Centripetal force question(involves satellite going around earth.

In summary: If so, it may be that the answer they want is an approximation and not the actual value. In summary, to determine how long it would take for a 500 kg satellite to make one complete revolution around the Earth, you can use the equation T = 2*pi*sqrt(r/a), where r is the combined radius of the Earth and the orbital altitude (6.38 x 10^3 km + 160 km) and a is the acceleration of gravity (9.8 m/s^2). The answer should be approximately 5166 seconds. It is important to use consistent units and to be mindful of significant figures when solving this problem.
  • #1
bilbobaggins
12
0

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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  • #2
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
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?
 
  • #4
yes but should reduce to v=sqrt(ar) so T=2*pi*sqrt(r/a) where R is in meters
 
  • #5
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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  • #6
i did and verified correct answer, but again, don't use km! Your answer:

6.28*sqrt(6540000/9.8)
 
  • #7
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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  • #8
well i can live with it, and its a lot better than 5000 off:uhh: seriously, think about sigfigs.
 
  • #9
denverdoc said:
well i can live with it, and its a lot better than 5000 off:uhh:

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?
 

1. What is centripetal force?

Centripetal force is a force that acts on an object in circular motion, pulling it towards the center of the circle.

2. How does centripetal force keep a satellite in orbit?

Centripetal force is responsible for keeping a satellite in orbit around the Earth. The force of gravity between the Earth and the satellite acts as the centripetal force, pulling the satellite towards the center of the Earth.

3. How is centripetal force calculated?

The magnitude of centripetal force can be calculated using the formula Fc=mv^2/r, where m is the mass of the object, v is the velocity, and r is the radius of the circular motion.

4. Can centripetal force be greater than the force of gravity?

Yes, it is possible for the centripetal force to be greater than the force of gravity. In this case, the object will move in a circular path with a smaller radius.

5. Can an object have centripetal force without moving in a circular path?

No, an object must be moving in a circular path in order to have centripetal force acting on it. If the object moves in a straight line, there is no centripetal force acting on it.

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