Derivation of Escape Velocity (1 Viewer)

tempco

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HSC 2003, Q17(b)

From this uniform circular orbit, the satellite can escape Earth's gravitational field when its kinetic energy is equal to the magnitude of the gravitational potential energy.

Use this relationship to calculate the escape velocity of the satellite.

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I know for a fact that escape velocity is root(2Gm/r). How do you derive that equation? I'm assuming kinetic energy is (mv^2)/r and the magnitude of the GPE is (GmM)/r... doesn't work out though :/
 

CrashOveride

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K.E. = (1/2)mv^2
GPE = GMm/r

Equating and re-arranging, v=sqrt(2GM/r)
So you can see it is inversely proportional to the distance between the planet and the object(so for escape velocity, this means radius of the planet) and proportional to mass of the planet.
 

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