ΔU = mgh only works near the surface
ΔU = mgh applies only where g is approximately constant, i.e. h ≪ R. For large h use ΔU = (−GMm/r_f) − (−GMm/r_i). Using mgh over a large height overestimates the change, because g is smaller up there.
h ≪ R: ΔU = mgh · large h: ΔU = GMm(1/r_i − 1/r_f)What this actually means
The decision rule is one comparison: is h small next to R = 6400 km? A 1000 m climb is, a 1000 km climb is not. Make that comparison explicitly in your working and the method mark is yours before any arithmetic.
See the size of the error. For a 3.0 kg mass raised 1000 m, mgh gives 29 kJ and the exact formula agrees. Raise it 1000 km and mgh claims 29 MJ while the correct answer is about 25 MJ, a 16% overestimate.
Overestimate, not underestimate, and you should be able to say why. mgh assumes the full surface value of g all the way up, but g weakens with height, so the real work needed is less than mgh suggests.
The two formulae also use different zeros. mgh measures from wherever you called the ground, −GMm/r measures from infinity. Never mix a mgh value and a −GMm/r value in the same energy equation.
Anything involving satellites, orbits, escape or another planet is automatically a full-formula question. mgh is only for laboratory-scale heights on the surface of a single planet.
Reaching for mgh whenever a height appears, without checking h against R.
Prove it — watch it be true
- Use the work-done two-marker tool for a small height above the surface and compare the reading with mgh.
- Now separate the markers by roughly a sixth of an Earth radius and compare again.
- Confirm the exact reading falls below the mgh estimate, and that the gap widens as the separation grows.