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View Full Version : Centripetal force question(involves satellite going around earth.


bilbobaggins
Apr18-07, 10:10 PM
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.:cry:

denverdoc
Apr18-07, 10:36 PM
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

bilbobaggins
Apr18-07, 11:06 PM
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

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

denverdoc
Apr18-07, 11:32 PM
yes but should reduce to v=sqrt(ar) so T=2*pi*sqrt(r/a) where R is in meters

bilbobaggins
Apr18-07, 11:47 PM
yes but should reduce to v=sqrt(ar) so T=2*pi*sqrt(r/a) where R is in meters

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.

denverdoc
Apr18-07, 11:59 PM
i did and verified correct answer, but again, don't use km!!! Your answer:

6.28*sqrt(6540000/9.8)

bilbobaggins
Apr19-07, 12:07 AM
i did and verified correct answer, but again, don't use km!!! Your answer:

6.28*sqrt(6540000/9.8)
I don't think that this answers good enough though. It's about 2 off.

denverdoc
Apr19-07, 12:19 AM
well i can live with it, and its a lot better than 5000 off:uhh: seriously, think about sigfigs.

bilbobaggins
Apr19-07, 12:21 AM
well i can live with it, and its a lot better than 5000 off:uhh:

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?

denverdoc
Apr19-07, 12:34 AM
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!

bilbobaggins
Apr19-07, 12:56 AM
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!

I still cannot get the answer of 5132.8...

denverdoc
Apr19-07, 01:28 AM
you're right there, it now becomes 5166 or so. This is the correct answer with all info provided. Is this online homework?