g is constant near the surface and equals free-fall acceleration
Near the Earth's surface g is approximately constant because h is negligible compared with R, so (R + h)² ≈ R². Its value, 9.81 N kg⁻¹, is numerically equal to the acceleration of free fall, 9.81 m s⁻².
g_h / g_s = (R / (R + h))² ≈ 1 when h ≪ RWhat this actually means
Syllabus LO (f) asks for both halves: that g is roughly constant near the surface, and that it equals the acceleration of free fall. Answer both or lose half the marks.
The constancy argument is a ratio, not a hand-wave. g_h/g_s = (R/(R+h))². Earth's radius is 6400 km, so at 1 km up the ratio is (6400/6401)² = 0.99969, i.e. 9.81 N kg⁻¹ to three significant figures. Quote a number if the question says 'show that'.
Why the two are equal: g = F/m from the definition, and F = ma from Newton's second law for a freely falling body, so a = F/m = g. Field strength and free-fall acceleration are the same number because they are the same quotient.
Follow that through to units. N kg⁻¹ and m s⁻² are dimensionally identical, since 1 N = 1 kg m s⁻². If a units MCQ appears, that is the whole question.
Careful with the word 'constant'. g is only approximately constant, and only near the surface. The moment a question gives you a height comparable with R (say 1000 km), the approximation dies and so does ΔU = mgh.
Small refinement worth knowing: a weighing scale reads slightly less at the equator than at the poles, because at the equator part of the weight is used up providing centripetal force for the Earth's rotation, so N = mg − F_c.
Calling g exactly constant, or forgetting to state that it equals the acceleration of free fall.
Prove it — watch it be true
- Open the g vs r graph with the Earth preset and zoom the probe to just above the surface.
- Step the probe up by a few kilometres and watch the g readout barely move.
- Now drag the probe out to several Earth radii and watch g collapse, showing where the constant-g approximation stops working.