Physics · Gravitation · NEET
Both are correct, but for different situations. The EXACT change when a mass m is raised from the surface (distance R from centre) to height h is ΔU = GMm[1/R − 1/(R+h)] = GMmh / [R(R+h)]. The formula mgh is only an APPROXIMATION that works when h is very small compared to R (so that g stays nearly constant). For NEET, if the question gives h in metres or a few km near the surface, use mgh. If h is comparable to R (like h = R or h = R/2), you MUST use the exact form.
Gravitational PE is defined as negative and equals −GMm/r, with zero at infinity. As you go up, r increases (from R to R+h), so −GMm/r becomes a smaller negative number, meaning it INCREASES (moves toward zero). So the change ΔU = U(final) − U(initial) comes out positive. Physically, you push against gravity while lifting, so you add energy to the mass. Higher position means more stored energy.
Use mgh only when h ≪ R (h is much smaller than Earth's radius, 6400 km). At heights of a few metres, a building, or even a few kilometres, g barely changes, so mgh is accurate. Once h becomes a large fraction of R — for example h = R/2 or h = R — g drops noticeably, and mgh over-estimates the true change. Then use ΔU = GMmh / [R(R+h)].
Start with ΔU = GMmh / [R(R+h)]. If h ≪ R, then (R+h) ≈ R, so the denominator ≈ R × R = R². This gives ΔU ≈ GMmh / R² = (GM/R²) m h. But GM/R² is exactly g at the surface, so ΔU ≈ mgh. This is why mgh is just a near-surface special case of the general formula.
The PE at a distance r is U(r) = −GMm/r (a single value, always negative). The CHANGE in PE is the difference between two positions: ΔU = U(final) − U(initial). This page is about the change when moving from the surface to height h. The change is what equals the work you must do against gravity, so it is the physically useful quantity in energy problems.
Taking gravitational potential energy at infinity to be zero, the change in PE (final − initial) of a mass m raised to height h above the earth's surface (radius R) is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ΔU = GMmh / [R(R+h)], where M is Earth's mass, R its radius, and h the height above the surface. For h ≪ R this reduces to mgh.
Yes. Moving farther from Earth's centre makes U = −GMm/r less negative, so the difference U_final − U_initial is positive. You add energy by lifting.
No. Gravity is a conservative force, so the change in PE depends only on the starting and ending heights, not on the route between them.
Only when h is much smaller than Earth's radius R (6400 km). At everyday heights g is nearly constant, so mgh is accurate. For h = R or larger, use the exact formula.
The change in gravitational PE equals the work done against gravity to lift the mass slowly (no change in kinetic energy). So ΔU = W_against gravity.
