal
UC(t) = −UB(t) ≡ U(t) from a waveform generator. We can assu-
me that |U(t)| fi U0. The electrons starting at different moments of
time are to be gathered together (focused) at the probe D, which is at
distance b from the gap C. In order to analyze this setup, answer the
following questions.
e A U(t) D B
C
a b
1) Assuming that U(t) ≡ 0, what is the time needed for the electrons
to travel from the gap B to the probe D?
2)FindthesametraveltimeassumingthatthevoltageU(t) ≡ U fi U0
is constant (your approximating expression should be a linear function
of U).
3) What functional equation should be satisfied for the waveform U(t)
in order to ensurethe focusing ofall the electrons at the probe D. Solve
this equation by assuming a fi b and |U(t)| fi U0.
4) The waveform generator yields a periodic signal of period T in such
a way that the profile U(t) is followed up to achieving some maximal
valueUm;afterthat,thevoltagedrops immediatelyto0andtheprocess
starts repeating. What is the fraction of electrons missing the time
focus at the probe D?

Solution:
3. Temporal focusing

1) Apparently, t0 = (a + b)/v0 = (a + b)
p
m/2U0e.
2) Apparently, t = a/va + b/vb ≈ t0 − [a(va − v0) + b(vb − v0)]/v2
0.
On the other hand, for constant voltage U(t), vb = v0. For the interval
between B and C, wehave ∆(mv2
/2) ≈ mv0(va − v0) = −Ue. So,
t = t0 + Uea/mv3
0 = t0 +
U
U0
a
2
r
m
2U0e
.
3) For the changing voltage, we can use the result of the previous ques-
tion, but the voltage value should be taken at the moment of electron
passing the gap. Also, the terminal velocity is changed, v1 − v0 is rela-
tedto the change of the kinetic energy e[U(t + ta) − U(t)]. The travel
time of the later electrons passingB at acertain moment oftime t > 0
should as much shorter as it was delayed,i.e.
t1 − t2 =
r
m
8U3
0 e
{a[U(0) − U(t)] + b[U(t + ta) − U(t)]} = t,
where the flight time ta ≈ a/v0 [since U fi U0]. Using the suggested
approximation, we obtain
r
m
8U3
0 e
b[U(t + a/v0) − U(t)]} = t.
If we seek a quadratic solution U(t) = At2
+ Bt,we get
2Ata(t + ta/2) + Bta =
t
b
r
8U3
0 e
m
.
So,
A =
r
8U3
0 e
m
/2tab =
r
8U3
0 e
m
p
2eU0/m
ab
=
4U2
0 e
mab
,
and B = −Ata. Finally we obtain
U(t) =
4U2
0 e
mab
t

t − a
p
m/2U0e

.
4) Lost are those electrons, which are in the interval between B and C,
when the voltage drops to 0. The duration of the time interval, when
these were emitted, is τ = a/v0. So, the asked fraction is given by
τ/T = a
T
p
m/2U0e.

Temporal focusing