# Circular Orbits and Motion - Satellites

1. Sep 25, 2007

### BlueSkyy

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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

2. Sep 25, 2007

### Kurdt

Staff Emeritus
What is Kepler's 3rd law? Have you learned Newton's form of Kepler's 3rd law for circular orbits?

3. Sep 25, 2007

### BlueSkyy

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...

4. Sep 25, 2007

### BlueSkyy

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...

5. Sep 25, 2007

### Kurdt

Staff Emeritus
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.