Circular Orbits and Motion - Satellites

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 8K views
BlueSkyy
Messages
34
Reaction score
0

Homework Statement



A spy satellite is in circular orbit around Earth. It makes one revolution in 6.02 hours. (Radius of the Earth=6.371 times 106 m)

(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?

Homework Equations



v = sqrt(G*M Earth / r)
Kepler's 3rd Law

The Attempt at a Solution



I found the angular velocity to be 2.899 rad/sec...
I really don't know where to start after that
 
Physics news on Phys.org
i just found Newton's form of Kepler's 3rd law on a different website - i will see if i can get it to work...
 
i'm still getting a ridiculous number for the distance...i assume i use Newton's form of Kepler's 3rd law but do i keep the period in hours or seconds? argh...
 
BlueSkyy said:
i'm still getting a ridiculous number for the distance...i assume i use Newton's form of Kepler's 3rd law but do i keep the period in hours or seconds? argh...

The SI units are seconds, so you will have to convert. Remember the distance will be the radius of the Earth plus the satellites orbital height.