PhysicsISC Class 11 · Gravitation

Acceleration Due to Gravity: Variation with Height, Depth and Latitude

Updated 17 October 2026 · Fact-checked

At the Earth's surface g = GM/R². At a height h, g_h = g R²/(R + h)² (about g(1 − 2h/R) for small h). At a depth d, g_d = g(1 − d/R). g is least at the equator because of rotation and greatest at the poles, and zero at the centre.

Notice g is zero at the centre, rises linearly to its maximum at the surface, then falls as 1/r², becoming g/4 at height h = R.

Illustration drawn for QuestionBuddy.

Understand Acceleration Due to Gravity and Its Variation

Relation with G. The weight of a body of mass m on the surface is mg = GMm/R², so g = GM/R² and M = gR²/G. Using M and the volume (4/3)πR³, the mean density of the Earth is ρ = 3g/(4πGR).

Above the surface. At a height h, g_h = GM/(R + h)² = g [R/(R + h)]². For h ≪ R, binomial expansion gives g_h ≈ g(1 − 2h/R). At h = R, g is g/4.

Below the surface. Taking the Earth as a uniform sphere, only the mass of the inner sphere of radius (R − d) attracts a body at depth d. Then g_d = g(1 − d/R), which decreases linearly to zero at the centre.

Latitude and rotation. A body at latitude λ moves in a circle of radius R cosλ with angular speed ω. The effective acceleration is approximately g_λ = g − Rω² cos²λ. It is least at the equator (λ = 0): g − Rω², and equal to g at the poles. Weight is mg_λ, so a body weighs slightly less at the equator.

Combining. Small-height and small-depth results can be compared: g at height h equals g at depth d if d = 2h. If the Earth rotated fast enough that Rω² = g, bodies would be weightless at the equator.

Remember. Mass is constant everywhere; weight changes with g. Gravitational acceleration does not depend on the mass of the falling body.

Acceleration Due to Gravity and Its Variation: formulas and key facts

g and G: $g = GM / R^{2} ∣ M = g R^{2} / G ∣ ρ = 3 g / (4 π GR)$; R = radius of the Earth.
Height: $g_{h} = g R^{2} / (R + h)^{2} ∣ g_{h} \approx g (1 - 2 h / R)$; The second form is for h much smaller than R.
Depth: $g_{d} = g (1 - d / R)$; Uniform-density Earth.
Latitude: $g_{λ} = g - R ω^{2} co s^{2} λ$; ω = 2π/(24 h), the Earth's angular speed.

Acceleration Due to Gravity and Its Variation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Weight at the top of a tall mountain

Suppose a mountaineer of mass 60 kg climbs a peak 8.0 km above sea level (take R = 6400 km and g = 9.8 m s⁻² at sea level).

Find the fall in her weight using the small-height formula.

Try it first, then show the solutionSolution

g_h ≈ g(1 − 2h/R) = 9.8(1 − 2 × 8/6400) = 9.8(1 − 0.0025) = 9.7755 m s⁻².
Weight at sea level = 60 × 9.8 = 588 N.
Weight at the peak = 60 × 9.7755 = 586.5 N, a fall of 1.5 N (0.25%).

Answer: About 1.5 N lighter; her mass is unchanged at 60 kg.

Memory hook: Mass is constant; only weight changes with g.

Tricks to remember Acceleration Due to Gravity and Its Variation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Twice for height
The fractional fall of g at height h is 2h/R, and at depth d it is d/R.
Use it when: Comparing height and depth results.
Weightless at the centre
g falls to zero at the centre; at the surface it is maximum for depth variation.
Use it when: Graphs of g against distance from the centre.

Acceleration Due to Gravity and Its Variation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: g at a depth · 1 mark · MCQBoard level

At a depth of R/2 below the Earth's surface the acceleration due to gravity is:
(a) g/4
(b) g/2
(c) 3g/4
(d) g

Try it first, then show the solutionSolution

g_d = g(1 − d/R) = g(1 − ½) = g/2.

Answer: (b) g/2

Example 2: g at a height · 3 marks · Short answerBoard level

Find g at a height of 3200 km above the Earth's surface (R = 6400 km, g = 9.8 m s⁻² at the surface).

Try it first, then show the solutionSolution

g_h = g [R/(R + h)]². (1 mark)
R + h = 9600 km, so [6400/9600]² = (2/3)² = 4/9. (1 mark)
g_h = 9.8 × 4/9 = 4.36 m s⁻². (1 mark)

Answer: 4.36 m s⁻²

Example 3: Height at which g is halved · 3 marks · Short answerHarder board level

At what height above the Earth's surface is g half its surface value? (R = 6400 km)

Try it first, then show the solutionSolution

[R/(R + h)]² = ½ → (R + h)/R = √2. (1 mark)
h = (√2 − 1)R = 0.4142 R. (1 mark)
h = 0.4142 × 6400 = 2651 km. (The small-height formula would fail here, since h is not much smaller than R.) (1 mark)

Answer: About 2650 km above the surface.

Example 4: Mass and density of a planet · 4 marks · Short answerHarder board level

Suppose a planet has radius 3.4 × 10⁶ m and surface gravity 3.7 m s⁻². Taking G = 6.67 × 10⁻¹¹ N m² kg⁻², find its mass and mean density.

Try it first, then show the solutionSolution

M = gR²/G = 3.7 × (3.4 × 10⁶)²/(6.67 × 10⁻¹¹). (1 mark)
(3.4 × 10⁶)² = 1.156 × 10¹³; 3.7 × 1.156 × 10¹³ = 4.277 × 10¹³; M = 4.277 × 10¹³/6.67 × 10⁻¹¹ = 6.41 × 10²³ kg. (1 mark)
V = (4/3)π(3.4 × 10⁶)³ = 4.189 × 3.930 × 10¹⁹ = 1.646 × 10²⁰ m³. (1 mark)
ρ = 6.41 × 10²³/1.646 × 10²⁰ ≈ 3.9 × 10³ kg m⁻³. (1 mark)

Answer: M ≈ 6.4 × 10²³ kg; ρ ≈ 3.9 × 10³ kg m⁻³.

Example 5: Effect of rotation · 5 marks · Long answerHOTS / competency

Take ω = 7.27 × 10⁻⁵ rad s⁻¹ and R = 6.4 × 10⁶ m. (i) Find the reduction in g at the equator. (ii) Find the loss in weight of a 60 kg person moving from a pole to the equator. (iii) Find the time of one rotation for which a body at the equator would be weightless (g = 9.8 m s⁻²).

Try it first, then show the solutionSolution

Rω² = 6.4 × 10⁶ × (7.27 × 10⁻⁵)² = 6.4 × 10⁶ × 5.285 × 10⁻⁹ = 0.0338 m s⁻². (2 marks)
Weight loss = 60 × 0.0338 = 2.03 N. (1 mark)
Weightless: Rω² = g → ω = √(g/R) = √(9.8/6.4 × 10⁶) = 1.237 × 10⁻³ rad s⁻¹. (1 mark)
Period = 2π/ω = 5.08 × 10³ s ≈ 85 minutes. (1 mark)

Answer: g falls by 0.034 m s⁻²; weight loss about 2.0 N; weightless if one day lasted about 85 minutes.

Where students lose marks in Acceleration Due to Gravity and Its Variation

The slips examiners see most often, and what to write instead.

Mistake: Mixing up height and depth formulas
Using g(1 − d/R) for a height.
Do this: Height: g R²/(R + h)² or g(1 − 2h/R). Depth: g(1 − d/R).
Usually costs: 1 to 2 marks.
Mistake: Forgetting to add R to h
Writing g_h = g R²/h².
Do this: The distance from the Earth's centre is R + h.
Usually costs: 2 marks.
Mistake: Saying mass changes with height
Writing that the mass of a body decreases at the top of a mountain.
Do this: Mass is constant; the weight changes.
Usually costs: 1 mark.

Exam tips

Write the distance from the centre of the Earth first: R + h at a height, R − d at a depth.
Say whether you use the exact formula or the small-height approximation.
Keep R and g as given in the question.
Check the answer against g at the surface: heights reduce g, depths reduce g, rotation reduces it slightly.

Practice questions on Acceleration Due to Gravity and Its Variation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The acceleration due to gravity at the centre of the Earth is:
(a) g
(b) g/2
(c) zero
(d) infinite
Hint
Use g_d = g(1 − d/R) with d = R.
Answer
(c) zero
Q2 · 2 marks · Short answerBoard level
A planet has mass one-tenth and radius half of the Earth's. Find g on it (g_E = 9.8 m s⁻²).
Hint
g ∝ M/R².
Answer
g = 9.8 × 0.1/0.25 = 3.92 m s⁻².
Q3 · 3 marks · Short answerHarder board level
A body weighs 700 N on the Earth's surface. What does it weigh at a height R/2 above, and at a depth R/2 below, the surface?
Hint
Height: [R/(R + h)]²; depth: (1 − d/R).
Answer
Height R/2: 700 × (1/1.5)² = 700 × 4/9 = 311 N. Depth R/2: 700 × ½ = 350 N.
Q4 · 3 marks · Short answerHarder board level
At what depth is the acceleration due to gravity equal to its value at a height of 5 km above the surface? (Use the small-height approximation.)
Hint
Equate g(1 − 2h/R) and g(1 − d/R).
Answer
1 − 2h/R = 1 − d/R → d = 2h = 10 km.
Q5 · 3 marks · Short answerHarder board level
A simple pendulum has a period of 2 s on the Earth's surface. What is its period at a height equal to R above the surface?
Hint
T ∝ 1/√g.
Answer
At h = R, g becomes g/4, so T doubles to 4 s.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the Earth were compressed to half its radius while keeping its mass (g = 9.8 m s⁻² at present).
(i) Find the new value of g. (1 mark)
(ii) Find the weight of a 50 kg person before and after. (1 mark)
(iii) By what factor would the mean density change? (1 mark)
(iv) How long would a ball dropped from 4.9 m take to land, compared with now? (1 mark)
Hint
g ∝ M/R².
Answer
(i) g′ = 4g = 39.2 m s⁻². (ii) 490 N before, 1960 N after. (iii) The volume falls by 2³ = 8, so the density rises 8 times. (iv) t = √(2h/g′) = √(9.8/39.2) = 0.5 s, against 1 s now.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.g in terms of G, M, R?Show answer
g = GM/R².
2.g at depth d?Show answer
g(1 − d/R).
3.Where is g least on the Earth's surface?Show answer
At the equator.
4.g at h = R?Show answer
g/4.

Acceleration Due to Gravity and Its Variation: frequently asked questions

Does g depend on the mass of the falling body?

No. The mass of the body cancels in F = ma, so every body falls with the same g in the same place, ignoring air.

Why is g least at the equator?

Because of the Earth's rotation, a body there moves in the largest circle and needs a centripetal force that reduces the effective g by Rω², and the Earth is slightly flattened too.

Is the linear fall of g with depth exact?

It is exact for a uniform sphere. The real Earth has a denser core, so the true variation is not perfectly linear; the syllabus uses the uniform model.