Satellite orbital speed homework

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To determine the altitude of a satellite with an orbital speed of 4.2 x 10^3 m/s, the gravitational force must equal the centripetal force. The relevant equations include a = (v^2)/r and F = ma, with the relationship v^2 = (Gmg)/r. The altitude can be calculated using the formula h = [(4.2x10^3)^2/a] - 6.38x10^6, where h represents the distance from the Earth's surface to the satellite. The mass of the Earth is 5.98 x 10^24 kg, and the radius of the Earth is 6.38 x 10^6 m. Understanding these principles allows for the successful calculation of the satellite's altitude.
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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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