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Simulated gravity-satellites in orbit

  • Thread starter kiro484
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  • #1
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Homework Statement


Ringworld is an artificial world constructed in the shape of a giant ring that rotates around a star similar to our sun. The radius of the ring is 1.53x10^11m (from the sun to the ring). The mass of ringworld is 2.1x10^27kg and its simulated gravitational acceleration is 9.73m/s^2. Determine how fast the inner surface of the ringworld is moving to experience a "gravitational" (no idea why this is put in quotations in the question) acceleration of 9.73 m/s^2.
Gravitational constant=6.67x10^-11


Homework Equations


Vcentripetal=2πr/T
Acentripetal=v^2/r or =4π^2(r)/T^2
F=Gm1m2/r^2
g=Gm/r^2
T1^2/R1^3=T2^2/R2^3
Any combination of the above also work.

The Attempt at a Solution


How do you find the mass of the central body and the orbital period? It seems impossible to do anything without knowing one or the another. I don't know where to start.

Any help is appreciated. Thanks.
 

Answers and Replies

  • #2
haruspex
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It says similar to the sun, so assume it has that same mass. The speed with which the ring is 'orbiting' its sun is what you are trying to calculate. But it isn't really orbiting it all, since its mass centre is at the sun. It is merely rotating around its sun.
I am a bit worried that it quotes the mass of the ring. Seems like they expect you to take into account its gravitational pull. Not sure how to get that when the point of interest is not on the axis of the ring. As I recall, you can't do that by taking the ring to be a line of zero width - the integral does not converge.
 
  • #3
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The acceleration due to gravity from the sun is less than 0.01 m/s^2, so it's not significant. The ring is a thousand times lighter, so its gravity isn't significant either. All the simulated "gravitiational" acceleration has to come from the acceleration due to the rotation of the ring. (the quotes are there because the acceleration isn't due to gravity). The computation would be the same if the ring was rotating on its own without a nearby sun.
 

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