Explore BrainMass

# Gravitation: Radius and Period of Orbits

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Question: Calculate the period of a satellite in an orbit just above the Earth's atmosphere (whose thickness may be neglected). Find also the periods for close orbits around the Moon and Jupiter.

© BrainMass Inc. brainmass.com September 27, 2022, 3:59 pm ad1c9bdddf

## SOLUTION This solution is FREE courtesy of BrainMass!

Please see the attachment file.

Thanks for using BrainMass.

When a satellite orbits a planet just above its surface, the gravitation force on it acts as the centripetal force.
Let the mass of the satellite is m, radius of orbit is equal to the radius of planet is R and mass of the planet is M. Then the force of gravitational attraction on the satellite will be

Now if the angular velocity of rotation of the satellite is w, then the centripetal force required on it will be

As the gravitational force is the centripetal force we have

Or
Or
Hence the time period of rotation of the satellite will be

Now for earth
M = kg
R = 6.37*106 m
Substituting we get

Thus the time period of the satellite of radius nearly equal to earth will be
5058 s = 1.405 hr.
Now for Moon
M = 7.36 kg
R = 1.74*106 m
Substituting we get

Thus the time period of the satellite of radius nearly equal to moon will be
1036 s = 17.27 min.
Now for Jupiter
M = 1.90 kg
R = 7.0*107 m
Substituting we get

Thus the time period of the satellite of radius nearly equal to moon will be
2.70*106 s = 748.95 hr = 31.21 earth days.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com September 27, 2022, 3:59 pm ad1c9bdddf>