College Physics
posted by Lelia .
A satellite in an elliptical orbit has a speed of 9.00km/s when it is at its closes approach to the Earth(perigee). The satellite is 7.00x10^6 m from the center of the Earth at this time. When the satellite is at its greatest distance from the center of the Earth (apogee), its speed is 3.66km/s. Find the distance from the satellite to the center of the Earth at apogee. (assume any energy losses are negligible.)

College Physics 
PhysicsPro
Post your work thus far and I'll continue/correct/guide

College Physics 
Lelia
Kfinal + Ugravfinal = Kinitial + Ugravinitial
which i ended up with
Vfinal^2 + [2GM/r] = Vinitial^2 + [2GM/r]
I believe I need to solve for r but this is where I got stuck. 
College Physics 
PhysicsPro
Well since it's an elliptical orbit you can set them the distances as functions of Kvff^3
<Kf^3>(2GM^4)=vf%
vf%<6.76[FL(3.33/Ki)]=d%
d%=<4.1882[Ug]
d=(1.333<pi^4>)(d%)
Tell me what you get 
College Physics 
PhysicsPro
Quick explanation because I have to go;
Kvff^3
Rearrange values so that kf^3=vf%
vf% is of 6.76 because of orbit while 3.33 is the continual. Solve for d%
d% the continual converted to elliptical motion is going to be roughly 4.1882 (4.188? I believe it's two, but shouldn't matter)
By Ugravinitial
and you have d when you apply the standard orbital motion functions.
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