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Centripetal Acceleration of Satellite

  • Thread starter Physics53
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  • #1
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Homework Statement


Hi, I am stuck on this question:
A satellite of mass 1200 kg is in orbit around the Earth at a distance of 22000 km from the centre of the Earth.
Calculate the magnitude of the centripetal acceleration of the satellite at this distance.

The Attempt at a Solution


I did:
mv^2/r= GMm/r^2
so v= square root of GM/r
v= square root of 6.67 x10^-11 x 6.0 x 10^24/ 22000 x 10^3
this gave me=
4.266x 10^-3 m/s but the answer at the back is 0.83 m/s^2, can someone please explain to where i went wrong.

Thanks[/B]
 

Answers and Replies

  • #2
vela
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Carefully reread the question and make sure you're answering it.
 
  • #3
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Carefully reread the question and make sure you're answering it.
Hi, I've reread it many time, but i still feel confused
 
  • #4
vela
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What's it asking you to find? What did you actually calculate?
 
  • #5
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What's it asking you to find? What did you actually calculate?
Oh I see what you mean, are you implying that I solved centripetal velocity instead of centripetal acceleration?
 
  • #6
vela
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Essentially. You solved for what's sometimes called the tangential velocity, not centripetal velocity.
 
  • #7
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Essentially. You solved for what's sometimes called the tangential velocity, not centripetal velocity.
So does this mean I consider gravity of Earth as my centripetal velocity and therefore:
g= GM/r^2, but i still have a query, if you could kindly clear this up for me, why can't we use the centripetal formula in this question.
 
  • #8
vela
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The work you did was correct, but you didn't actually solve for the quantity you were asked for. (The fact that the units were different is a big clue.)

I was just pointing out the phrase "centripetal velocity" doesn't really make sense. The velocity you found is tangent to the satellite's path; it doesn't point inward. The acceleration, however, does, so that's why it's referred to as centripetal acceleration.
 

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