Satellite orbital speed homework

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SUMMARY

The discussion focuses on calculating the altitude of a satellite with an orbital speed of 4.2 x 10^3 m/s. Using the mass of Earth (5.98 x 10^24 kg) and the radius of Earth (6.38 x 10^6 m), participants derive the necessary equations to find the altitude. The centripetal acceleration formula, a = (v^2)/r, is applied, leading to the conclusion that the gravitational force must equal the centripetal force for the satellite to maintain its orbit. The final altitude formula derived is h = [(4.2 x 10^3)^2/a] - 6.38 x 10^6.

PREREQUISITES
  • Understanding of centripetal acceleration and force
  • Familiarity with gravitational force equations
  • Basic algebra skills for manipulating equations
  • Knowledge of Earth's mass and radius
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  • Research gravitational force and centripetal force equations
  • Learn about orbital mechanics and satellite motion
  • Study the implications of varying satellite speeds on altitude
  • Explore the concept of geostationary orbits and their calculations
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1. A satellite has an orbital speed of 4.2 x 10^3 m/s. What is its altitude above Earth's surface?
mass of earth= 5.98x10^24kg
radius of Earth = 6.38x10^6m

2. acceleration and force are both centripetal
a=(v^2)/r
F=ma
possibly v^2=(Gmg)/r

3. Need help with algebra
h= distance from Earth surface to satellite
I got h= [(4.2x10^3)^2/a] - 6.38x10^6
 
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Use the fact that to remain in a circular orbit, the gravitational force must equal the centripetal force.

You do not need to know acceleration of the satelite
 

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