Value of g on Other Planets (Mass, Radius, Density)
Physics · Gravitation · NEET
The acceleration due to gravity on any planet is g = GM/R², where M is the planet's mass and R its radius. So g grows with mass and falls fast with radius (as 1/R²). If you know density (ρ) instead, use g = (4/3)πGRρ, which shows g is directly proportional to both radius and density. Memory hook: "More mass pulls harder, more radius pushes it away by the square."
g = GM/R² depends only on the planet's mass and radius. A planet with 1/10 the mass and 1/2 the radius has g' = (1/10)/(1/2)² × g = 0.4g = 3.92 m/s² (NEET 2024), because halving the radius quarters R².
Your doubts, answered
Does g on a planet depend on the mass of the object falling?
No. In g = GM/R², M is the mass of the PLANET, not the object. The falling object's own mass cancels out (weight W = mg has the object mass m, but g itself does not). This is why a feather and a stone fall with the same g in a vacuum. Only the planet's mass and radius decide g.
Why does a bigger planet not always have a bigger g?
Because g depends on TWO things: mass (M) in the top and radius squared (R²) in the bottom. A bigger planet has more mass (raises g) but also a bigger radius (lowers g as 1/R²). Which effect wins depends on how the mass and radius compare. Example: a planet with 1/10 the mass and 1/2 the radius has g' = (1/10)/(1/2)² × g = 0.4g, so it is SMALLER even though it could look 'similar' in size.
How do I find g if I am only given density and radius?
Use g = (4/3)πGRρ. This comes from putting M = volume × density = (4/3)πR³ρ into g = GM/R². The R³ over R² leaves one R on top, so g is directly proportional to R and to ρ. This is the fastest route whenever a NEET question gives you density instead of mass.
If a planet's radius doubles (same mass), what happens to g?
g becomes one-fourth. Since g = GM/R² and M is fixed, doubling R multiplies R² by 4, so g drops to g/4. This inverse-square behaviour is the single most tested idea: any change in radius affects g by its square.
Does g change if the value of big G changes?
Yes. Since g = GM/R², g is directly proportional to G. If G were made 10 times larger, g on that planet would also become 10 times larger. This is the trick behind the NEET 2018 question: g depends on G and the planet's own mass, not on the Sun's mass.
⚠️ The NEET trap ✗ Diameter is half, so g becomes half; students plug the diameter ratio straight into g and divide by 2. ✓ g depends on RADIUS SQUARED, and radius = diameter/2. If diameter is halved, radius is halved, so R² becomes 1/4, which puts a factor of 4 (not 2) in the denominator. Then multiply by the mass ratio. 🧠 Always convert diameter to radius first, then SQUARE it. NTA loves giving 'diameter' to catch students who forget the square.
Real NEET questions
2024
The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
A · 9.8 m/s²
B · 4.9 m/s²
C · 3.92 m/s² ✓
D · 19.6 m/s²
Solution: Use g = GM/R². For the planet, M' = M/10 and diameter is half, so radius R' = R/2. Then g' = G(M/10) / (R/2)² = G(M/10) / (R²/4) = (4/10) × GM/R² = 0.4 g. With g = 9.8 m/s², g' = 0.4 × 9.8 = 3.92 m/s². Key step: halving the diameter halves the radius, and squaring gives 1/4, which becomes ×4 in the numerator.
2018
If the mass of the Sun were ten times smaller and the universal gravitational constant G ten times larger, which statement is NOT correct?
A · Time period of a simple pendulum on Earth would decrease
B · Walking on the ground would become more difficult
C · Raindrops will fall faster
D · 'g' on the Earth will not change ✓
Solution: Earth's surface gravity is g = G × M_earth / R_earth². It depends on G and the EARTH's mass, not on the Sun's mass. Making G ten times larger makes g ten times larger, so g DOES change. Therefore the statement 'g will not change' is the incorrect one. With larger g: pendulum period T = 2π√(L/g) decreases, walking is harder, and raindrops fall faster. So option D is the false statement and is the answer.
Solved Gravitation NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
g = GM/R², where G is the universal gravitational constant, M is the planet's mass and R is its radius. If density ρ is given instead, use g = (4/3)πGRρ.
How does g depend on mass, radius and density?
g is directly proportional to mass M, inversely proportional to radius squared (1/R²), and (using density) directly proportional to both radius R and density ρ. So g ∝ M, g ∝ 1/R², and g ∝ Rρ.
Why is g on the Moon about 1/6 of Earth's g?
The Moon has much less mass and a smaller radius than Earth. Putting the Moon's mass and radius into g = GM/R² gives roughly g_moon ≈ 1.6 m/s², which is about one-sixth of Earth's 9.8 m/s².
Does an object's weight change on another planet?
Yes. Weight W = mg. The object's mass m stays the same everywhere, but g differs from planet to planet, so weight changes. On a planet with smaller g you weigh less.
Is this topic important for NEET?
Yes. Surface g on planets is a repeat favourite in NEET Gravitation, appearing in 2018, 2024 and related years. The questions are quick ratio calculations, so they are easy marks if you remember g = GM/R² and to square the radius.
