The gravitational potential at a point above the surface of earth is -5.12 × 10⁷ J/kg and the acceleration due to gravity at that point is 6.4 m/s². Assume that the mean radius of earth to be 6400 km. The height of this point above the earth's surface is: (a) 1600 km (b) 1200 km (c) 540 km (d) 1000 km

Answer: (a). V = -GM/r and g = GM/r², so r = |V|/g. r = (5.12×10⁷)/(6.4) = 8×10⁶ m = 8000 km h = 8000 - 6400 = 1600 km So the answer is (a) 1600 km.

The gravitational potential at a point above the surface of earth is… | JEE Main 2024 (30 Jan, Shift 1)

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Gravitational Potential & Energy

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

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Answer: (a)

$V = -\frac{GM}{r}$ and $g = \frac{GM}{r^2}$ , so $r = \frac{|V|}{g}$ .
$r = \frac{5.12\times10^7}{6.4} = 8\times10^6\ \text{m} = 8000\ \text{km}$
$h = 8000 - 6400 = 1600\ \text{km}$
So the answer is (a) 1600 km.

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