Finding Aphelion Distance After Satellite Explosion

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To find the aphelion distance after a satellite explosion, first derive the velocity expression for a satellite in circular orbit around a planet of mass M, using the orbit radius r. Following this, calculate the satellite's time period. Upon explosion, one fragment (mass m) becomes stationary, allowing the use of momentum conservation to determine the instantaneous velocity of the other fragment (mass 4m), which is found to be 5/4 sqrt(GM/r). The explosion causes the mass 4m to enter an elliptical orbit, and the task is to find its distance from the Earth's center at the aphelion, which is presumed to be 25/7 times the radius r, with the perihelion occurring at radius r. The conservation of energy and angular momentum will be key to solving for the aphelion distance.
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Hi, how does one go about finding the aphelion distance when all the information you've got is the satellite's velocity at one point in time?? Here's the Q:
1) first you're to find an expression for the velocity of a satellite in circular orbit about planet of mass M. The orbit radius = r. Simple enough
2) then you find its time period - simple enough
3) then it says, the satellite explodes; breaking into two pieces. One of mass 4m and another of mass m. The mass m becomes stationary with respect to the Earth (i.e. v=0). Find the instantaneous v of mass 4m now - consv of momentum Q. Simple enough.
(Answer btw is 5/4 sqrt. (GM/r)--- for next Q. )
This is where i get lost.
4) it says that the broken satellite of mass 4m, because of increase in speed, now undergoes elliptical motion. We're supposed to find its distance away from the center of the Earth when it reaches the other end of the ellipse.
Answer's (25/7)r
Now I presume its talking about the aphelion distance. Cause at the instance of explosion its going to be at radius r, and again I'm guessing that that's the perihelion. So the 'other end' could be the aphelion? Regardless, I just don't know how to get about the Q. Help me out pls.
 
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HINT: Both the energy and angular momentum of the fragment will be conserved.
 
Yea? I'll give it a try.
 
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