Gravitation: The Force That Ties Apples, Planets and Satellites Together

Master Newton's law of gravitation, variation of g, escape and orbital velocity, and Kepler's laws with clear formulas and a worked example for JEE.

By the PadhoDost Team · 📖 8 min read · Updated 4 August 2026

Part of JEE (Main + Adv) prep

🧠 The Moon is always falling

Throw a ball horizontally and it curves down and lands. Throw it faster and it lands farther. Newton's leap of imagination: throw it so fast that as it falls, the Earth curves away beneath it by the same amount, and it never lands, it orbits. The Moon is doing exactly this. It is perpetually 'falling' toward Earth but moving sideways fast enough to keep missing. The same force that pulls the apple down holds the Moon in its path: gravitation.

Newton's law of universal gravitation

m₁ m₂ F F r F = G·m₁·m₂ / r²
Every two masses pull on each other — stronger with more mass, weaker with distance².

Every particle of matter in the universe attracts every other particle. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. It acts along the line joining the two masses and is always attractive.

F = G m1 m2 / r^2, where G = 6.674 x 10^-11 N m^2 kg^-2 is the universal gravitational constant. G is the same everywhere in the universe; g (acceleration due to gravity) is not.
QuantityFormulaNote
Force of gravitationF = G m1 m2 / r^2Always attractive
Acceleration due to gravityg = GM / R^2About 9.8 m/s^2 at Earth's surface
g at height hg_h = g R^2/(R+h)^2 ~ g(1 - 2h/R)Decreases as you go up
g at depth dg_d = g(1 - d/R)Becomes zero at the centre
Gravitational potential energyU = -G M m / rZero at infinity, negative elsewhere
Escape velocityv_e = sqrt(2GM/R) = sqrt(2gR)About 11.2 km/s for Earth
Orbital velocityv_o = sqrt(GM/r)About 7.9 km/s near Earth
Time period of orbitT = 2*pi*sqrt(r^3/GM)Gives Kepler's third law
Total energy in orbitE = -G M m / 2rKE = +GMm/2r, PE = -GMm/r

How g changes

  • With height: g decreases; for small h, fractional drop is about 2h/R.
  • With depth: g decreases linearly and is zero at Earth's centre.
  • With latitude/rotation: g' = g - R*omega^2*cos^2(lambda), so g is maximum at the poles and minimum at the equator.
  • Shape of Earth: since Earth bulges at the equator, g is slightly larger at the poles for this reason too.

Escape velocity and orbital velocity

📝 Worked example: escape velocity from Earth

Given: g = 9.8 m/s^2 and Earth's radius R = 6.4 x 10^6 m.

Formula: v_e = sqrt(2gR).

Substitute: v_e = sqrt(2 x 9.8 x 6.4 x 10^6).

Inside the root: 2 x 9.8 = 19.6; 19.6 x 6.4 x 10^6 = 1.2544 x 10^8.

Take the square root: v_e = sqrt(1.2544 x 10^8) = 1.12 x 10^4 m/s.

Answer: v_e = 11.2 km/s, and notice it does NOT depend on the mass of the escaping object.

Kepler's laws and orbital energy

Three laws + one energy rule

  • Law of Orbits: every planet moves in an ellipse with the Sun at one focus.
  • Law of Areas: the line from Sun to planet sweeps equal areas in equal times (this is conservation of angular momentum; planets move faster when nearer the Sun).
  • Law of Periods: T^2 is proportional to a^3, where a is the semi-major axis.
  • Bound orbit energy: total energy E = -GMm/2r is negative; a more negative energy means a smaller, tighter orbit.
⚠️ Common JEE traps: (1) escape velocity is independent of the object's mass and its direction of projection. (2) A geostationary satellite must have a period of 24 hours, orbit over the equator, and move west-to-east, giving a height of about 36,000 km. (3) Remember v_e = sqrt(2) x v_o for the same radius.

⚡ Quick check

The escape velocity from Earth's surface is about 11.2 km/s. What is the orbital velocity of a satellite in a circular orbit just above Earth's surface?

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