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(D) The gravitational potential at a height $h$ from the surface of the Earth is given by $V_h = -\frac{GM}{R+h} = -5.4 	imes 10^7\, J kg^{-1}$.The a

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$2600$

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(D) The gravitational potential at a height $h$ from the surface of the Earth is given by $V_h = -\frac{GM}{R+h} = -5.4 	imes 10^7\, J kg^{-1}$.The acceleration due to gravity at the same height is $g_h = \frac{GM}{(R+h)^2} = 6.0\, m s^{-2}$.We can relate these two expressions as $g_h = \frac{GM}{(R+h)^2} = \frac{1}{R+h} \left( \frac{GM}{R+h} ight) = \frac{-V_h}{R+h}$.Rearranging for $(R+h)$,we get $R+h = \frac{-V_h}{g_h}$.Substituting the given values: $R+h = \frac{-(-5.4 	imes 10^7)}{6.0} = \frac{5.4 	imes 10^7}{6.0} = 0.9 	imes 10^7 = 9.0 	imes 10^6\, m$.Converting the radius of the Earth to meters: $R = 6400\, km = 6.4 	imes 10^6\, m$.Now,$h = (R+h) - R = 9.0 	imes 10^6 - 6.4 	imes 10^6 = 2.6 	imes 10^6\, m$.Converting back to kilometers: $h = 2600\, km$.

At a height of $h$ $km$ from the surface of the Earth,the gravitational potential and the value of $g$ are $-5.4 \times 10^7\, J kg^{-1}$ and $6.0\, m s^{-2}$ respectively. Take the radius of the Earth as $6400\, km$.

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