Satellite Physics Problem: Calculating Velocity for Orbit Maintenance

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To calculate the velocity required for a satellite to maintain its orbit, the centripetal acceleration must equal the gravitational acceleration. The radius of the satellite's orbit is given as 6.37 x 10^6 m. The initial assumption is to use the formula ac = v^2/r, leading to the query about using 9.8 m/s² for gravitational acceleration. However, this value is only accurate at Earth's surface, and the gravitational acceleration must be recalculated for the satellite's altitude. The discussion emphasizes the need to understand gravitational force and its variation with distance from Earth's center.
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Homework Statement



a satellite will remain in orbit such that the centripetal acceleration is equal to the acceleration due to gravity. The radius of the satellites orbit is 6.37*10^6m. What speed must the satellite travel to stay in orbit?

Homework Equations



ac=v^2/r

The Attempt at a Solution



Would I just take the sqrt of ac*r to get the velocity?
 
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This is a good start. But do you know ac?
 
Won't ac be 9.8? Since it is equal to acc due to gravity?
 
Coco12 said:
Won't ac be 9.8? Since it is equal to acc due to gravity?

That's only valid at the surface of the Earth ... do you know how to find it at any distance from the center of the earth? That is what you need!
 
How do u find it?
 
Newton's Universal Law of Gravitation ... F = G mM/r^2.
 
mM stands for?
 
Mass of earth, mass of satellite.
 
Coco12 said:
The radius of the satellites orbit is 6.37*10^6m.
Compare that to the radius of the Earth. Seems that the satellite is just skimming the surface. Is that the intention?
 
  • #10
I think so because we did not learn the formula: Newtons gravitational force yet. We only learned about ac and FC and simple harmonic motion. So I don't think the teacher will give us something that we never used yet??
 
  • #11
In that case use acceleration = 9.8 m/s/s
 
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