105A - Set 6 - Solution
(Grades are out of 150)
1. (36pt) Assume Earth’s orbit to be circular and that the Sun’s mass suddenly decreases
by half. What orbit does the Earth then have? Will the Earth escape the solar system?
Answer: The potential energy is
GM m⊕
U =− (1)
r⊕
The kinetic every is m⊕ v 2 /2 where v = ωr which we can find from by equating the
gravitational force to the centripetal force
GM m⊕
m⊕ ω 2 r⊕ = 2
(2)
r⊕
so
GM
ω2 = 3
(3)
r⊕
and then before the Sun’s mass decreased we have:
1 GM m⊕ 1 GM 2 GM m⊕
E = T + U = m⊕ ω 2 r⊕
2
− = m⊕ 3 r⊕ −
2 r⊕ 2 r⊕ r⊕
1 GM m⊕ 1
= − = U (4)
2 r⊕ 2
So E = T + U = U/2 so T = −U/2.
If the sun’s mass suddenly goes to half its original value, T remains unchanged but U
is halved. So the Energy after EA :
1 1 1
EA = TA + UA = T + U = − U + U = 0 (5)
2 2 2
So The energy is 0, so the orbit is a parabola. For a parabolic orbit, the earth will
escape the solar system.
2. (38pt) Two particles move about each other in circular orbits under the influence of
gravitational forces, with a period τ . Their motion is suddenly stopped, and they are
then√released and allowed to fall into each other. Prove that they collide after a time
τ /(4 2). Hint 1: find τ for circular motion.
Hint 2: when you get to r̈ equal to some thing and its hard to solve it, I suggest to
multiply by ṙ - an alternative way is to think about the energy.
Answer: Since we are dealing with gravitational forces, the potential energy between
the particles is
k
U (r) = − (6)
r
1
, and, after reduction to the equivalent one-body problem, the Lagrangian is
m 2 k
L= ṙ + r2 θ̇2 + (7)
2 r
where µ is the reduced mass. The equation of motion for r is
k l2 k
µr̈ = µrθ̇2 − 2
= 2
− 2 (8)
r µr r
We can also use the fact that the motion is circular r = r0 , i.e.,
k
µω 2 r0 = µθ̇2 r0 = (9)
r02
Which means that the
k
ω2 = (10)
r03 µ
So the period τ is r
2π r03 µ
τ= = 2π (11)
ω k
Now if the particles stopped, the angular momentum goes to zero. So plugging this in
eq. (8) we get
k
µr̈ = − 2 (12)
r
Then we use the hist and multiply by ṙ
k
µr̈ṙ = − ṙ (13)
r2
which is equivalent to writing
d 2 2k
(ṙ ) = + (14)
dt µr
So then:
2k
ṙ2 = + + C, (15)
µr
where C is constant and is determined from the boundary condition on ṙ, which is
when ṙ = 0 then r = r0 since initially the particles are not moving at all. So we find
that C = −2k/r0 /µ so s r
dr 2k 1 1
ṙ = = − , (16)
dt µ r r0
We could now proceed to solve this differential equation for r(t), but since in fact we’re
interested in solving for the time difference corresponding to given boundary values of
r, it’s easier to invert this equation and solve for t(r):
Z 0 r Z 0r
dr µ rr0
t= dr = dr (17)
r0 dt 2k r0 r0 − r
2
(Grades are out of 150)
1. (36pt) Assume Earth’s orbit to be circular and that the Sun’s mass suddenly decreases
by half. What orbit does the Earth then have? Will the Earth escape the solar system?
Answer: The potential energy is
GM m⊕
U =− (1)
r⊕
The kinetic every is m⊕ v 2 /2 where v = ωr which we can find from by equating the
gravitational force to the centripetal force
GM m⊕
m⊕ ω 2 r⊕ = 2
(2)
r⊕
so
GM
ω2 = 3
(3)
r⊕
and then before the Sun’s mass decreased we have:
1 GM m⊕ 1 GM 2 GM m⊕
E = T + U = m⊕ ω 2 r⊕
2
− = m⊕ 3 r⊕ −
2 r⊕ 2 r⊕ r⊕
1 GM m⊕ 1
= − = U (4)
2 r⊕ 2
So E = T + U = U/2 so T = −U/2.
If the sun’s mass suddenly goes to half its original value, T remains unchanged but U
is halved. So the Energy after EA :
1 1 1
EA = TA + UA = T + U = − U + U = 0 (5)
2 2 2
So The energy is 0, so the orbit is a parabola. For a parabolic orbit, the earth will
escape the solar system.
2. (38pt) Two particles move about each other in circular orbits under the influence of
gravitational forces, with a period τ . Their motion is suddenly stopped, and they are
then√released and allowed to fall into each other. Prove that they collide after a time
τ /(4 2). Hint 1: find τ for circular motion.
Hint 2: when you get to r̈ equal to some thing and its hard to solve it, I suggest to
multiply by ṙ - an alternative way is to think about the energy.
Answer: Since we are dealing with gravitational forces, the potential energy between
the particles is
k
U (r) = − (6)
r
1
, and, after reduction to the equivalent one-body problem, the Lagrangian is
m 2 k
L= ṙ + r2 θ̇2 + (7)
2 r
where µ is the reduced mass. The equation of motion for r is
k l2 k
µr̈ = µrθ̇2 − 2
= 2
− 2 (8)
r µr r
We can also use the fact that the motion is circular r = r0 , i.e.,
k
µω 2 r0 = µθ̇2 r0 = (9)
r02
Which means that the
k
ω2 = (10)
r03 µ
So the period τ is r
2π r03 µ
τ= = 2π (11)
ω k
Now if the particles stopped, the angular momentum goes to zero. So plugging this in
eq. (8) we get
k
µr̈ = − 2 (12)
r
Then we use the hist and multiply by ṙ
k
µr̈ṙ = − ṙ (13)
r2
which is equivalent to writing
d 2 2k
(ṙ ) = + (14)
dt µr
So then:
2k
ṙ2 = + + C, (15)
µr
where C is constant and is determined from the boundary condition on ṙ, which is
when ṙ = 0 then r = r0 since initially the particles are not moving at all. So we find
that C = −2k/r0 /µ so s r
dr 2k 1 1
ṙ = = − , (16)
dt µ r r0
We could now proceed to solve this differential equation for r(t), but since in fact we’re
interested in solving for the time difference corresponding to given boundary values of
r, it’s easier to invert this equation and solve for t(r):
Z 0 r Z 0r
dr µ rr0
t= dr = dr (17)
r0 dt 2k r0 r0 − r
2
