# Satellite and gravitation

1. Jan 15, 2014

### peripatein

Hi,
1. The problem statement, all variables and given/known data
A satellite with mass m orbits a planet of mass M in a circular path with radius r and velocity v. Due to some internal technical failure, the satellite breaks into two, similar parts with mass m/2 each. In the satellite's frame of reference, it appears the two parts move radially, in opposite directions, along the line connecting the original satellite and the planet's center, each with velocity v0/2. I am expected to show that right after the technical failure, each of the two parts has a total energy equal to -3GM/16r and angular momentum equal to (m/2)√(GMr), wrt the planet's center.

2. Relevant equations

3. The attempt at a solution
The total energy of each of the two parts should be, I believe: Etot = mv02/16 - GmM/(2r). Now, isn't angular momentum preserved despite the failure? However, why isn't the angular momentum zero if the two parts are moving in opposite directions?

2. Jan 15, 2014

### tiny-tim

hi peripatein!
yes

they are moving in opposite directions in the rest-frame of the satellite

imagine a rod moving sideways, and two beads moving with equal speeds away from each other along it …

the actual motions of the beads are both diagonally forwards

3. Jan 15, 2014

### peripatein

The velocity of the the CoM remains v, right? And I know that wrt that CoM, each part moves at +-v0/2, right? Ergo, the velocity of each part wrt to the planet's center should be the sum of the CoM's velocity (which I know) and the velocity of the respective part wrt the CoM (which I also know), correct?

4. Jan 15, 2014

### tiny-tim

yes.

5. Jan 15, 2014

### peripatein

Don't I need some sort of relation between v and v0? I am unable to obtain that through conservation of angular momentum though.

6. Jan 16, 2014

### tiny-tim

i think vo and v are suposed to be the same

7. Jan 16, 2014

### haruspex

No. v is the original tangential velocity. v0 is the radial velocity with which the two move apart.

8. Jan 16, 2014

### haruspex

They also have the KE of the original tangential velocity, v.

9. Jan 16, 2014

### tiny-tim

i think vo and v are supposed to be the same

10. Jan 16, 2014

### haruspex

Ah, you mean equal, which is not quite the same as same .