Satellite & Gravity Homework: Find Period of Revolution

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SUMMARY

The discussion focuses on calculating the period of revolution for a communications satellite in a circular orbit at a height of 3.59 x 10^7 meters above Earth's surface. Using the provided values, including Earth's radius (6.37 x 10^6 meters), mass (5.98 x 10^24 kg), and the gravitational constant (6.67 x 10^-11 m^3 kg^-1 s^-2), the expected period of revolution is determined to be 24 hours. The relationship between centripetal acceleration and gravitational force is emphasized, leading to the calculation of angular velocity (omega) and its connection to the orbital period.

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  • Understanding of gravitational force and centripetal acceleration
  • Familiarity with the formula for angular velocity (omega)
  • Knowledge of the gravitational constant and its application in orbital mechanics
  • Basic proficiency in algebra for manipulating equations
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Homework Statement



It is proposed to place a communications satellite in a circular orbit rounf the equator at a height og 3.59 x 10^7m above the Earth's surface. Find the period of revolution of the satellite in hours and comment on result.

Values given:
Radius of Earth: 6.37 x 10^6m
Mass of Earth: 5.98 x 10^24kg
Gravitational Constant: 6.67x10^-11 m^3 kg^-1 s^-2

Homework Equations



Not sure which one you can you, but i think two different ones have to be used

The Attempt at a Solution



This is the only question which i couldn't answer, and when i tried several ways out, it got very far away from result. the answer should be :24 hrs
 
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You know that the centripetal accelration is provided by the gravitational force.

So,

mr(omega)^2 = GMm/r^2

SO you can calculate omega, the angular velocity.

Then you should have an equation relating the period tn the angular velocity.
 
thx, i'll try it out. If i get wrong result, i'll post my working to see what's wrong
 

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