Definition
Acceleration due to gravity
The acceleration produced by gravitational force on a body in free fall near Earth's surface: g = G M / R² ≈ 9.8 m/s², where M and R are Earth's mass and radius.
-- NCERT Class 11 Physics, Ch. 7, p. 133Acceleration Due to Gravity
8 MCQs4 revision cards9-step worked example
Source: NCERT GravitationPYQ coverage: NEET 2022, 2024Official key: NTA-verifiedLast updated: 26 Sep 2026
Acceleration Due to Gravity, explained for NEET
Acceleration due to gravity is the acceleration a freely falling body experiences near a massive object. On Earth's surface, g ≈ 9.8 m/s² and connects to Newton's law of gravitation: equating F = GMm/R² with mg gives g = GM/R² (NCERT Class 11 Physics Chapter 7, page 131).
The trap that costs marks: when a question gives altitude as a fraction of R — say h = R — students reach for the linear approximation g(1 − 2h/R) and get nonsensical answers. That approximation is only valid when h ≪ R. The exact formula is:
g_h = g (R / (R + h))²
For h = R, the exact result is g/4, while the linear formula gives g(1 − 2) = −g — physically meaningless. NEET has tested this in 2024 and 2025.
Variation with depth follows a different law. Assuming uniform density:
g_d = g (1 − d/R)
This IS genuinely linear — g decreases uniformly and reaches zero at Earth's centre. The common confusion is mixing up which formula is linear and which is inverse-square: altitude is inverse-square, depth is linear.
Why this matters for you: g-variation questions appear roughly every 2–3 years. They carry medium negative-marking risk because the linear-vs-inverse-square trap generates plausible-looking wrong options. If you can write down the correct formula within 5 seconds of reading the stem, the question is free marks.
Watch-out: problems sometimes give height as a multiple of R (h = R/2, h = 2R) precisely to punish the linear approximation. Always check whether h is comparable to R before choosing your formula.
Can you answer these Acceleration Due to Gravity MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The acceleration due to gravity on Earth's surface is related to the universal gravitational constant by which expression?
Show answer and why every option is right or wrong
Answer: B. From Newton's law of gravitation applied at Earth's surface: mg = GMm/R², cancelling m gives g = GM/R² (NCERT Class 11 Physics Chapter 7, page 131).
Why A is wrong: A gives g = GM/R, which has dimensions of m²/s² (energy per unit mass), not acceleration. This is actually the gravitational potential magnitude, not g.
Why C is wrong: C gives g = GM²/R, which has dimensions of kg·m²/s² — not acceleration. This confuses the roles of mass and distance in Newton's law.
Why D is wrong: D gives g = G/MR², which has dimensions of 1/(kg·s²) — not acceleration. This inverts the mass dependence.
MCQ 2Easy RecallPractice
At what location does the acceleration due to gravity become zero, assuming Earth has uniform density?
Show answer and why every option is right or wrong
Answer: D. At the centre, g_d = g(1 − d/R) = g(1 − 1) = 0. At infinite distance, g_h = g(R/(R+h))² → 0 as h → ∞. Both locations give zero g (NCERT Class 11 Physics Chapter 7, page 134).
Why A is wrong: A is wrong. At the surface, d = 0 and h = 0, so g equals its full surface value g = GM/R².
Why B is wrong: B accounts for only one location. g also vanishes at Earth's centre under the depth formula.
Why C is wrong: C accounts for only one location. g also approaches zero at infinite distance under the altitude formula.
MCQ 3Easy RecallPractice
The variation of g with depth below Earth's surface (uniform density) is:
Show answer and why every option is right or wrong
Answer: A. g_d = g(1 − d/R). This is a linear function of d, decreasing from g at the surface to 0 at the centre (NCERT Class 11 Physics Chapter 7, page 133).
Why B is wrong: B suggests g ∝ 1/d, which would make g blow up near the surface (d → 0). The depth formula is linear, not inverse.
Why C is wrong: C suggests an inverse-square dependence on depth. That's incorrect — the inverse-square behaviour applies to altitude above the surface, not depth below it. This is the core altitude-vs-depth confusion trap.
Why D is wrong: D suggests exponential decay. No gravitational formula for uniform-density spheres involves exponentials. This has no basis in the derivation.
MCQ 4Direct ApplicationPractice
A body weighs 720 N on Earth's surface. What is its weight at a height equal to half the radius of Earth (h = R/2) above the surface?
Show answer and why every option is right or wrong
Answer: C. g_h = g(R/(R + R/2))² = g(R/(3R/2))² = g(2/3)² = 4g/9. Weight = 720 × 4/9 = 320 N. The exact inverse-square formula must be used because h = R/2 is not small compared to R (NCERT Class 11 Physics Chapter 7, page 133).
Why A is wrong: A (480 N) comes from forgetting the square in the inverse-square law: g × R/(R + R/2) = 2g/3, so 720 × 2/3 = 480 N (trap: forgetting the square in the inverse-square law).
Why B is wrong: B (240 N) is 720/3: it takes h/(R + h) = 1/3 as the factor, putting the height where the radius belongs in the ratio, and without the square. The factor is (R/(R + h))² = 4/9.
Why D is wrong: D (180 N) is the weight at h = R, where (R/2R)² = 1/4 — but the question states h = R/2, not h = R.
MCQ 5Direct ApplicationPractice
At what height above Earth's surface does the acceleration due to gravity reduce to 25% of its surface value? (Express answer in terms of Earth's radius R.)
Show answer and why every option is right or wrong
Answer: D. g_h/g = (R/(R+h))² = 0.25 = 1/4. So R/(R+h) = 1/2, giving R + h = 2R, hence h = R (NCERT Class 11 Physics Chapter 7, page 133).
Why A is wrong: A (h = R/2) gives g_h = g(R/(3R/2))² = g(4/9) ≈ 0.44g, which is 44%, not 25%. This likely comes from an incorrect linear-proportion reasoning.
Why B is wrong: B (h = 4R) gives g_h = g(R/(5R))² = g/25 = 4%, far below 25%. This comes from confusing 'g becomes 1/4' with 'distance becomes 4R' — neglecting that distance from centre is R+h, not h.
Why C is wrong: C (h = 2R) gives g_h = g(R/(3R))² = g/9 ≈ 11%, not 25%. This overshoots because the inverse-square fall-off is steep.
MCQ 6Direct ApplicationPractice
A mine shaft reaches a depth of R/4 below Earth's surface (R = Earth's radius, uniform density assumed). The acceleration due to gravity at the bottom of the shaft is:
Show answer and why every option is right or wrong
Answer: B. g_d = g(1 − d/R) = g(1 − (R/4)/R) = g(1 − 1/4) = 3g/4 (NCERT Class 11 Physics Chapter 7, page 133).
Why A is wrong: A (g/4) comes from taking d/R = 1/4 as the factor itself instead of subtracting it from 1. The depth formula is g(1 − d/R) = 3g/4.
Why C is wrong: C (g/2) would require d = R/2. For d = R/4, the reduction factor is 3/4, not 1/2.
Why D is wrong: D (4g/5) comes from using the altitude ratio R/(R + d) = 4/5 without squaring — and the altitude formula is the wrong one for depth; inside the Earth g falls linearly, g(1 − d/R).
MCQ 7CalculationPractice
A body weighs W on Earth's surface. It is taken to a height h = R above the surface. Separately, the same body is taken to a depth d below the surface where it has the same weight as at height h. Find d in terms of R. (Assume uniform density.)
Show answer and why every option is right or wrong
Answer: C. At height h = R: g_h = g(R/(R+R))² = g/4. At depth d: g_d = g(1 − d/R). Set equal: 1 − d/R = 1/4, so d/R = 3/4, giving d = 3R/4. This requires correctly applying the altitude formula first, then matching via the depth formula.
Why A is wrong: A (d = R/2) comes from using R/(R + h) without squaring: g × R/(2R) = g/2, then 1 − d/R = 1/2 gives d = R/2. The altitude formula needs the square.
Why B is wrong: B (d = R/4) comes from naively setting d = h/4 or confusing 'g reduces to 1/4' with 'd = R/4'. The depth formula gives g(1 − 1/4) = 3g/4 at d = R/4, which does NOT equal g/4.
Why D is wrong: D (d = 2R/3) comes from writing g_h ≈ g/(1 + 2h/R) = g/3, a small-height approximation that fails at h = R, and then 1 − d/R = 1/3. At h = R the exact factor is (R/2R)² = 1/4.
MCQ 8CalculationPractice
The ratio of g at a depth of R/2 to g at a height of R/2 above the surface is: (Assume uniform density.)
Show answer and why every option is right or wrong
Answer: A. At depth R/2: g_d = g(1 − 1/2) = g/2. At height R/2: g_h = g(R/(3R/2))² = g(4/9) = 4g/9. Ratio = (g/2)/(4g/9) = (g/2)(9/4g) = 9/8. Both formulas must be applied correctly and independently.
Why B is wrong: B (8/9) is the reciprocal of the correct answer — computed as g_h/g_d instead of g_d/g_h. Check which quantity is in the numerator.
Why C is wrong: C (2/1) comes from computing g at depth as g/2 correctly but using the linear altitude approximation: g(1 − 2×R/(2R)) = g(1 − 1) = 0, which is nonsensical and abandoned, then guessing the ratio is 2 (trap: treating g at altitude as linear in h).
Why D is wrong: D (1/2) likely comes from only computing g_d = g/2 and assuming g_h = g at h = R/2 (i.e., ignoring the altitude reduction entirely), giving ratio = (g/2)/g = 1/2.
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Acceleration Due to Gravity: quick recall before you leave
How do you solve a Acceleration Due to Gravity question? A worked example
Pattern: Weight at altitude — direct application of g_h = g(R/(R+h))² (observed in NEET 2024 and 2025).
- 1
Given
A body weighs 800 N on the surface of Earth. Earth's radius R is given. Find the body's weight at a height h = R above the surface.
- 2
Required
Weight at height h = R above the surface.
- 3
Concept
Acceleration due to gravity decreases with altitude. Since h = R is NOT small compared to R, the exact inverse-square formula must be used. The linear approximation g(1 − 2h/R) is invalid here.
- 4
Formula
g_h = g × (R / (R + h))²
- 5
Substitution
g_h = g × (R / (R + R))²
g_h = g × (R / 2R)²
g_h = g × (1/2)² - 6
Calculation
g_h = g × 1/4 = g/4
Weight at height R = 800 × (1/4) = 200 N.
Note on exact values: The integers 800, 1, 2, and 4 are exact (counting/problem-defined values) and do not limit significant figures. - 7
Final answer
Weight at height h = R is 200 N (one-quarter of surface weight).
- 8
Common trap
Using the linear approximation: g(1 − 2R/R) = g(1 − 2) = −g. This gives a negative, physically meaningless result. The linear formula is only valid when h ≪ R. For h = R, you MUST use the exact formula g(R/(R+h))².
- 9
Similar NEET-style question
A satellite orbits at height h = 3R above Earth's surface. What fraction of surface gravity does it experience?
Approach: g_h = g(R/(R+3R))² = g(1/4)² = g/16. The satellite experiences 1/16 of surface gravity.
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What to remember before solving Acceleration Due to Gravity questions
Which Acceleration Due to Gravity formulas do you need for NEET?
1 formula — click to collapse
Newton's law of gravitation
Attractive force between any two masses. Inverse-square central force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | force | N |
| G | grav constant = 6.674e-11 | N*m^2/kg^2 |
| m1, m2 | masses | kg |
| r | centre-to-centre distance | m |
Valid when
- Point masses or spherically symmetric distributions
- r > sum of body radii (else use shell theorem)
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 4 question patterns from its other lessons.
Acceleration Due to Gravity questions from past NEET papers
2 questions from NEET 2022, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
119.6 m s⁻²
29.8 m s⁻²
34.9 m s⁻²
43.92 m s⁻²
NTA Answer: Option 4(revised_final)
NEET 2022
1180 N/kg
20.05 N/kg
350 N/kg
420 N/kg
NTA Answer: Option 3(final)
Sources
NCERT refs: Class 11 Physics Chapter 7, p.131
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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