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You can use V^2=V_0^2+2a(delta(y))

we know that V_0 = 0, delta(y)=h and a=GM/R^2

substituting we get V^2=2GMh/R^2

so V=sqrt(2GMh/R^2)

You can also do this using the gravitational definition of potential energy

If GMm/R^2=force of gravity then we know the integral of this is equal its potential energy.

so PE= -GMm/R evaluated from r to r+h setting this equal to kinetic energy (1/2*mv^2) and evaluating our integral we get

v=sqrt[2GM(1/R-1/(R+h))]

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12y ago
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√(gm/r)

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gm/r

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Q: An object of mass m is dropped from height h above a planet of mass M and radius R?
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