Field-strength ratios: never plug in G
When two situations share the same M, divide the equations: g₂/g₁ = (r₁/r₂)². If instead the DENSITY is fixed, M ∝ r³ so g ∝ r.
g₂/g₁ = (r₁/r₂)² · same density ⇒ g = (4/3)πGρr ∝ rWhat this actually means
This is the single highest-yield technique in the topic and it is almost always an MCQ. Spot it when the question gives you RELATIONSHIPS rather than numbers: 'at a height 2R', 'a sphere of twice the diameter', 'the surface value is g_s, find it at...'.
Write g = GM/r² twice and divide. G and M vanish and you are left with a pure ratio you can do in your head. A meteorite at height 2R above a planet of radius R is at r = 3R from the CENTRE, so a/g_s = (R/3R)² = 1/9.
That centre business is where most marks die. Heights are measured from the surface, but r in the formula is measured from the centre. Always convert: r = R + h, every single time.
The same-density variant is a different beast. If ρ is fixed then M = ρ(4/3)πr³, so g = GM/r² = (4/3)πGρr, i.e. g ∝ r, not 1/r². Double the diameter at the same density and you DOUBLE g. Students who reflexively reach for the inverse square get this exactly backwards.
Combined versions exist: 'the acceleration of free fall on Earth is 6 times that on the Moon and Earth's density is 5/3 of the Moon's, find the radius ratio'. Use g ∝ ρr, so r_E/r_M = (g_E/g_M)(ρ_M/ρ_E) = 6 × 3/5 = 3.6.
Using height above the surface as r instead of r = R + h, and assuming g ∝ 1/r² in a fixed-density comparison.
Prove it — watch it be true
- With the Earth preset, note the g readout at the surface.
- Drag the test mass probe to r = 3R and check g has fallen to one ninth of the surface value.
- Switch to a denser or larger planet preset and compare surface g values to see how M and R trade off.