Energy in Gravtational Field?

Question

I've been trying this all night. Can someone please help?
Two particles of masses M1=63000kg and M2=7200kg are initially at rest an infinite distance apart. As a result of the gravitational force the two masses move towards each other. Calculate the speed of M1 when their separation distance is 41.3km.

I've tried using conservation of potential energy, but I get stuck. Can anyone please help?

Answer 1

It is basically a 2=body problem. As there is no external force on system, the subsequent motion due to gravitational force does not change the position of center of mass as per second law of motion. The bodies simply move towards each other such that center of mass remains at rest.

let r1, v1, and r2, v2 be magnitudes of linear distance from the center of mass and speeds of two bodies. Also let “center of mass” of the system be at origin.
Centre of mass r(cm) = (-m1r1+m2r2)/(m1+m2)
But r(cm) = 0 >> m1r1=m2r2
Similarly for velocity >> v(cm) gives m1v1=m2v2
m1a1 = - m2a2
F12 = - F21
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in nutshell, reduced mass u(m) = m1m2/(m1+m2) as if 1-body plays role in it.
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Initial gravitational energy of the system = 0 (separated by infinite distance)
Start from rest > initial kinetic energy = 0
Initial mechanical energy = 0
Let vr be approach velocity (v1+v2) or relative velocity when 2 are at distance (r) apart. Applying conservation of mechanical energy> KEi+PEi = KEf+PEf
0+0 = 0.5 u(m) * vr^2 - GM1M2/r >>>> reduced mass used
vr^2 = 2 GM1M2/r*u(m) = [2 GM1M2/r][(M1+M2)/M1M2]
vr = [2G(M1+M2)/r]^1/2
v1+v2 = [2G(M1+M2)/r]^1/2
also M1v1 = M2 v2
v1 + (M1/M2) v1 = [2G(M1+M2)/r]^1/2
v1 = {M2}[2G /r*(M1+M2)]^1/2
v1 = 7200*[2*6.67*10^-11 /41.3*10^3*(63000+7200)]^1/2
v1 = 1.54*10^-6 m/s

Answer 2

Speed relative to what?

Lets assume its relative to M2, so that we are using the rest frame of M2.

Calculate the gravitational potential energy at infinity. Then calculate it at 41.3 km. The difference is the kinetic energy of M1.

Answer 3

1

Answer 4

Wait. Time out.

If they are an infinite distance apart, then the formulas for gravity do not work. (ADDED: In fact, gravity doesn't work). I think there is something missing in the problem as stated.

ADDED: Hope you didn't lose too much sleep!