Asked 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.)
Answered by PhysicsPro
Post your work thus far and I'll continue/correct/guide
Answered by 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.
Answered by 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
Answered by 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.
There are no AI answers yet. The ability to request AI answers is coming soon!