The gravitational potential at a point above | vedclass

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(A) Let $R_E$ be the radius of the Earth and $h$ be the height of the point above the surface. The distance from the center of the Earth is $r = R_E +

Vedclass · Accepted Answer

$1600$

Vedclass · Answer

(A) Let $R_E$ be the radius of the Earth and $h$ be the height of the point above the surface. The distance from the center of the Earth is $r = R_E + h$.Gravitational potential $V = -\frac{G M_E}{r} = -5.12 	imes 10^7 \,J/kg$ ... $(i)$Acceleration due to gravity $g' = \frac{G M_E}{r^2} = 6.4 \,m/s^2$ ... $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{V}{g'} = \frac{-G M_E / r}{G M_E / r^2} = -r$$r = -\frac{V}{g'} = -\frac{-5.12 	imes 10^7}{6.4} = 0.8 	imes 10^7 \,m = 8000 \,km$Since $r = R_E + h$, we have $h = r - R_E = 8000 \,km - 6400 \,km = 1600 \,km$.

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

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