Gravitational potential at a point above surface of earth is -5.12 × 10^7 mathrm{~J} / mathrm{kg} a

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7.Gravitation

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The gravitational potential at a point above the surface of earth is $-5.12 \times 10^7 \mathrm{~J} / \mathrm{kg}$ and the acceleration due to gravity at that point is $6.4 \mathrm{~m} / \mathrm{s}^2$. Assume that the mean radius of earth to be $6400 \mathrm{~km}$. The height of this point above the earth's surface is :

A

$1600 \mathrm{~km}$

B

$540 \mathrm{~km}$

C

$1200 \mathrm{~km}$

D

$1000 \mathrm{~km}$

(JEE MAIN-2024)

Solution

$ -\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7} \ldots . \text { (i) } $

$ \frac{G M_E}{\left(R_E+h\right)^2}=6.4 \ldots \text { (ii) }$

By $(i)$ and $(ii)$

$\Rightarrow h=16 \times 10^5 \mathrm{~m}=1600 \mathrm{~km}$

Standard 11

Physics

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