PhysicsCBSE Class 11 · Gravitation

Acceleration due to Gravity and its Variation Class 11

Updated 17 October 2026 · Fact-checked

At the surface g = GM/R² = (4/3)πGρR. At height h, g(h) = gR²/(R + h)² ≈ g(1 − 2h/R) for h much smaller than R. At depth d, g(d) = g(1 − d/R) for a uniform Earth, and g is zero at the centre. The mass of a body does not change with place, but its weight mg does.

Notice g is zero at the centre, rises linearly to its maximum at the surface, then drops as 1/r² to g/4 at 2R.

Illustration drawn for QuestionBuddy.

Understand Acceleration due to Gravity and its Variation

Surface value. For a body of mass $m$ at the surface of a planet of mass $M$ and radius $R$ , Newton's law gives $m g = \frac{GM m}{R ^{2}}$ , so
$ $g = \frac{GM}{R ^{2}} = \frac{4}{3} π GρR,$ $
where $ρ$ is the mean density. For the Earth, $g \approx 9.8 m/s^{2}$ . A planet with the same density has $g \propto R$ ; with different values, $g \propto M / R^{2}$ .

**Above the surface (height $h$ ).** The distance from the centre is $R + h$ :
$ $g_{h} = \frac{GM}{( R + h ) ^{2}} = g (\frac{R}{R + h})^{2} \approx g (1 - \frac{2 h}{R}) (h ≪ R) .$ $
At $h = R$ , $g_{h} = g /4$ ; at $h = 2 R$ , $g /9$ . The approximation is good for $h$ up to a small fraction of $R$ and gets poor at large heights.

**Below the surface (depth $d$ ).** Using the shell theorem (a uniform shell exerts no force inside it), only the sphere of radius $R - d$ attracts, so for uniform density
$ $g_{d} = g (1 - \frac{d}{R}) = g \frac{R - d}{R},$ $
which falls linearly to zero at the centre. A body at depth $d$ has weight $m g (1 - d / R)$ .

Comparing height and depth. For small $h$ , $g$ falls twice as fast with height ( $2 h / R$ ) as with depth ( $d / R$ ). The same weight at height $h$ and depth $d$ gives $d \approx 2 h$ .

Mass versus weight. Mass is a fixed amount of matter; weight is the gravitational force and changes with $g$ (on the Moon it is about one-sixth of its value on the Earth).

Acceleration due to Gravity and its Variation: formulas and key facts

Surface gravity: $g = \frac{GM}{R ^{2}} = \frac{4}{3} π GρR$; Planet mass M, radius R.
At height h: $g_{h} = g (\frac{R}{R + h})^{2}$; Exact.
Small height: $g_{h} \approx g (1 - \frac{2 h}{R})$; For $h ≪ R$ .
At depth d: $g_{d} = g (1 - \frac{d}{R})$; Uniform density.
Same g at height and depth: $d \approx 2 h$; Small h.
Weight: $W = m g$; Changes with place.

How to solve Acceleration due to Gravity and its Variation questions

Decide whether the point lies outside (use the inverse-square form) or inside (use the linear form); mass stays constant while weight changes with g.

1Write the surface value g (or GM/R²) and the radius R.
2Outside: replace R by R + h (distance from the centre) and use the inverse-square ratio.
3Inside: use (R − d)/R, valid for uniform density.
4For two planets, form ratios using g ∝ M/R² or g ∝ ρR.
5Remember mass is unchanged; compute weight as mg at the new g.
6For small h, use the binomial approximation only when h is much smaller than R.

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.

Weighing at a hill station

Suppose a hill station is 2.0 km above sea level and a person weighs 600 N at sea level (R = 6400 km).

Find the weight there using g(1 − 2h/R).

Try it first, then show the solutionSolution

2h/R = 4/6400 = 6.25 × 10⁻⁴.
W = 600 × (1 − 6.25 × 10⁻⁴) = 599.6 N.

Answer: About 599.6 N.

Memory hook: The drop is tiny: g falls by 0.06%.

A deep mine

Suppose a mine shaft is 3.2 km deep and a sample weighs 400 N at the surface.

Estimate its weight at the bottom.

Try it first, then show the solutionSolution

d/R = 3.2/6400 = 5 × 10⁻⁴.
W = 400 × (1 − 5 × 10⁻⁴) = 399.8 N.

Answer: About 399.8 N.

Memory hook: Depth reduces g linearly.

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.

Inverse-square outside, linear inside
Outside: (R/(R + h))². Inside: (R − d)/R.
Use it when: Choosing the formula.
Twice as fast up
Up: 2h/R. Down: d/R.
Use it when: Quick comparison.
g ∝ ρR
Same density: g is proportional to radius.
Use it when: Planets of equal density.
Mass fixed, weight varies
On the Moon you weigh 1/6, your mass is the same.
Use it when: Conceptual questions.

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 on Mars · 2 marks · Very short answerBoard level

Mars has mass 6.4 × 10²³ kg and radius 3.4 × 10⁶ m (G = 6.67 × 10⁻¹¹ N m²/kg²). Find g on its surface.

Try it first, then show the solutionSolution

g = GM/R² = 6.67 × 10⁻¹¹ × 6.4 × 10²³/(3.4 × 10⁶)². (1)
= 4.27 × 10¹³/1.156 × 10¹³ ≈ 3.7 m/s². (1)

Answer: About 3.7 m/s².

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

Find g at a height 2R above the Earth's surface (g = 9.8 m/s² at the surface), and the weight there of a 60 kg person.

Try it first, then show the solutionSolution

Distance from the centre = 3R, so g_h = g/9. (1)
g_h = 9.8/9 = 1.09 m/s². (1)
Weight = 60 × 1.09 = 65.3 N, compared with 588 N on the surface; the mass is still 60 kg. (1)

Answer: 1.09 m/s²; 65.3 N.

Example 3: Height and depth for half of g · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

Depth: g(1 − d/R) = g/2 gives d = R/2 = 3200 km. (1)
Height: (R/(R + h))² = 1/2 gives R + h = √2 R. (1)
h = (√2 − 1) × 6400 = 0.414 × 6400 ≈ 2650 km. (1)

Answer: Depth 3200 km; height about 2650 km.

Example 4: A planet of different density · 4 marks · Long answerHarder board level

A planet has radius 1.5 times the Earth's and mean density 0.8 times the Earth's. Using g = 9.8 m/s² for the Earth, find (a) g on the planet, (b) the ratio of its mass to the Earth's, and (c) the ratio of the escape speeds.

Try it first, then show the solutionSolution

g = (4/3)πGρR ∝ ρR, so g/g_E = 0.8 × 1.5 = 1.2. (1)
g = 1.2 × 9.8 = 11.76 m/s². (1)
M ∝ ρR³ = 0.8 × 3.375 = 2.7 times the Earth's mass. (1)
Escape speed ∝ R√ρ = 1.5 × √0.8 = 1.34. (1)

Answer: (a) 11.8 m/s²; (b) 2.7; (c) 1.34.

Example 5: Equal weights above and below · 4 marks · Long answerHOTS / competency

A body has the same weight at a height h = 5 km above the Earth's surface as at a depth d inside. (a) Show that d ≈ 2h for h ≪ R. (b) Find d for h = 5 km (R = 6400 km). (c) Check the result with the exact formula for height and find the weight of a 70 kg person at 5 km height (g = 9.8 m/s²).

Try it first, then show the solutionSolution

(a) g(1 − 2h/R) = g(1 − d/R) gives d = 2h. (1)
(b) d = 10 km. (1)
(c) Exact: g_h = g(6400/6405)² = 0.99844 g, so 1 − d/R = 0.99844 and d = 0.00156 × 6400 ≈ 10.0 km. (1)
Weight at 5 km = 70 × 9.8 × 0.99844 ≈ 684.9 N. (1)

Answer: d = 2h = 10 km; weight ≈ 685 N.

Where students lose marks in Acceleration due to Gravity and its Variation

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

Mistake: Using h instead of R + h in the inverse-square law.
Measuring from the surface.
Do this: Distance from the centre is R + h.
Usually costs: The answer mark.
Mistake: Applying g(1 − d/R) to a point above the surface.
Mixing the two cases.
Do this: Outside uses inverse square; inside uses linear.
Usually costs: Method mark.
Mistake: Saying mass decreases at height.
Confusing mass and weight.
Do this: Mass is constant; weight changes.
Usually costs: Concept mark.
Mistake: Using g(1 − 2h/R) for h comparable to R.
Overusing the binomial approximation.
Do this: Use the exact formula when h is large.
Usually costs: 1 mark.

Exam tips

Write whether the point is above or below the surface before choosing the formula.
Quote the data you assume (g, R, G) in the first line.
Show the ratio form for planet comparisons rather than recomputing from scratch.
Mention that mass is unchanged when you quote weight.
Keep units consistent (km or m) in h/R.

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 · MCQHarder board level
Two planets have the same mean density and radii in the ratio 2 : 1. The ratio of their surface gravities is
(a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 8 : 1
Hint
g ∝ ρR.
Answer
(b) 2 : 1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Inside a uniform Earth, g decreases linearly with depth.
Reason (R): At depth d only the sphere of radius R − d attracts the body, and its mass is proportional to (R − d)³.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are 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.
Hint
g ∝ M(r)/r² ∝ r.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A 60 kg person goes to the Moon where g is one-sixth of the Earth's (g = 9.8 m/s²). Find the mass and weight on the Moon.
Hint
Mass does not change.
Answer
Mass = 60 kg; weight = 60 × 9.8/6 = 98 N.
Q4 · 3 marks · Short answerHarder board level
A planet has 8 times the Earth's mass and twice the Earth's radius. Find the ratio of its surface gravity and of its escape speed to the Earth's.
Hint
g ∝ M/R², ve ∝ √(M/R).
Answer
g = 8/4 = 2 times; ve = √(8/2) = 2 times.
Q5 · 4 marks · Long answerHOTS / competency
A 50 kg man could stand in a deep tunnel through a uniform Earth. Find his weight (a) at a depth of 1600 km and (b) at a height of 1600 km above the surface (R = 6400 km, g = 9.8 m/s²). (c) At what depth would his weight be the same as in (b)?
Hint
Use the linear law inside and the inverse-square law outside.
Answer
(a) 50 × 9.8 × (1 − 1/4) = 367.5 N. (b) g_h = g(6400/8000)² = 0.64 g, so weight = 313.6 N. (c) 1 − d/R = 0.64, so d = 0.36 R = 2304 km.

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 at height h?Show answer
g(R/(R + h))².
2.g at depth d?Show answer
g(1 − d/R).
3.g at the centre?Show answer
Zero.
4.g and density?Show answer
g = (4/3)πGρR.
5.g at h = R?Show answer
g/4.
6.Weight on the Moon?Show answer
About one-sixth of that on Earth.

Acceleration due to Gravity and its Variation: frequently asked questions

Why is g zero at the centre of the Earth?

The attractions from all the surrounding matter cancel by symmetry.

Does g depend on the mass of the falling body?

No, the body's mass cancels from F = mg.

Is the depth formula exact for the real Earth?

No; the Earth's density is not uniform, so the real variation differs.

Why does weight change with latitude?

Because of the Earth's shape and rotation; this syllabus mainly uses the variation with height and depth.

What is the SI unit of g?

m/s², equivalent to N/kg.