Problem 1:
Assume for this problem the Earth is a sphere of radius R and mass M. An object of mass m enters Earth’s atmosphere at distance R’ > R from Earth’s center with speed v at angle a from the radial direction. Ignoring any friction or air resistance, what angle b (from the radial direction) will it hit Earth’s surface at?
Solution:
 | Concepts, principles, relations
that apply to the problem: Motion in a central potential, energy and angular momentum conservation |
| Why do they apply? In a central potential energy and angular momentum are conserved. |
| How do they apply? At a distance r > R from the center of Earth the object has potential energy –GMm/r. Energy conservation: (1/2)mvR’2 – GmM/R’ = (1/2)mvR2 – GmM/R, vR2 = vR’2 + 2(R’ – R)GM/(RR’). Angular momentum conservation: mR’vR’sina = mRvRsinb, sinb = sina (R’/R)(vR’/vR). sinb = (R’/R)sinavR’/(vR’2 + 2(R’ – R)GM/(R'R ))1/2. Let R' = aR. Then sinb = asinavR’/(vR’2 + 2(a - 1)GM/(aR))1/2. We need 0 < sinb < 1 for the object to hit the Earth. If this condition is not fulfilled than the object reaches its distance of closest approach (sinb = 1) and continues in its orbit. |
| Details of the calculation: None |
Problem 2:
A particle of mass m moves in
a circular orbit of radius r = a under the influence of the central attractive force
f(r)
= -g exp(-br)/r2,
where g and b are positive constants.
(a) What is the effective
potential energy, Ueff(M,r), for radial motion in terms of r and the angular
momentum M? (Your answer may contain an integral.)
(b) For what values of b will
this orbit be stable?
(c) What is the frequency of
small radial oscillation about these stable circular orbits?
Solution:
| Concepts, principles, relations
that apply to the problem: Motion in a central potential, F = f(r)(r/r), f(r) = -¶U/¶r Energy E and angular momentum M are conserved. M = mr2df/dt, E = (1/2)m(dr/dt)2 + M2/(2mr2) + U(r) = (1/2)m(dr/dt)2 + Ueff(r) . md2r/dt2 = -¶Ueff(r)/¶r. |
| Why do they apply? The force is given. The particle moves under the action of a central force. |
| How do they apply? (a) U(r) = -ò¥rf(r)dr = ò¥r(g exp(-br)/r2)dr. Ueff(r) = M2/(2mr2) + ò¥r(g exp(-br)/r2)dr. (b) For a circular orbit of radius a we need d2r/dt2|a = 0, therefore ¶Ueff(r)/¶r|a = 0, -f(a) = M2/ma3, g exp(-ba)/a2 = M2/ma3. For the orbit to be stable we need a restoring force. Let r = a + r, r >> r. We need md2r/dt2 = -ar. But md2r/dt2 = -¶Ueff/¶r = -¶2Ueff/¶r2|r=0r. We need ¶2Ueff/¶r2 > 0 at r = 0. ¶2Ueff/¶r2|r=0 = (¶2Ueff/¶r2)|r=a = ¶(-M2/(mr3) - f(r))/¶r|a = 3M2/(ma4) - g exp(-ba)[2/a3 + b/a2]. We need 3M2/(ma4) - g exp(-ba)[2/a3 + b/a2] > 0, or 3g exp(-ba)/a3 - g exp(-ba)[2/a3 + b/a2] > 0. We therefore need 1/a - b > 0, b < 1/a. |
| Details of the calculation: (c) md2r/dt2 = -¶2Ueff/¶r2|r=0r = -[g exp(-ba)/a2][1/a - b]r. w = (g exp(-ba)/(ma2)][1/a - b])1/2. In the limit b = 0 we get the result for the Kepler orbit, w = (g/(ma3))1/2, the period Tosc of small radial oscillations is equal to the orbital period Torbit = 2p(g/(ma3))-1/2. |
Problem 3:
This question is about an elliptical “transfer orbit” B from an inner
circular orbit A to an outer circular orbit B. The transfer starts at point P
and is completed at point Q. The transfer orbit is an ellipse which is tangent
to A at point P and tangent to C at point Q.
(a) Derive a formula for the relationship between v and r for circular orbits.
Is the speed in orbit C greater or less than the speed in orbit A?
(b) For the transfer, should the satellite speed up or slow down at point P?
(c) For the transfer, should the satellite speed up or slow down at point Q?
Solution:
| Concepts, principles, relations
that apply to the problem: Motion in a potential of the form V(r) = -a/r |
| Why do they apply? The satellite moves in the gravitational potential of Earth. Since the mass of Earth, M, is much greater than the mass of the satellite, m, we assume that the CM of the system is at the center of Earth. |
| How do they apply? (a) The acceleration of the satellite in a circular orbit is a = GM/r2 = v2/r. The speed is v = (GM/r)1/2. The speed in orbit C is less than the speed in orbit A. (b) The transfer orbit is an elliptical orbit. Its semi-major axis is greater that the radius of orbit A and less than the radius of orbit B. From part (a) we have T = ½mv2 = ½GMm/r = -½U. E = T + U = ½U = -½GMm/r. We then can write r = GMm/(2|E|). This generalizes for an elliptical Kepler orbit to (rmin + rmax)/2 = GMm/(2|E|). To increase the semi-major axis we have to decrease |E|, we have to make E less negative. We have to increase the kinetic energy and therefore speed the satellite up at point A. (c) By the same argument, we again have to increase the speed of the satellite at point B. |
| Details of the calculation: None |
Problem 4:
A comet of mass m approaches the solar system with a velocity v0, and if it had not been attracted towards the sun, it would have missed the sun by a distance d. Calculate its minimum distance z from the sun as it passes through the solar system. Make and state any reasonable simplifying assumptions.
Solution:
| Concepts, principles, relations that apply to the
problem: Conservation of energy and angular momentum |
| Why do they apply? For the Kepler problem these conservation laws hold. Key assumptions: Neglect other planets, assume m << Msun, neglect relativity. |
| How do they apply? At r à ¥ , L = mv0d. L = constant. At the distance of closest approach r = z, L = mvmaxz. mv0d = mvmaxz, vmax = v0d/z. E = mv02/2 = constant = mvmax2/2 – GMm/z. Therefore z2 + 2GMz/v02 – d2 = 0, z = (G2M2/v04 + d2)1/2 – GM/v02. |
| Details of the calculation: None |
Problem 5:
(a) Derive the relationship
between the impact parameter b and the scattering angle
q
for Rutherford scattering of a projectile of mass m by a fixed target particle of mass M.
(b)
If 4 MeV alpha particles are incident on a gold foil (Z = 79), calculate
the impact parameter that would give a deflection of 10 degrees.
(c)
Explain what modifications of this calculation must be made if the gold
atom is allowed to recoil during the collision.
Solution:
| Concepts, principles, relations that apply to the
problem: Motion in a central potential, elastic scattering |
| Why do they apply? For the case of the repulsive Coulomb potential U(r) = a/r, we are asked to find the relationship between the impact parameter and the scattering angle. |
| How do they apply? In a central potential the motion is in a plane and M and E are constant.
we find
Let q be the angle between the incident and the scattered direction and f0 be the angle between r( z = -¥) and rmin. Then
and
determines umax. We have q = p - 2f0 for a repulsive potential. If
then
,
b = ( a/2E)cot(q/2). |
| Details of the calculation: (b) E = 4 MeV = 4x106 eV(1.6x10-19J/eV) = 6.4 x 10-13 J a = 1/(4pe0)(2)(79)(1.6x10-19 C)2 = (9.0x109 Nm2/C2)(158)(2.56x10-38 C2) = 3640x10-29 Nm2 = 3.6x10-26Jm q = p - 2f0 = 10o. cot 5o = 11.43 b = (3.6 x10-26Jm)(11.43)/2(6.4 x 10-13J) = 3.2 x 10-13 m. (c) If the target is allowed to recoil, then we solve for the motion of the CM and the motion about the CM, i.e. the relative motion. The CM moves with constant velocity V = m1v1/(m1+ m2). The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass m moving in a central potential U(r). The reduced mass of the alpha particle-gold nucleus system is m = m1m2/(m1 + m2). The energy of the fictitious particle is E, the energy of particle 1 is E' = (m/m1)E. Given the scattering angle q in the lab frame we need to find the scattering angle q' in the CM frame. Then we use b = (a/2E')cot(q'/2) to find the impact parameter b for a given q. |
Problem 6:
A particle of mass m moves in a central force field such that its potential
energy is given by V = krn, where r is the distance from the center
of force and k and n are constants.
(a)
Write down the Lagrangian for this system and determine the equations of
motion in polar coordinates.
(b)
Show that angular momentum is conserved for the system.
(c)
Find an expression for the total energy of the system that depends only
on the radial variable.
(d)
Find the conditions (sign and magnitude of n and k) for a stable circular
orbit by investigating the particle at stable equilibrium.
Solution:
| Concepts, principles, relations that apply to the
problem: Motion in a central potential, Lagrange's equations, stable orbits |
| Why do they apply? We are asked to write down the Lagrangian, find an expression for the total energy, and investigate the stability of orbits. |
| How do they apply? (a) L = T - U = (1/2)m[(dr/dt)2 + r2(df/dt)2] - krn. The generalized coordinates are r and f. The coordinate f is cyclic. Lagrange's equations yield the equations of motion. (d/dt)[mr2(df/dt)] = 0. md2r/dt2 = M2/mr3 - nkrn-1. (b) (d/dt)[mr2(df/dt)] = 0. M = mr2(df/dt) = constant, angular momentum is conserved. (c) E = T+U = (1/2)m(dr/dt)2 + M2/(2mr2) + krn = (1/2)m(dr/dt)2 + Ueff(r). This expression looks like the total energy of a particle moving in one dimension in a potential Ueff(r). |
| Details of the calculation: (d) For a circular orbit at with radius a we need ¶Ueff(r)/¶r|a = 0. ¶Ueff(r)/¶r = nkrn – 1 - M2/(mr3). nkan – 1 = M2/(ma3), an+2 = M2/(mnk), (n ¹ 0). For a stable orbit of radius a we need a restoring force. Let r = a + r, r << r. We need md2r/dt2 = -ar = -¶Ueff/¶r = -(¶2Ueff/¶r2)|r=0r, or ¶2Ueff/¶r2 > 0 near r = 0. ¶Ueff(r)/¶r = nkrn – 1 - M2/(mr3). ¶Ueff(r)/¶r = nk(a + r)n – 1 - M2/[m(a + r)3] = nkan-1(1 + r/a)n–1 - [M2/(ma3)](1 + r/a)-3 = nkan-1(1 + (n-1)r/a) - [M2/(ma3)](1 - 3r/a). ¶2Ueff/¶r2|r=0 = n(n-1)kan-2 + 3M2/(ma4) > 0. mn(n – 1)kan+2 + 3M2 > 0. mn(n – 1)k[M2/(mnk)] + 3M2 > 0. (n – 1)M2 + 3M2 > 0. (n – 1) + 3 > 0. n + 2 > 0. Requirements for stable circular orbit with radius a for potential V(r) = krn are: n > -2 k = M2/(nman+2). When n < 0 then k < 0, and when n > 0 then k > 0. |
Problem 7:
Find the maximum time a comet (C) of mass m following a parabolic trajectory around the Sun (S) can spend within the orbit of the Earth (E). Assume that the Earth’s orbit is circular and in the same plane as that of the comet.
Solution:
| Concepts, principles, relations that apply to the
problem: The Kepler problem |
| Why do they apply? Both the Earth and the comet move in the central potential V(r) = -a/r. Let the radius of the circular orbit of the Earth be a. We need to find the time a comet following a parabolic orbit can spend between rmin and a. |
| How do they apply? The total energy of a particle following a parabolic orbit is zero. |
| E = ½mc(dr/dt)2 +
Ueff(r) = 0, Ueff(r) = U(r) + M2/(2mcr2)
= -mca/r
+ M2/(2mcr2). At r = rmin we have dr/dt = 0 and mca/rmin = M2/(2mcrmin2). M2 = 2mc2armin. We have found the relationship between M and rmin.  .The time
the comet spend between rmin and a is |
| Details of the calculation: Let x = rmin/a. t = C(1+2x)(1-x)1/2. Let us find an extremum for t. dt/dx = 2C(1-x)1/2 - (1/2)C(1+2x)(1-x)-1/2 = 0. 2(1-x) - (1/2)(1+2x) = 0. 3x = 3/2, x = 1/2. Since for rmin > a the time t spend within Earth's orbit is zero, the extremum is a maximum. The comet following a parabolic orbit spends the maximum time between rmin and a when rmin = a/2. Then tmax = (1/a)1/2(4/3)a3/2. Since (a3/a)1/2 = T/2p (Kepler's third law), we have tmax = 2T/(3p). tmax = (2*365 days)/(3p) = 77 days. |
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