JEE Main Physics · Chapter 6

Gravitation

Newton’s law of gravitation, variation of g, potential energy, escape and orbital speed, satellites and Kepler’s laws.

NCERT Class 11JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

Gravitation is the attractive force between any two masses. Newton's law of gravitation explains falling bodies, the motion of the Moon and planets, and the orbits of satellites. The chapter covers the variation of g, gravitational potential and potential energy, escape and orbital speeds, satellite energy and Kepler's laws. JEE Main asks one question almost every year — often on g at a height/depth, orbital speed or satellite energy.

What JEE asks most
  • Variation of g with height, depth and latitude.
  • Escape velocity and orbital velocity; their relation.
  • Energy of a satellite and energy needed to change orbits.
  • Kepler's third law (T² ∝ r³).

2. Newton's Law of Gravitation

F = Gm₁m₂/r²    G = 6.67 × 10⁻¹¹ N m² kg⁻²
  • The force is always attractive, acts along the line joining the masses, and forms an action–reaction pair.
  • It is independent of the medium and obeys superposition.
  • A uniform spherical shell attracts outside masses as if all its mass were at the centre, and exerts zero force on a mass inside it.

3. Acceleration due to Gravity

At Earth's surface g = GM/R² ≈ 9.8 m s⁻².

SituationFormula
Height hgh = g R²/(R + h)² ≈ g(1 − 2h/R) for h ≪ R
Depth dgd = g(1 − d/R); zero at the centre
Latitude λ (rotation)gλ = g − ω²R cos²λ
R g ∝ r (inside)g ∝ 1/r² (outside) gr
g increases linearly inside a uniform Earth and falls as 1/r² outside; it is largest at the surface.

g is maximum at the poles and minimum at the equator (because of Earth's rotation and its slightly flattened shape).

4. Gravitational Potential and Energy

  • Gravitational potential at distance r from mass M: V = −GM/r (zero at infinity).
  • Potential energy of two masses: U = −Gm₁m₂/r. For many masses, add U for every pair.
  • Gravitational field E = −dV/dr = GM/r² (towards the mass).
  • Inside a uniform solid sphere: V = −GM(3R² − r²)/(2R³); at the centre V = −1.5 GM/R.

5. Escape Speed, Orbits and Kepler's Laws

QuantityFormulaEarth value
Escape speedve = √(2GM/R) = √(2gR)≈ 11.2 km s⁻¹
Orbital speed (near surface)vo = √(GM/r) = √(gR)≈ 7.9 km s⁻¹
Relationve = √2 vo
PeriodT = 2π√(r³/GM)≈ 84 min near surface
KE, PE, total energyK = GMm/2r; U = −GMm/r; E = −GMm/2r
  • Total energy E = −K and U = 2E. The binding energy of a satellite is +GMm/2r.
  • Geostationary satellite: period 24 h, orbit in the equatorial plane, height about 36 000 km, moves west to east. Used for communication. Polar satellites orbit at low heights for remote sensing.
  • Astronauts in orbit feel weightless because they and their craft fall freely together.
  • Kepler's laws: (1) orbits are ellipses with the Sun at one focus; (2) the line from the Sun sweeps equal areas in equal times (angular momentum conservation); (3) T² ∝ a³.

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 energyU = −GMm/r
Escape speedve = √(2gR) ∝ R√ρ
Orbital speedvo = √(GM/r)
Satellite energyE = −GMm/2r
Kepler IIIT² ∝ r³

7. Mind Map

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

8. Tips and Tricks

Shortcuts
  • g decreases faster with height than with depth: for small values, a height h gives the same fall as a depth 2h.
  • For same-density planets, g ∝ R and ve ∝ R.
  • To move a satellite from radius r₁ to r₂, energy needed = GMm/2 × (1/r₁ − 1/r₂).
  • Kepler ratio questions: T₂/T₁ = (r₂/r₁)3/2.
Mistakes students make
  • Using h measured from Earth's centre instead of from the surface (r = R + h).
  • Using g(1 − 2h/R) when h is not small compared with R.
  • Forgetting the negative sign of gravitational potential energy.

9. Important One-liners

  • Escape speed does not depend on the mass of the body or the direction of projection.
  • g is zero at the centre of the Earth.
  • A geostationary satellite has a period of 24 hours.
  • Total energy of a bound satellite is negative.
  • Kepler's second law follows from conservation of angular momentum.
  • Orbital speed does not depend on the satellite's mass.

10. Solved Numericals: Average to Advanced

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

Average level

N1Average

Find the value of g at a height equal to half the radius of the Earth.

Show answer & solution

Concept: gh = gR²/(R + h)² (h is not small, so use the exact form).

  1. R + h = 1.5R.
  2. gh = g/(1.5)² = g/2.25 = 4g/9.
  3. = 40/9 ≈ 4.44 m s⁻².

Answer: about 4.44 m s⁻²

N2Average

Find g at a depth equal to half the radius of the Earth.

Show answer & solution

Concept: gd = g(1 − d/R).

  1. d/R = 1/2.
  2. gd = 10 × (1 − 0.5) = 5 m s⁻².

Answer: 5 m s⁻²

N3Average

A person weighs 600 N on Earth. What will the person weigh on the Moon, where g is one-sixth of Earth's value? What is the person's mass there?

Show answer & solution

Concept: mass is unchanged; weight = mg.

  1. Mass = 600/10 = 60 kg (same everywhere).
  2. Weight on Moon = 60 × 10/6 = 100 N.

Answer: 100 N; 60 kg

N4Average

A planet is 4 times as far from its star as Earth is from the Sun (assume the star has the Sun's mass). How many Earth-years does it take to complete one orbit?

Show answer & solution

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

  1. T/TE = (4)3/2.
  2. = (√4)³ = 2³ = 8.

Answer: 8 years

N5Average

Find the gravitational force between two 100 kg spheres whose centres are 1 m apart.

Show answer & solution

Concept: F = Gm₁m₂/r².

  1. F = 6.67 × 10⁻¹¹ × 100 × 100/1².
  2. = 6.67 × 10⁻⁷ N — tiny, which is why we do not notice attraction between everyday objects.

Answer: 6.67 × 10⁻⁷ N

Advanced level

N6Advanced

The escape speed from Earth is 11.2 km s⁻¹. Find the escape speed from a planet with the same density but twice the radius.

Show answer & solution

Concept: ve = √(2GM/R) with M = (4/3)πR³ρ ⇒ ve = R√(8πGρ/3) ∝ R√ρ.

  1. Same ρ, so ve ∝ R.
  2. ve,planet = 2 × 11.2 = 22.4 km s⁻¹.

Answer: 22.4 km s⁻¹

N7Advanced

A satellite orbits at a height equal to Earth's radius. Find its orbital speed and period.

Show answer & solution

Concept: v = √(GM/r) with GM = gR² and r = 2R.

  1. v = √(gR²/2R) = √(gR/2) = √(10 × 6.4 × 10⁶/2) = √(3.2 × 10⁷).
  2. v ≈ 5657 m s⁻¹ ≈ 5.66 km s⁻¹.
  3. T = 2πr/v = 2π × 1.28 × 10⁷/5657 ≈ 1.42 × 10⁴ s.
  4. ≈ 3.95 hours.

Answer: about 5.66 km s⁻¹; about 3.95 h

N8Advanced

How much energy is needed to launch a 100 kg satellite from Earth's surface into a circular orbit at height R? (Ignore Earth's rotation.)

Show answer & solution

Concept: energy needed = Eorbit − Esurface.

  1. At the surface (at rest): E₁ = −GMm/R.
  2. In orbit at r = 2R: E₂ = −GMm/(2 × 2R) = −GMm/4R.
  3. ΔE = GMm/R (1 − ¼) = ¾ mgR.
  4. = 0.75 × 100 × 10 × 6.4 × 10⁶ = 4.8 × 10⁹ J.

Answer: 4.8 × 10⁹ J

N9Advanced

At what height above Earth's surface does g decrease by 1%? At what depth does it decrease by 1%?

Show answer & solution

Concept: for small h, Δg/g = 2h/R; for depth, Δg/g = d/R.

  1. Height: 2h/R = 0.01 ⇒ h = R/200 = 6400/200 = 32 km.
  2. Depth: d/R = 0.01 ⇒ d = R/100 = 64 km.
  3. Note the depth is twice the height for the same small change.

Answer: 32 km high; 64 km deep

N10Advanced

With what speed must a body be thrown vertically upward from Earth's surface so that it just reaches a height equal to R? (Ignore air resistance.)

Show answer & solution

Concept: energy conservation with U = −GMm/r (not mgh, because h is large).

  1. ½mv² − GMm/R = 0 − GMm/2R.
  2. ½v² = GM/2R ⇒ v² = GM/R = gR.
  3. v = √(10 × 6.4 × 10⁶) = 8000 m s⁻¹ = 8 km s⁻¹.

Answer: 8 km s⁻¹

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

The time period of a geostationary satellite is:

  1. 12 h
  2. 24 h
  3. 84 min
  4. 365 days
Show solution

It must keep pace with Earth's rotation, so its period is 24 h.

Answer: (B)

Q2Easy

The ratio of escape speed to orbital speed near Earth's surface is:

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

ve = √(2gR), vo = √(gR); ratio √2.

Answer: (C)

Q3Medium

Kepler's second law (equal areas in equal times) is a consequence of conservation of:

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

Areal velocity dA/dt = L/2m. Gravity is a central force (zero torque about the Sun), so L is constant.

Answer: (C)

Q4Hard

If Earth shrank so that its radius decreased by 1% with no change in mass, g at its surface would:

  1. Decrease by 1%
  2. Increase by 1%
  3. Increase by 2%
  4. Decrease by 2%
Show solution

g = GM/R² ⇒ Δg/g = −2ΔR/R = −2(−1%) = +2%.

Answer: (C)

Q5Medium

The value of g is greatest at:

  1. The equator
  2. The poles
  3. The centre of the Earth
  4. A height of 100 km
Show solution

At the poles there is no centrifugal effect of rotation and the radius is slightly smaller, so g is largest.

Answer: (B)

Q6Numerical

The total mechanical energy of a satellite is −2 MJ. Find the magnitude of its gravitational potential energy in MJ.

Show solution

U = 2E = −4 MJ; magnitude 4 MJ (and K = +2 MJ).

Answer: 4

12. Quick Revision Summary

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