Homework Help: Centripetal force question(involves satellite going around earth.

1. Apr 18, 2007

bilbobaggins

1. The problem statement, all variables and given/known data
Determine how long it would take for the satellite to make 1 complete revolution around earth
A 500 kg satellite is orbiting earth, radius of earth is 6.38 x 10^3 km, and acceleration of gravity as the orbital altitude of 160km is the same as earth basically 9.8.

2. Relevant equations
fc=mv^2/v
fc= 4pi^2*m*r/ t^2

3. The attempt at a solution
I've tried using both above equations. usually i started with energy equations like eg= mgh and ek= 1/2mv^2 to find velocity. Then I would put it into the fc= mv^2/v and find force then i would put that info into fc= 4pi^2*m*r/ t^2 to find the time. But none of it came out with the right answer which is 5132.8 seconds, I have tried everything I could think of. If you could help me please do.

Last edited: Apr 18, 2007
2. Apr 18, 2007

denverdoc

BEWARE of your units. You need to use 6.38E6+160E3 for radius.

Your first eqn is fine v^2/r=a
Period=circumference/v

3. Apr 18, 2007

bilbobaggins

I have used the right units, it's 6380 km + 160 km right?

4. Apr 18, 2007

denverdoc

yes but should reduce to v=sqrt(ar) so T=2*pi*sqrt(r/a) where R is in meters

5. Apr 18, 2007

bilbobaggins

can you plug the numbers in for me, im still confused as heck. Remember the answer has to be 5132.8 seconds, which i just cant find how to get to. With those I got to 114.74, actually i got that multiple times, but it's not the right answer.

Last edited: Apr 18, 2007
6. Apr 18, 2007

denverdoc

i did and verified correct answer, but again, don't use km!!! Your answer:

6.28*sqrt(6540000/9.8)

7. Apr 18, 2007

bilbobaggins

I don't think that this answers good enough though. It's about 2 off.

Last edited: Apr 18, 2007
8. Apr 18, 2007

denverdoc

well i can live with it, and its a lot better than 5000 off:uhh: seriously, think about sigfigs.

9. Apr 18, 2007

bilbobaggins

Okay heres what i tried to do
500kg*9.8^2/6540000 = .0073
then i plugged it into
t=sqrt of 4pi^2*500kg*6540000m/.0073

This seems like it should be right but it just isn't! darnit. this is making me really angry I wonder if my teacher possibly posted the wrong answer?

Last edited: Apr 18, 2007
10. Apr 18, 2007

denverdoc

Really if you're woried about the result, give it a rest. i am sure if we didn't make the assumption that a=9.8 and instead did the more rigorous approach of using,

G*M/r^2 ,

with our revised r of 6540000, we'd nail it. It was your suggestion so I thought perhaps something in the problem suggested it was OK. and we need not not even know GM
but simply mult 9.8*(6.38E6/6.54E6)^2 which of course could have been included in my expression 2*pi*sqrt(xxxxxx).

But get the mass of 500kg out of there, cancels out everywhere!

Last edited: Apr 18, 2007
11. Apr 18, 2007

bilbobaggins

I still cannot get the answer of 5132.8...

12. Apr 19, 2007

denverdoc

you're right there, it now becomes 5166 or so. This is the correct answer with all info provided. Is this online homework?