Formula
Variation of g with altitude
At height h above Earth's surface: g_h = g (R/(R+h))² ≈ g(1 - 2h/R) for h << R. Decreases with altitude.
-- NCERT Class 11 Physics, Ch. 7, p. 133g Variation Altitude
8 MCQs3 revision cards9-step worked example
Source: NCERT GravitationPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
g Variation Altitude, explained for NEET
The trap that costs marks: You see "find g at height h = R above the surface" and instinctively write g_h = g(1 − 2h/R). That gives g(1 − 2) = −g — a negative acceleration due to gravity. The answer is obviously wrong, but under exam pressure, students pick the distractor built from exactly this mistake. This trap has appeared in NEET 2024 and 2025.
The actual formula. Gravitational acceleration at altitude h above a spherical Earth of radius R is:
g_h = g × (R / (R + h))²
This is an inverse-square dependence on the distance from Earth's centre, not a linear decrease from the surface. The derivation is direct: at the surface, g = GM/R²; at height h, the distance from the centre is (R + h), so g_h = GM/(R + h)² = g × R²/(R + h)² (NCERT Class 11 Physics Chapter 7, page 133).
When does the approximation work? The linear form g_h ≈ g(1 − 2h/R) is a binomial expansion valid only when h ≪ R. For a satellite 200 km above Earth (R ≈ 6400 km), h/R ≈ 0.03 — the approximation is fine. For h = R/2, h = R, or h = 2R, it fails badly.
Quick checkpoint: at h = R, the exact formula gives g_h = g(R/2R)² = g/4. The linear approximation gives g(1 − 2) = −g. If your answer is negative, you used the wrong formula.
Watch-out for NEET stems: Questions often state height as a fraction of R (e.g., "at a height equal to half the radius"). Convert to h = R/2 and substitute into the exact formula. Do not default to the approximation unless the problem explicitly states h ≪ R.
Can you answer these g Variation Altitude 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 at a height h above the Earth's surface is given by which expression? (R = radius of Earth, g = surface gravity)
Show answer and why every option is right or wrong
Answer: A. A is correct. The exact formula for gravitational acceleration at altitude h is g_h = g(R/(R + h))², derived from GM/(R + h)² = g × R²/(R + h)² (NCERT Class 11 Physics Chapter 7, page 133).
Why B is wrong: B gives the formula for g at depth d inside the Earth (with d in place of h), not at altitude above the surface.
Why C is wrong: C is the binomial approximation valid only when h << R. It is not the general expression and fails for significant altitudes.
Why D is wrong: D uses (R − h) in the denominator, which has no physical basis. Distance from Earth's centre at altitude h is (R + h), not (R − h).
MCQ 2Easy RecallPractice
The approximate formula g_h ≈ g(1 − 2h/R) for variation of g with altitude is valid under which condition?
Show answer and why every option is right or wrong
Answer: C. C is correct. The expression g(1 − 2h/R) is a first-order binomial expansion of g(R/(R + h))², which requires h/R << 1 for the higher-order terms to be negligible.
Why A is wrong: A is wrong. When h = R, the approximation gives g(1 − 2) = −g, a physically meaningless negative value. The exact formula gives g/4.
Why B is wrong: B is wrong. When h >> R, the approximation breaks down completely (gives large negative values). The exact formula must be used.
Why D is wrong: D is wrong. When h = 2R, the approximation gives g(1 − 4) = −3g, which is absurd. The exact formula gives g(R/3R)² = g/9.
MCQ 3Easy RecallPractice
At what altitude above the Earth's surface does the acceleration due to gravity become zero, according to g_h = g(R/(R + h))²?
Show answer and why every option is right or wrong
Answer: B. B is correct. As h → ∞, R/(R + h) → 0, so g_h → 0. The gravitational acceleration never reaches exactly zero at any finite altitude.
Why A is wrong: A is wrong. At h = R, g_h = g(R/2R)² = g/4, which is one-quarter of surface gravity — not zero.
Why C is wrong: C is wrong. At h = 2R, g_h = g(R/3R)² = g/9, which is still a non-zero value.
Why D is wrong: D is wrong. At h = R/2, g_h = g(R/(3R/2))² = g(2/3)² = 4g/9, which is nearly half of surface gravity.
MCQ 4Direct ApplicationPractice
A body weighs 63 N on the surface of the Earth. What is its weight at a height equal to half the radius of the Earth above the surface?
Show answer and why every option is right or wrong
Answer: C. C is correct. At h = R/2: g_h = g(R/(R + R/2))² = g(R/(3R/2))² = g(2/3)² = 4g/9. Weight = 63 × 4/9 = 28 N.
Why A is wrong: A results from using g_h = g/9, as if h = 2R instead of h = R/2. Check the substitution: h = R/2 gives (R + h) = 3R/2, not 3R.
Why B is wrong: B results from taking h/(R + h) = 1/3 as the factor, putting the height where the radius belongs and leaving out the square: 63/3 = 21 N. The factor is (R/(R + h))² = 4/9.
Why D is wrong: D results from forgetting to square the ratio: g × R/(R + h) = 2g/3, and 63 × 2/3 = 42 N (trap: treating g variation as linear in h instead of inverse-square).
MCQ 5Direct ApplicationPractice
The acceleration due to gravity at a height of 3200 km above the Earth's surface is (take R = 6400 km, g = 9.8 m/s²):
Show answer and why every option is right or wrong
Answer: B. B is correct. h = 3200 km = R/2. g_h = 9.8 × (6400/9600)² = 9.8 × (2/3)² = 9.8 × 4/9 ≈ 4.36 m/s².
Why A is wrong: A results from g × R/(R + h) = 9.8 × 2/3 = 6.53, which omits the squaring in (R/(R+h))² (trap: linear vs inverse-square).
Why C is wrong: C results from using the linear approximation g(1 − 2h/R) = 9.8(1 − 1) = 0. This is the classic trap of applying the binomial approximation outside its validity range (h is not << R here).
Why D is wrong: D results from taking h/(R + h) = 1/3 as the factor, putting the height where the radius belongs: 9.8/3 ≈ 3.27. The factor is (R/(R + h))² = 4/9.
MCQ 6Direct ApplicationPractice
At what height above the Earth's surface does the acceleration due to gravity reduce to 1/9 of its surface value? (R = radius of Earth)
Show answer and why every option is right or wrong
Answer: D. D is correct. Set g/9 = g(R/(R + h))². Then (R/(R + h))² = 1/9, so R/(R + h) = 1/3, giving R + h = 3R, hence h = 2R.
Why A is wrong: A is wrong. At h = R/2, g_h = g(2/3)² = 4g/9, not g/9. This is a common error from confusing the fraction in the result with the height fraction.
Why B is wrong: B is wrong. At h = R, g_h = g(R/2R)² = g/4, not g/9. This error comes from solving R/(R + h) = 1/3 incorrectly as h = R instead of h = 2R (trap: linear reasoning).
Why C is wrong: C is wrong. At h = 3R, g_h = g(R/4R)² = g/16, not g/9. This comes from adding an extra R — the distance from centre is (R + h) = 4R, not 3R.
MCQ 7CalculationPractice
A body weighs W on the surface of the Earth. At what height above the surface will its weight become W/16? Express your answer in terms of R (radius of Earth).
Show answer and why every option is right or wrong
Answer: D. D is correct. Step 1: W/16 = W × (R/(R + h))², so (R/(R + h))² = 1/16. Step 2: R/(R + h) = 1/4, giving R + h = 4R, so h = 3R.
Why A is wrong: A is wrong. This comes from forgetting the square root: setting R/(R + h) = 1/16 gives R + h = 16R and h = 15R. The factor 1/16 is (R/(R + h))², so R/(R + h) = 1/4.
Why B is wrong: B is wrong. This comes from setting R + h = 4R (correct) but then reporting R + h = 4R as the height instead of h = 3R. The question asks for h (height above surface), not the distance from Earth's centre.
Why C is wrong: C is wrong. At h = 2R, g_h = g(R/3R)² = g/9, which gives weight W/9, not W/16. This results from a square-root or algebra error.
MCQ 8CalculationPractice
The ratio of acceleration due to gravity at a height R above the Earth's surface to that at a depth R/2 below the Earth's surface is: (R = radius of Earth, assume uniform density)
Show answer and why every option is right or wrong
Answer: A. A is correct. At height h = R: g_h = g(R/2R)² = g/4. At depth d = R/2: g_d = g(1 − d/R) = g(1 − 1/2) = g/2. Ratio: (g/4)/(g/2) = 1/2 = 1:2.
Why B is wrong: B comes from incorrectly using the same formula for both cases — either applying the depth formula to the altitude case or vice versa. The two situations use different formulas: inverse-square for altitude, linear for depth.
Why C is wrong: C reverses the ratio — gives g_h/g_d = 2 instead of 1/2. This may come from swapping numerator and denominator or from computing g_d/g_h instead of the asked g_h/g_d.
Why D is wrong: D comes from applying the altitude formula at depth: g(R/(R + d))² = 4g/9 at d = R/2, giving (g/4)/(4g/9) = 9/16. Inside the Earth g falls linearly, g_d = g(1 − d/R).
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
g Variation Altitude: quick recall before you leave
How do you solve a g Variation Altitude question? A worked example
- 1
Given
A body weighs 900 N on the surface of the Earth. Find its weight at a height equal to the radius of the Earth (h = R) above the surface. Take R = 6400 km.
- 2
Required
Weight at height h = R.
- 3
Concept
Gravitational acceleration decreases with altitude as an inverse-square function of the distance from Earth's centre (NCERT Class 11 Physics Chapter 7, page 133). Weight is proportional to local g.
- 4
Formula
g_h = g × (R / (R + h))²
W_h = W × (R / (R + h))² - 5
Substitution
h = R, so (R + h) = 2R.
W_h = 900 × (R / 2R)²
W_h = 900 × (1/2)² - 6
Calculation
W_h = 900 × 1/4 = 225 N
Note: the factor 4 in the denominator is an exact counting integer (2² = 4) and does not affect significant-figure count. - 7
Final answer
W_h = 225 N
The exact constant 4 (from 2² in the denominator) and the integer ratio R/2R = 1/2 do not limit significant figures. The precision is governed by the given weight (900 N, which we treat as exact for this problem since it is a clean given value). - 8
Common trap
Using the linear approximation g(1 − 2h/R) here gives g(1 − 2) = −g, implying negative weight. This is the high-frequency trap for this topic — the approximation requires h ≪ R, and h = R violates that condition completely.
- 9
Similar NEET-style question
A satellite orbits at a height equal to 3 times the radius of the Earth. What fraction of the surface gravity does it experience? (Answer: g_h = g(R/4R)² = g/16.)
---
What to remember before solving g Variation Altitude questions
Which g Variation Altitude formulas do you need for NEET?
1 formula — click to collapse
g variation with altitude
Gravitational acceleration decreases with altitude above Earth's surface.
| Symbol | Quantity | SI Unit |
|---|---|---|
| g_h | g at height h | m/s^2 |
| g | surface g | m/s^2 |
| R | Earth radius | m |
| h | altitude | m |
Valid when
- Static observer at altitude
- Earth treated as uniform sphere
Where do students lose marks on g Variation Altitude?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student treats g(h) as linear in h. Actual: g(R/(R+h))² (inverse-square).
When it triggers
Question asks for g at significant altitude (e.g. R/2 above surface).
How to avoid
Use g_h = g (R/(R+h))². Linear approximation g(1 - 2h/R) only valid for h << R.
Root cause: formula misuse
Correction
Use g_h = g(R/(R+h))² (inverse-square). Linear approximation g(1-2h/R) is only valid for h << R. For h = R/2, the exact formula gives g_h = (2/3)² g = 4g/9, not g(1-1) = 0.
More in Gravitation: 2 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
g Variation Altitude questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys. — click to collapse
How does NEET ask about g Variation Altitude?
1 recurring pattern from past papers — click to collapse
Body weighs W on surface; find weight at height h above surface. Use g_h = g(R/(R+h))^2.
Direct ApplicationEasy
Common distractors
uses linear decrease with h
Treats g as linear in h instead of inverse-square
Sources
NCERT refs: Class 11 Physics Chapter 7, p.133
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →