Centripetal force question(involves satellite going around earth.

AI Thread Summary
To determine the time for a satellite to complete one revolution around Earth, the radius must be calculated correctly, combining Earth's radius and the altitude. The correct formula involves using centripetal acceleration and gravitational force, with the radius expressed in meters. Participants in the discussion emphasize the importance of unit consistency and suggest using the gravitational constant for more accurate results. Despite various attempts, one user struggles to arrive at the expected answer of 5132.8 seconds, leading to frustration and speculation about potential errors in the problem itself. Ultimately, the discussion highlights the necessity of careful calculations and unit conversions in orbital mechanics.
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Homework Statement


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.

Homework Equations


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

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:
 
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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
 
denverdoc said:
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?
 
yes but should reduce to v=sqrt(ar) so T=2*pi*sqrt(r/a) where R is in meters
 
denverdoc said:
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, I am still confused as heck. Remember the answer has to be 5132.8 seconds, which i just can't 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.
 
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i did and verified correct answer, but again, don't use km! Your answer:

6.28*sqrt(6540000/9.8)
 
denverdoc said:
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.
 
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well i can live with it, and its a lot better than 5000 off:rolleyes: seriously, think about sigfigs.
 
denverdoc said:
well i can live with it, and its a lot better than 5000 off:rolleyes:

Okay here's 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?
 
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  • #10
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!
 
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  • #11
denverdoc said:
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...
 
  • #12
you're right there, it now becomes 5166 or so. This is the correct answer with all info provided. Is this online homework?
 
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