Gravitation for JEE Main

Gravitation is one of the most formula-driven chapters in JEE Main physics, and that is good news. The same dozen results come back in different combinations: how g changes, how fast a satellite moves, how much energy it has, and how its period scales with its orbit. Learn where each formula comes from and which ones are exact and which are approximations, and the chapter becomes dependable marks.

The syllabus covers the universal law of gravitation, the variation of g with altitude and depth, Kepler's laws, gravitational potential energy and potential, escape velocity, and the orbital velocity, period and energy of satellites.

Newton's law of gravitation

F = Gm₁m₂/r², attractive, along the line joining the two masses, with G = 6.67 × 10⁻¹¹ N m² kg⁻². Forces from several masses add as vectors.

Two results from the shell theorem make spheres easy:

• Outside a uniform spherical shell or solid sphere, it acts as if all its mass were at the centre.
• Inside a uniform shell, the gravitational field is zero.

Acceleration due to gravity and how it varies

At the surface, g = GM/R². Using M = (4/3)πR³ρ, g = (4/3)πGρR, so for the same density g is proportional to R.

The depth result follows from the shell theorem: at depth d only the inner sphere of radius (R − d) pulls, and its mass scales as (R − d)³.

Field and potential

The field is the force per unit mass; the potential is the potential energy per unit mass, taken as zero at infinity. Gravitational potential is always negative.

At the centre of a solid sphere the potential is −3GM/(2R), which is 1.5 times the surface value.

Potential energy: U = −GMm/r. The familiar mgh is an approximation for small heights. The exact work needed to lift a mass m from the surface to height h is GMmh/[R(R + h)] = mgh/(1 + h/R).

Escape speed

Give a body just enough kinetic energy to reach infinity with zero speed:
½mv² − GMm/R = 0, so v_e = √(2GM/R) = √(2gR), about 11.2 km/s for Earth.

It does not depend on the mass of the body or the direction of launch (ignoring air). Because M ∝ ρR³, v_e ∝ R√ρ.

Satellites

For a circular orbit of radius r = R + h, gravity supplies the centripetal force: GMm/r² = mv²/r.

So KE = −E and PE = 2E. A satellite with more negative total energy is more tightly bound. The binding energy is GMm/2r.

Near the surface, v_e/v₀ = √2. A body in a circular orbit needs its speed raised by a factor of √2 to escape.

A geostationary satellite orbits above the equator, from west to east, with a period of 24 hours. Its orbit radius is about 42,000 km from Earth's centre, about 36,000 km above the surface. Astronauts in orbit feel weightless because they and their craft fall freely together, not because gravity vanishes.

Kepler's laws

1. Planets move in ellipses with the Sun at one focus.
2. The line joining a planet to the Sun sweeps equal areas in equal times. This is conservation of angular momentum, since gravity exerts no torque about the Sun. So v_perihelion × r_perihelion = v_aphelion × r_aphelion.
3. T² ∝ a³, where a is the semi-major axis (the radius for a circular orbit).

Worked problems

Problem 1: small height and depth. At what height does g fall by 1%? At what depth? (R = 6,400 km)

Height: 2h/R = 0.01, so h = R/200 = 32 km.
Depth: d/R = 0.01, so d = R/100 = 64 km.

Problem 2 (numerical answer): escape speed on another planet. A planet has the same density as Earth and twice its radius. Earth's escape speed is 11.2 km/s. Find the planet's escape speed in km/s.

v_e ∝ R√ρ. Same density, double radius, so v_e = 2 × 11.2 = 22.4 km/s.

Problem 3: launching into orbit. How much energy is needed to put a 200 kg satellite into a circular orbit at height R above Earth? (g = 10 m/s², R = 6.4 × 10⁶ m; ignore Earth's rotation)

On the surface: E₁ = −GMm/R. In orbit (r = 2R): E₂ = −GMm/4R.
Energy needed = GMm/R − GMm/4R = (3/4) GMm/R = (3/4) mgR.
= 0.75 × 200 × 10 × 6.4 × 10⁶ = 9.6 × 10⁹ J.

Problem 4: Kepler's third law. A satellite's orbit radius is increased to four times. What happens to its period?

T² ∝ r³. The radius is 4 times, so T² is 64 times and T is 8 times.

Practice set

1. Find the ratio of escape speed to orbital speed for a satellite close to Earth's surface.
2. A satellite's kinetic energy is K. Find its potential energy and total energy.
3. At what height above the surface does g become g/4?
4. At what depth does g become g/2? At what height? (answer in terms of R)
5. Find the ratio of the gravitational potential at the centre of a uniform solid sphere to that at its surface.
6. Two satellites have periods in the ratio 1 : 8. Find the ratio of their orbit radii.
7. A planet has 1/9 of Earth's mass and half its radius. Find g on its surface (g on Earth = 10 m/s²).
8. A planet moves at 30 km/s at its closest point to the Sun, at distance r. Its farthest distance is 1.5r. Find its speed there.

Answers

1. √2.
2. PE = −2K; total = −K.
3. R²/(R + h)² = 1/4 gives R + h = 2R, so h = R.
4. Depth: 1 − d/R = 1/2, so d = R/2. Height: (R + h)² = 2R², so h = (√2 − 1)R ≈ 0.414R.
5. (−3GM/2R)/(−GM/R) = 3 : 2.
6. r³ ∝ T². The T² ratio is 1 : 64, so the r³ ratio is 1 : 64 and the radii are 1 : 4.
7. g ∝ M/R² = (1/9)/(1/4) = 4/9, so g = 40/9 ≈ 4.4 m/s².
8. v₁r₁ = v₂r₂: 30 × r = v × 1.5r, so v = 20 km/s.

What to do next

• Derive escape speed, orbital speed and satellite energy from one equation each.
• Memorise which g formula is exact and which is an approximation.
• Solve 20 problems mixing variation of g, satellite energy and Kepler's laws.
• Revise circular motion for the orbit equation and work, energy and power for potential energy, then move on to elasticity and fluids.