# Period of a satellite orbiting Earth

1. Aug 27, 2011

### fawk3s

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

A satellite is orbiting Earth 4200 km from Earth's surface. Considering the trajectory to be a circle, calculate the period of the orbiting satellite.

2. Relevant equations

a=v2/r
F=GmM/r2
c=2r*PI
v=l/t

3. The attempt at a solution

Well, the textbook gives the mass of the Earth to be M=5,98*1024 kg and the radius of Earth RE=6,38*106 m

So the radius of the orbit is R=RE+4200 km=10,58*106 m

Now off to find the gravitational acceleration for the satellite at the given distance (which is also the centripetal acceleration in this case)

F=GmM/R2
a=F/m

so

a=GM/R2
a=3,56 m/s2

a=v2/R => v=sqrt(a*R)
v=6,14*103 m/s

c=2R*PI
c=66,44*106

And now for the period

v=c/T => T=c/v
T=10820 s

The textbook says it ought to be 8300 s. Where did I go wrong? Have double checked the numbers, so dont think its a calculation error.

fawk3s

2. Aug 27, 2011

### vela

Staff Emeritus
I get the same answer you do. The book's answer is wrong.

3. Aug 27, 2011

### Lobezno

This post may be redundant, as I'm not sure exactly what you did on the second half, but...

If you use the Newtonian variation of Kepler's third law, the number pops out straight away!

4. Aug 28, 2011

Thanks guys.