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 :

Solution

Physics sir · 4 steps

Divide the potential by g: GM cancels and you get the distance from the earth's centre directly.

1. Two formulas. At a distance r from the earth's centre, the potential is V = -GM/r and the field is g = GM/r². Both contain GM, which you do not know.
2. Take the ratio. Divide the size of V by g. The GM cancels and one power of r is left.

   |V|/g = (GM/r)/(GM/r²) = r

3. Put in numbers. r = (5.12 × 10⁷)/6.4 = 8 × 10⁶ m = 8000 km. This r is measured from the centre of the earth, not from the surface.
4. So. Height above the surface is h = r - R = 8000 - 6400 = 1600 km, which is option (a).

The idea to keep. When you are given both V and g at the same point, |V|/g gives r straight away. Always remember r is from the centre, so subtract R to get the height.

Answer from the NTA final answer key. Solution by PrepWiser.