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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 :

Question

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 :

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Solution

The gravitational potential energy (V) at a point in space due to Earth is given by the formula:

V = -GM/r

where: G is the gravitational constant (6.674 x 10^-11 N(m/kg)^2), M is the mass of the Earth (5.972 x 10^24 kg), r is the distance from the center of the Earth to the point in space.

The acceleration due to gravity (g) at that point is given by the formula:

g = GM/r^2

We can solve the second equation for M:

M = gr^2/G

Substitute M from the second equation into the first equation:

V = -g*r

Solve for r (distance from the center of the Earth to the point in space):

r = -V/g

Given that V = -5.12 x 10^7 J/kg and g = 6.4 m/s^2, we can find r:

r = -(-5.12 x 10^7 J/kg) / (6.4 m/s^2) = 8 x 10^6 m = 8000 km

The height h above the Earth's surface is the difference between r and the radius of the Earth (R):

h = r - R = 8000 km - 6400 km = 1600 km

So, the height of the point above the Earth's surface is 1600 km.

This problem has been solved

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