The acceleration due to gravity at the Earth's surface depends on
- (a)its mass only.
- (b)its radius only.
- (c)both its mass and radius.
- (d)either its mass or its radius.
Correct — C, both its mass and radius. Newton's law of gravitation gives the force on a body of mass m at the surface of a planet of mass M and radius R as G times M times m divided by R squared. Divide by m and the acceleration comes out as g equals GM divided by R squared — the body's own mass has cancelled, but the planet's mass and its radius are both still in the expression, one on top and one squared underneath. That is why the value on the Moon is roughly a sixth of the Earth's: the Moon has far less mass, which pulls g down, but it is also much smaller, which pushes g up, and the two effects only partly cancel. Neither quantity alone will do.
- (a)its mass only. — Take two planets of equal mass and different sizes and the larger one has the weaker surface gravity, because the surface sits further from the centre. Radius enters the formula as a square, so it matters more sharply than mass does.
- (b)its radius only. — Two bodies of the same size but different densities have different masses and therefore different surface gravity. Radius alone fixes nothing.
- (d)either its mass or its radius. — The word either makes it an alternative when the formula makes it a requirement. Both quantities appear in g equals GM by R squared at the same time, and knowing only one of them leaves g undetermined.
The expression g equals GM divided by R squared explains most of what a candidate needs about surface gravity. Double the mass of a planet and g doubles; double its radius and g falls to a quarter. Double both and g halves, which is the standard examination twist. Above the surface g falls off with distance from the centre, and below the surface it falls again because only the mass inside the radius pulls, reaching zero at the centre.
The stem says at the Earth's surface, which fixes the setting and points at the plain formula. Worth adding is the reason the measured value is not the same everywhere on that surface: the Earth is not a perfect sphere but bulges at the equator, so the equatorial surface is further from the centre and g there is smaller than at the poles, and the rotation of the Earth removes a further slice at low latitudes. Those are refinements on a value that the formula in the answer already delivers to good accuracy.
- Acceleration due to gravity at a planet's surface is g equals GM divided by R squared.
- The falling body's own mass cancels out, which is why all bodies fall together in a vacuum.
- Radius enters as a square, so a change in size affects g more sharply than the same fractional change in mass.
- A planet with twice the mass and twice the radius of another has half its surface gravity.
- Measured g is slightly larger at the poles than at the equator, because the Earth is not a perfect sphere and because it rotates.
Cancel the falling body first; whatever survives on the right is what g depends on.
- Reading either as both; option (d) is the careless pick on this stem.
- Forgetting that R is squared, which makes size the more sensitive of the two.
- Assuming g must be larger on a larger planet.
As a which-quantities-matter item like this one, or as a short numerical comparing two planets whose mass and radius are given as multiples.
Assertion (A): The weight of a body decreases with the increase of latitude on earth. Reason (R): The earth is not a perfect sphere.
- (a) Both A and R are individually true and R is the correct explanation of A
- (b) Both A and R are individually true but R is NOT the correct explanation of A
- (c) A is true but R is false
- (d) A is false but R is true
Answer(d) A is false but R is true
The same dependence taken one step further. Because the Earth bulges at the equator, the surface there is further from the centre and g is smaller, so weight rises rather than falls as one moves towards the poles — a direct consequence of R sitting in the denominator.
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
Answer(c) g₁/2
The formula put to work. Doubling the mass doubles g and doubling the radius divides it by four, so the surface gravity halves — a result that is impossible to reach if either quantity is left out.
CDS_GK_2022_I_Q212022An object weighs 9 N on the surface of the Earth. What would be its weight, when measured on the surface of a planet where the acceleration due to gravity is 9 times that on the surface of the Earth ?
- (a) The weight would remain the same
- (b) The weight would be equal to 1 N
- (c) The weight would become 9 times
- (d) The weight will be reduced to 1⁄9 N
Answer(c) The weight would become 9 times
The same year's first session, testing what follows once g is known. Weight is mass times g, so a place where g is nine times as large multiplies the weight by nine while the mass is untouched.
- practice — not a real PYQ
A planet has twice the mass and half the radius of the Earth. Its surface gravity, compared with the Earth's, is
- (a)the same
- (b)twice as large
- (c)four times as large
- (d)eight times as large
Answer(d) eight times as large — doubling M doubles g and halving R multiplies it by four, so the two effects together give a factor of eight.
- practice — not a real PYQ
A body is taken from the Earth to the Moon. Which one of the following remains unchanged?
- (a)Its weight
- (b)Its mass
- (c)The acceleration due to gravity acting on it
- (d)The force of gravity on it
Answer(b) Its mass — mass is a property of the body, while weight is the force on it and changes with the local value of g.