A planet has a mass M₁ and radius R₁. The value of acceleration due to gravity on its surface is g₁. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?
- (a)g₁
- (b)2g₁
- (c)g₁/2
- (d)g₁/4
Correct — C, g₁/2. The acceleration due to gravity at the surface of a planet is g = GM/R², where G is the universal gravitational constant, M the mass of the planet and R its radius. For planet 2 the mass is 2M₁ and the radius is 2R₁, so g₂ = G(2M₁)/(2R₁)² = G(2M₁)/(4R₁²) = (1/2) × GM₁/R₁² = g₁/2. Doubling the mass alone would have doubled g, but doubling the radius divides it by four because the radius is squared, and the net effect of the two changes together is to halve it.
- (a)g₁ — This assumes the two doublings cancel. They would only cancel if the radius entered the formula to the first power. Because it is squared, the radius change is the stronger of the two, so g must fall rather than stay the same.
- (b)2g₁ — This tracks the mass and forgets the radius altogether. A more massive planet does pull harder, but if it is also bigger you stand further from its centre, and that distance works against you twice over.
- (d)g₁/4 — This applies the radius change and forgets the mass. Dividing by four is what the doubled radius does on its own; the doubled mass then multiplies the result back by two, leaving a half rather than a quarter.
Newton's law of gravitation gives the force between two masses as F = Gm₁m₂/r². Applying it to a body of mass m resting on a planet's surface, and equating that force to mg, cancels m out and leaves g = GM/R². The mass of the object being attracted therefore never appears — which is why all bodies fall at the same rate in a vacuum. What does appear is the planet's own mass, in direct proportion, and its radius, in inverse square proportion. That is why the Moon, with about one eightieth of the Earth's mass but only about a quarter of its radius, still manages a surface gravity of about one sixth of the Earth's rather than one eightieth.
Problems of this shape are pure exponent bookkeeping, and the safe method is to write the formula, substitute the multipliers as symbols and simplify — never to reason in words about which effect is 'stronger'. Here the mass multiplier is 2 on top and the radius multiplier is 2² = 4 underneath, giving 2/4 = 1/2. The same discipline handles every variant: mass tripled and radius tripled gives g/3; mass unchanged and radius halved gives 4g; mass doubled and radius unchanged gives 2g. Note also that if the two planets were made of the same material, doubling the radius would multiply the mass eightfold, not twofold — so this question is describing a less dense second planet, which is consistent with its weaker surface gravity.
- The acceleration due to gravity at a planet's surface is g = GM/R², independent of the mass of the falling body.
- The universal gravitational constant G is about 6·67 × 10⁻¹¹ N m² kg⁻² and is the same everywhere.
- On Earth g is about 9·8 m s⁻²; on the Moon it is about one sixth of that.
- Mass is the same everywhere, but weight, which is mg, changes with the value of g at the location.
Two on top against four underneath gives one half — option (c).
- Treating mass and radius as symmetric. The radius is squared, so a given factor applied to it counts twice as hard.
- Bringing the mass of the falling body into the answer; it cancels out of g entirely.
- Assuming a bigger planet must have a stronger surface gravity. It depends on how the mass and radius change together.
NDA asks how g changes when a planet's mass or radius is scaled, or compares the surface gravity of two bodies given their mass and size ratios.
The mass of a body on Earth is 100 kg (acceleration due to gravity, gₑ = 10 m/s²). If acceleration due to gravity on the Moon = gₑ/6, then the mass of the body on the moon is
- (a) 100/6 kg
- (b) 60 kg
- (c) 100 kg
- (d) 600 kg
Answer(c) 100 kg
The companion distinction to this card — g changes from world to world, but the mass of the body being weighed does not, which is exactly why the falling body's mass cancels out of g = GM/R².
Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be
- (a) 16 F
- (b) 1 F
- (c) 4 F
- (d) None of the above
Answer(a) 16 F
- practice — not a real PYQ
If the radius of the Earth were halved while its mass stayed the same, the acceleration due to gravity at its surface would become
- (a)half its present value
- (b)twice its present value
- (c)four times its present value
- (d)unchanged
Answer(c) four times its present value — g varies as the inverse square of the radius.
- practice — not a real PYQ
A planet has three times the mass and three times the radius of the Earth. Its surface value of g, compared with the Earth's, is
- (a)three times
- (b)nine times
- (c)one-third
- (d)one-ninth
Answer(c) one-third — the factor is 3 divided by 3², which is 1/3.