Circular motion in planets and satellites

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Homework Statement



A spaceship in outerspace has a donut shaped with a 500 m outer radius. The inhabitants stand with their heads towards the center and their feet on an outside rim. Over what time interval with the spaceship have to complete one rotation on its axis to make a bathroom scale have the same reading for the person in space as when on Earth's surface?

Homework Equations



Possibly G*m1*m2/r^2

G= 6.67*10^-11 N*m^2

Or

4*pi^2*R/ T^2

The Attempt at a Solution



Honestly I'm completely lost here. I'm not sure which formula to use to be able to solve for this.
 

Answers and Replies

  • #2
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Possibly G*m1*m2/r^2
Hmmm ---- possibly. What are m1 and m2 here?
 
  • #3
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Correct me if I'm wrong, but you're looking for the period in which it rotates so that the acceleration is equal to the acceleration due to Earth's gravity?
 
  • #4
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Gravity will be negligible. There is another effect that the scale will detect. Hint: the inhabitants rotate together with the space station.
 
  • #5
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Hmmm ---- possibly. What are m1 and m2 here?
There were no values given for any mass, so I'm not sure.
 
  • #6
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Mass doesn't matter.. It will cancel out.
 
  • #7
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If you want to use Gm1m2/r2, it will give you the acceleration at earth's surface provided you use earth's mass for one of the mass values.
to make a bathroom scale have the same reading for the person in space as when on Earth's surface?
Or, you can use previous knowledge of that value.
what time interval
one rotation on its axis
What do those two parts of the question bring to mind?
 
  • #8
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If you want to use Gm1m2/r2, it will give you the acceleration at earth's surface provided you use earth's mass for one of the mass values.

Or, you can use previous knowledge of that value.

What do those two parts of the question bring to mind?

Do we have to find the period here? If so, how do we do that in this problem? Do we set something equal to mv^2/r?
 
  • #9
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Do we set something equal to mv^2/r?
Well, that would be equal to the force created, but we only need the acceleration, correct? And as Force = (Acceleration)(Mass), then without (Mass), v2/r = ?
 
  • #10
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Well, that would be equal to the force created, but we only need the acceleration, correct? And as Force = (Acceleration)(Mass), then without (Mass), v2/r = ?
The question asks to solve for the length of the time interval the spaceship would have to complete one rotation on its axis, so isn't that finding the period, not the acceleration?
 
  • #11
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You are correct, but to do so we need the (tangential) velocity at which it rotates. (I think. Somebody PLEASE correct me if I am wrong before I make it worse lol).
 
  • #12
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PLEASE correct me
For all participants: is there a relationship between period (or angular velocity) and acceleration?
 
  • #13
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My train of thought is that we can use the acceleration to find tangential velocity, and then use that and the circumference of the spaceship to find the time, or the period. Probably wrong, that's just what I'm going with.
 
  • #14
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I'll restate my last question: What is the relationship between angular velocity and acceleration?
 
  • #15
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The question is the period of rotation of the spaceship about its axis not its period along its orbit.

Over what time interval with the spaceship have to complete one rotation on its axis
 
  • #16
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Are these not the same?
 
  • #17
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But I'm gonna step out now, definitely gonna keep an eye on this post though. Sorry op, not trying to hijack your post lol
 
  • #18
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My train of thought is that we can use the acceleration to find tangential velocity, and then use that and the circumference of the spaceship to find the time, or the period. Probably wrong, that's just what I'm going with.
Instead of the tangential velocity at the outside of the spaceship, determine the angular velocity from the expected centripetal acceleration.
 
  • #19
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A spaceship in outerspace
There are no orbits involved at all.
 

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