NEET UG Physics · Chapter 6

Gravitation

Law of gravitation, variation of g, gravitational potential energy, escape and orbital speed, Kepler’s laws.

NCERT Class 11NEET: 1–2 Qs per paper 10 min read

1. Chapter at a Glance

Gravitation is the universal attraction between all masses. The same force that makes an apple fall keeps the Moon in orbit and holds satellites around the Earth. This chapter covers Newton's law of gravitation, acceleration due to gravity and how it changes, gravitational potential energy, escape speed, orbital motion of satellites and Kepler's laws. NEET asks one or two questions from it every year.

What NEET asks most
  • Variation of g with height and depth.
  • Escape speed and orbital speed, and their ratio √2.
  • Kepler's third law: T² ∝ r³.
  • Energy of a satellite (kinetic, potential, total, binding).
  • Weightlessness in orbit.

2. Newton's Law of Gravitation and Kepler's Laws

F = Gm₁m₂/r²    G = 6.67 × 10⁻¹¹ N m² kg⁻²
  • The force is always attractive, acts along the line joining the centres, is independent of the medium between the bodies, and obeys superposition.
  • A uniform spherical shell attracts outside bodies as if its mass were at its centre, and exerts no net force on a body inside it.
  • G is a universal constant; g is not — it changes from place to place.

Kepler's laws of planetary motion

  1. Law of orbits: every planet moves in an ellipse with the Sun at one focus.
  2. Law of areas: the line joining the planet to the Sun sweeps equal areas in equal times — the planet moves fastest nearest the Sun (perihelion). This follows from conservation of angular momentum.
  3. Law of periods: T² ∝ a³, where a is the semi-major axis.

3. Acceleration due to Gravity

At the Earth's surface, g = GM/R² ≈ 9.8 m s⁻² (M ≈ 6 × 10²⁴ kg, R ≈ 6400 km).

SituationFormulaNEET Note
Height hgh = gR²/(R + h)²≈ g(1 − 2h/R) if h ≪ R
Depth dgd = g(1 − d/R)Zero at the centre
Latitudeg is least at equator, greatest at polesDue to rotation and shape
R (surface) g ∝ rg ∝ 1/r² gr
g increases linearly from zero at the centre to its maximum at the surface, then decreases as 1/r² outside.

4. Gravitational Potential and Potential Energy

  • Gravitational potential at distance r from mass M: V = −GM/r (zero at infinity, negative everywhere else).
  • Gravitational potential energy of mass m: U = −GMm/r. Near the surface, the change is ≈ mgh only when h ≪ R.
  • Change in PE when a mass is raised from the surface to height h: ΔU = GMm[1/R − 1/(R + h)] = mgh/(1 + h/R).
  • Escape speed: the least speed that lets a body leave the Earth for ever: ve = √(2GM/R) = √(2gR) ≈ 11.2 km s⁻¹. It does not depend on the body's mass or the direction of projection.

5. Earth Satellites

Quantity (orbit radius r = R + h)FormulaNEET Note
Orbital speedvo = √(GM/r) = √(gR²/r)Near surface ≈ 7.9 km s⁻¹
PeriodT = 2πr/vo = 2π√(r³/GM)Near surface ≈ 84 min
Kinetic energyK = GMm/2rPositive
Potential energyU = −GMm/rU = −2K
Total energyE = −GMm/2rE = −K (bound)
Binding energy+GMm/2rEnergy needed to free it

Near the surface, ve = √2 vo. Orbital speed and period do not depend on the satellite's mass. Astronauts feel weightless because they and their spacecraft fall freely around the Earth together.

6. Formula Master Table

QuantityFormula
Gravitational forceF = Gm₁m₂/r²
g at surfaceg = GM/R² = (4/3)πGρR
g at height / depthg(R/(R + h))²; g(1 − d/R)
Potential / PEV = −GM/r; U = −GMm/r
Escape speedve = √(2GM/R) = √(2gR)
Orbital speedvo = √(GM/r)
Satellite energyE = −GMm/2r
Kepler IIIT² ∝ r³

7. Master Summary Table

ConceptKey pointNEET Note
G vs gG universal; g variesG = 6.67 × 10⁻¹¹ SI
Height vs depthg falls both ways from the surfaceFalls faster with height
Escape speed√(2gR) ≈ 11.2 km s⁻¹Independent of mass
Orbital speed√(GM/r)ve = √2 vo near surface
Satellite energyE = −K, U = −2KNegative ⇒ bound
KeplerEllipse, equal areas, T² ∝ r³Equal areas ⇒ L conserved

8. Mind Map

Gravitation
Law
  • F = Gm₁m₂/r²
  • Shell theorem
  • Kepler's laws
g
  • Height, depth
  • Latitude
Energy
  • V = −GM/r
  • U = −GMm/r
  • Escape speed
Satellites
  • Orbital speed
  • Period
  • Weightlessness

9. Tips and Tricks

Shortcuts
  • For small heights, a height h reduces g as much as a depth 2h.
  • For planets of the same density, g ∝ R and ve ∝ R. In general ve ∝ √(M/R).
  • g becomes g/4 at height h = R, and g/n² at h = (n − 1)R.
  • Satellite energies: K : U : E = 1 : −2 : −1.
Common NEET traps
  • Using the approximation 1 − 2h/R when h is not small.
  • Measuring the orbit radius from the surface instead of from the Earth's centre.
  • Thinking escape speed depends on the angle of projection.

10. Important One-liners

  • The value of g is maximum at the poles and minimum at the equator.
  • g is zero at the centre of the Earth.
  • Escape speed from the Earth is about 11.2 km s⁻¹.
  • A planet moves fastest when it is closest to the Sun.
  • The total energy of an orbiting satellite is negative.
  • Orbital speed does not depend on the mass of the satellite.
  • Weightlessness in orbit is due to free fall, not absence of gravity.

11. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take g = 10 m s⁻², R = 6400 km, G = 6.67 × 10⁻¹¹ N m² kg⁻². Click to reveal each answer and working.

Average level

N1Average

Find the gravitational force between two 50 kg spheres whose centres are 0.5 m apart.

Show answer & solution

Concept: F = Gm₁m₂/r².

  1. m₁m₂ = 2500 kg²; r² = 0.25 m².
  2. F = 6.67 × 10⁻¹¹ × 2500/0.25 = 6.67 × 10⁻⁷ N.

Answer: 6.67 × 10⁻⁷ N

N2Average

Find the value of g at a height of 3R above the Earth's surface.

Show answer & solution

Concept: gh = gR²/(R + h)² — use the exact form because h is large.

  1. R + h = 4R.
  2. gh = g/16 = 10/16 = 0.625 m s⁻².

Answer: 0.625 m s⁻²

N3Average

What is the weight of a 60 kg miner at a depth of R/4 below the Earth's surface?

Show answer & solution

Concept: gd = g(1 − d/R).

  1. gd = 10 × (1 − ¼) = 7.5 m s⁻².
  2. Weight = 60 × 7.5 = 450 N.

Answer: 450 N

N4Average

Find the orbital speed of a satellite just above the Earth's surface and the escape speed from the surface.

Show answer & solution

Concept: vo = √(gR); ve = √(2gR).

  1. gR = 10 × 6.4 × 10⁶ = 6.4 × 10⁷.
  2. vo = √(6.4 × 10⁷) = 8000 m s⁻¹ = 8 km s⁻¹.
  3. ve = √2 × 8 ≈ 11.3 km s⁻¹.

Answer: 8 km s⁻¹; about 11.3 km s⁻¹

N5Average

A planet orbits a star at 9 times the distance at which another planet takes 1 year. How long does the farther planet take to go round once?

Show answer & solution

Concept: T ∝ r3/2 (Kepler's third law).

  1. T₂/T₁ = 93/2 = (√9)³ = 27.
  2. T₂ = 27 years.

Answer: 27 years

Advanced level

N6Advanced

A satellite circles the Earth at a height of 3R. Find its orbital speed and period.

Show answer & solution

Concept: v = √(gR²/r) with r = 4R.

  1. v = √(gR/4) = √(1.6 × 10⁷) = 4000 m s⁻¹.
  2. T = 2πr/v = 2π × 4 × 6.4 × 10⁶/4000.
  3. T ≈ 4.02 × 10⁴ s ≈ 11.2 hours.

Answer: 4 km s⁻¹; about 11.2 h

N7Advanced

A planet has 8 times the Earth's mass and 2 times its radius. Find the escape speed from it (ve for Earth = 11.2 km s⁻¹) and the value of g on it.

Show answer & solution

Concept: ve ∝ √(M/R); g ∝ M/R².

  1. ve ratio = √(8/2) = 2 ⇒ ve = 22.4 km s⁻¹.
  2. g ratio = 8/2² = 2 ⇒ g = 20 m s⁻².

Answer: 22.4 km s⁻¹; 20 m s⁻²

N8Advanced

A 200 kg satellite orbits at a height equal to R. Find its total energy and binding energy.

Show answer & solution

Concept: E = −GMm/2r with r = 2R and GM = gR².

  1. E = −gR²m/(4R) = −mgR/4.
  2. = −200 × 10 × 6.4 × 10⁶/4 = −3.2 × 10⁹ J.
  3. Binding energy = +3.2 × 10⁹ J.

Answer: −3.2 × 10⁹ J; 3.2 × 10⁹ J

N9Advanced

At what height and at what depth does g become one-fourth of its surface value?

Show answer & solution

Concept: height: (R/(R + h))² = ¼; depth: 1 − d/R = ¼.

  1. Height: R/(R + h) = ½ ⇒ h = R = 6400 km.
  2. Depth: d/R = ¾ ⇒ d = 4800 km.

Answer: 6400 km high; 4800 km deep

N10Advanced

A 10 kg body is raised from the Earth's surface to a height equal to R. Find the increase in its potential energy. Why is mgh wrong here?

Show answer & solution

Concept: ΔU = GMm(1/R − 1/2R) = GMm/2R = mgR/2.

  1. ΔU = 10 × 10 × 6.4 × 10⁶/2 = 3.2 × 10⁸ J.
  2. mgh would give 6.4 × 10⁸ J — double the true value.
  3. mgh assumes g stays constant, but g falls to g/4 at this height.

Answer: 3.2 × 10⁸ J

12. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

The escape speed of a body from the Earth depends on:

  1. The mass of the body
  2. The direction of projection
  3. The mass and radius of the Earth
  4. The height of the building it is thrown from, only
Show solution

ve = √(2GM/R) depends only on the planet's mass and radius (and the distance from its centre), not on the body's mass or direction.

Answer: (C)

Q2Easy

Kepler's second law is a consequence of the conservation of:

  1. Energy
  2. Linear momentum
  3. Angular momentum
  4. Mass
Show solution

Gravity is a central force, so it exerts no torque about the Sun; angular momentum is conserved, giving constant areal velocity.

Answer: (C)

Q3Medium

The ratio of escape speed to orbital speed for a satellite close to the Earth's surface is:

  1. 1
  2. √2
  3. 2
  4. 1/√2
Show solution

ve = √(2gR), vo = √(gR); ratio √2.

Answer: (B)

Q4Medium

If the radius of a planet's orbit around the Sun is made four times, its period becomes:

  1. 4 times
  2. 8 times
  3. 16 times
  4. 2 times
Show solution

T ∝ r3/2 = 43/2 = 8.

Answer: (B)

Q5Medium

The kinetic energy of a satellite is 5 MJ. Its total energy is:

  1. 5 MJ
  2. −5 MJ
  3. −10 MJ
  4. 10 MJ
Show solution

For a satellite, E = −K = −5 MJ (and U = −10 MJ).

Answer: (B)

Q6Medium

Astronauts in an orbiting spaceship feel weightless because:

  1. There is no gravity in space
  2. They and the ship fall freely with the same acceleration
  3. The ship's engines cancel gravity
  4. Their mass becomes zero
Show solution

Gravity still acts (g is only slightly less at the orbit), but everything in the ship accelerates together towards the Earth, so no normal force is felt.

Answer: (B)

Q7Hard

At what height above the Earth's surface does g become 64% of its surface value? (R = 6400 km)

  1. 1600 km
  2. 3200 km
  3. 800 km
  4. 2400 km
Show solution

(R/(R + h))² = 0.64 ⇒ R/(R + h) = 0.8 ⇒ R + h = 1.25R ⇒ h = R/4 = 1600 km.

Answer: (A)

Q8Hard

Two planets have equal density. Planet A has twice the radius of planet B. The ratio of their surface g values, gA : gB, is:

  1. 1 : 2
  2. 2 : 1
  3. 4 : 1
  4. 1 : 1
Show solution

g = (4/3)πGρR ∝ R for equal density, so gA : gB = 2 : 1.

Answer: (B)

13. Quick Revision Summary

Remember before the exam
  • F = Gm₁m₂/r²; g = GM/R² ≈ 9.8 m s⁻².
  • g(h) = g(R/(R + h))²; g(d) = g(1 − d/R); zero at the centre.
  • ve = √(2gR) ≈ 11.2 km s⁻¹; vo = √(GM/r); ve = √2 vo.
  • Satellite: K = GMm/2r, U = −2K, E = −K.
  • Kepler: ellipses, equal areas, T² ∝ r³.
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