An object of mass 1 kg is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, g=10 ms⁻² and radius of earth =6400 km ]

Solution

Physics sir · 4 steps

At a height of 3R, g is far from constant, so drop mgh and use the -GMm/r formula.

1. Why not mgh. The formula mgh only works when h is tiny compared with R, because it assumes g stays constant. Here h = 3R, so the distance from the centre goes from R to 4R and g drops a lot on the way.

   Δ U = -GMm/4R - (-GMm/R) = 3/4 GMm/R = 3/4 mgR

2. Replace GM. On the surface g = GM/R², so GM = gR². That turns GMm/R into mgR, which uses only the given data.
3. Put in numbers. mgR = 1 × 10 × 6.4 × 10⁶ = 6.4 × 10⁷ J. Then Δ U = 3/4 × 6.4 × 10⁷ = 4.8 × 10⁷ J.
4. So. The gain in potential energy is 48 MJ, option (a).

The idea to keep. For a height h comparable to R, use Δ U = mgh/(1 + h/R). With h = nR this becomes n/(n+1) mgR, so for h = 3R you get 3/4mgR.

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