at which height acceleration due to gravity becomes one forth compare to earth surface
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g = G*M/(r)^2
g_prime = G*M/(r+h)^2
g/g_prime = (r+h)^2/(r)^2
sqrt(g/g_prime) = (r+h)/r
substituting the given values in the above equation:
sqrt(g/g/4)=sqrt(4)=2=(r+h)/r
2*r = r+h
h = r
Therefore at a height equal to radius of the earth(6400 km), earth’s acceleration due to gravity will be one fourth of it’s value on the surface of the earth.
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