The height above the Earth's surface at which the value of acceleration due to gravity reduces to half its value on the Earth's surface is h. Assume the Earth to be a sphere of radius 6400 Km. Find h.

Calculate acceleration due to gravity at a height of $2000 k m$ above the Earth's surface. Mass of Earth = $6 \times 10^{24} k g$ , radius of Earth = $6.4 \times 10^{6}$ m and (Gravitational constant, G $= 6.7 \times 10^{- 11} N m^{2} k g^{- 2}$ )

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The gravitational potential at height h above the earth's surface is −5.12×107J/kg and acceleration due to gravity at this point is 6.4ms−2. If radius of the earth is 6400km, the value of h is :-Mass of object m=1 kg

The correct option is B 1600kmGravity potential energy at a height h, P.E=−GMmR+hAcceleration due to gravity at that point g=GM(R+h)2⟹ P.E=−mg(R+h)So, −5.12×107=−1×6.4×(R+h)So, R+h=8000000 m=8000 kmOr 6400+h=8000⟹ h=1600 km

The gravitational potential at height $h$ above the earth's surface is $- 5.12 \times 10^{7} J / k g$ and acceleration due to gravity at this point is $6.4 m s^{- 2}$ . If radius of the earth is $6400 k m$ , the value of $h$ is :-
Mass of object $m = 1 k g$