Gravitation

One law for the falling apple and the orbiting Moon · Class 11 Physics

1The big idea

The single most important thought in this chapter is that the force pulling an apple to the ground and the force holding the Moon in its orbit are the same force, obeying the same law. Newton's leap was to realise that the Earth's pull does not stop at the treetops — it reaches out, thinning with distance, all the way to the Moon and beyond. Every particle of matter attracts every other particle. That is universal gravitation.

Once you accept one compact formula for the pull between two masses, everything else in this chapter — why g is what it is, why it drops as you climb a mountain and as you dig a mine, how fast a rocket must go to break free, why a higher satellite moves slower, and Kepler's beautiful planetary laws — all of it follows by pure reasoning. Almost nothing here needs to be memorised if you can rebuild it.

2The universal law of gravitation

Two point masses m₁ and m₂ separated by a distance r attract each other along the line joining them with a force

Newton's law of gravitation F = G m₁ m₂r²

Read the formula slowly, because its shape carries the physics:

Proportional to each mass. Double either body and the pull doubles — that is why the force is symmetric: the Earth pulls you exactly as hard as you pull the Earth (Newton's third law), yet only you visibly move, because you have the smaller mass.
Inverse-square in distance. Move twice as far apart and the force falls to a quarter, not a half. This 1/r² falloff is the fingerprint of something spreading out over the surface of a sphere, and it is what makes closed orbits possible.
Always attractive, and directed along the line of centres. There is no gravitational "push".

G is the universal gravitational constant, the same everywhere in the cosmos:

G = 6.67 × 10⁻¹¹ N m² kg⁻²

It is fantastically small, which is why you feel no tug toward the person next to you — gravity only becomes commanding when one of the masses is planet-sized.

Remember this

The law is written for point masses, yet we happily use it for the Earth. That is allowed because of a deep result Newton proved (the shell theorem): a uniform solid sphere pulls on anything outside it as if its entire mass were squeezed to a point at its centre. This is the licence to write r = distance to the Earth's centre, not to its surface.

3From G to g — weighing the Earth

Now specialise the law. Put a small mass m at the surface of the Earth (mass M, radius R). Treating the Earth as a point of mass M at its centre, the pull on m is

F = G M mR²

But we also call this pull the weight, F = m g. Equate the two and the little mass m cancels:

Acceleration due to gravity m g = G M mR² ⟹ g = G MR²

Two things worth noticing. First, g does not depend on the falling object's mass — a feather and a cannonball accelerate identically in vacuum, exactly Galileo's discovery, now explained. Second, this equation lets us weigh the Earth: rearranged as M = g R²/G, and with g = 9.8 m s⁻², R ≈ 6.4 × 10⁶ m and the measured G, it gives M ≈ 6 × 10²⁴ kg. The very same relation, written as G M = g R² ≈ 4 × 10¹⁴ m³ s⁻², is a shortcut we will lean on again and again — whenever you meet G M for the Earth, quietly replace it by g R².

4How g changes: height, depth, latitude

The value 9.8 m s⁻² is a surface value. Leave the surface — up or down — and it changes, but by different laws in the two directions.

Going up (height h)

At height h your distance from the centre is R + h, so

g_h = G M(R + h)² = g R²(R + h)² = g(1 + h/R)²

For heights small compared with the Earth (h ≪ R), expand by the binomial approximation (1 + x)⁻² ≈ 1 − 2x:

Small height g_h ≈ g (1 − 2hR)

Going down (depth d)

Digging into the Earth is subtler. By the shell theorem, once you are at depth d the outer shell of material above you pulls equally in all directions and contributes nothing. Only the inner sphere of radius (R − d) pulls you. Assuming uniform density, that inner sphere's mass scales as its volume, i.e. as (R − d)³:

g_d = G M(R−d)³/R³(R − d)² = G MR³(R − d)

Since g = G M/R², this tidies to a clean linear law:

Depth g_d = g (1 − dR)

So g falls straight to zero at the centre of the Earth — you would be weightless there, pulled equally in every direction. Notice the contrast: near the surface height reduces g twice as fast as the same depth does (the factor 2 in the height law versus 1 in the depth law). And g is greatest exactly at the surface.

Inside the Earth g grows in direct proportion to r (zero at the centre); outside it falls off as 1/r². The peak sits exactly at the surface, r = R.

Moving to a different latitude

There is a third, smaller effect: the Earth spins. A body sitting at latitude λ is carried around the Earth's axis on a circle of radius R cos λ, so part of the true gravitational pull is "used up" providing the centripetal acceleration for that circular motion, leaving a slightly reduced effective weight:

Latitude (rotation effect) g_λ ≈ g − ω² R cos²λ

At the equator (λ = 0, cos λ = 1) the reduction is largest; at the poles (λ = 90°, cos λ = 0) there is no reduction at all. This is one reason you weigh a touch more at the poles than at the equator, and why rockets are launched eastward from near-equatorial sites — the spin gives them a free head start.

5Gravitational potential energy & potential

At school you learned U = m g h. That formula is only a near-surface approximation, valid when g is effectively constant. The honest, universal expression comes from asking: how much work must be done, against gravity, to carry a mass m from infinitely far away in to a distance r from a body of mass M? Adding up (integrating) the pull over that journey gives

Gravitational potential energy U = − G M mr

The minus sign is not a typo and it matters enormously. We choose U = 0 at infinity (the natural "no interaction" reference). Because gravity is attractive, a bound mass is in an energy well — you must add energy to climb out toward infinity — so its energy is below zero. The closer in (smaller r), the deeper the well and the more negative U.

Dividing out the test mass gives the gravitational potential V — the potential energy per unit mass, a property of the field alone:

V = Um = − G Mr

Reassuringly, the old m g h is hiding inside this. Raise a mass from R to R + h and the change in U, for small h, works out to m g h — so nothing you learned earlier was wrong, it was just the flat-ground corner of a bigger picture.

6Escape velocity

How fast must you hurl an object straight up so that it never falls back — so it just barely reaches infinity? "Just barely" means it arrives at infinity with zero speed left over: zero kinetic energy and, by our convention, zero potential energy. So its total energy must be zero. Set the total energy at launch (surface) equal to that:

½ m v_e² − G M mR = 0

The launch mass m cancels — a wonderful result, meaning escape velocity does not depend on what you throw. Solving,

Escape velocity v_e = √(2 G MR) = √(2 g R)

where in the last step we again used G M = g R². Put in the Earth's numbers:

v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹

Remember this

Escape velocity ≈ 11.2 km s⁻¹ for Earth, independent of the projectile's mass and of the direction of launch (ignoring air resistance and the Earth's spin). It depends only on the planet: a more massive or more compact world holds on harder. It is not the speed a satellite needs — that is smaller, as we see next.

7Satellites: orbital speed, energy, geostationary

Orbital velocity

A satellite in a circular orbit of radius r is not "beyond gravity" — gravity is precisely the centripetal force curving it into a circle. So set the gravitational pull equal to the required centripetal force m v₀²/r:

G M mr² = m v₀²r ⟹ v₀ = √(G Mr)

Two consequences leap out. Just above the surface (r ≈ R) this gives v₀ = √(g R) ≈ 7.9 km s⁻¹ — the "first cosmic speed". And comparing with escape velocity, v_e = √2 · v₀: you must go about 41% faster to escape than merely to orbit. The other consequence surprises many: because v₀ ∝ 1/√r, a higher satellite moves slower.

Total energy of an orbiting satellite

Add the kinetic and potential energies. The kinetic energy is ½ m v₀² = ½ · G M m/r (using the orbital-speed result), and the potential energy is − G M m/r:

Total energy of a satellite E = 12G M mr − G M mr = − G M m2 r

The total energy is negative — the signature of a bound system. Notice two elegant facts: the kinetic energy is exactly half the magnitude of the potential energy, and the total is minus the kinetic energy. To lift a satellite to a higher orbit you must make E less negative, i.e. add energy — yet it ends up moving slower. When E reaches zero the satellite is no longer bound: that is escape.

Geostationary satellites

A geostationary satellite hangs over the same spot on the equator, appearing motionless in the sky — perfect for communications and TV relay. For that it must (i) orbit in the equatorial plane, (ii) travel west-to-east like the Earth's spin, and (iii) have an orbital period of exactly one day. From T² = 4π²r³/G M (derived next) with T = 24 h, the required radius comes out to about 42,000 km from the Earth's centre — an altitude of roughly 36,000 km above the surface, where the satellite drifts along at a leisurely ~3.1 km s⁻¹. We compute this in the worked example.

8Kepler's three laws

Decades before Newton, Kepler distilled Tycho Brahe's naked-eye planetary data into three empirical laws. Newton's gravity then explained all three.

The orbit is an ellipse with the Sun at one focus. The two shaded areas are equal, so they are swept in equal times — the planet must move faster near the Sun (perihelion) than far from it (aphelion). a is the semi-major axis.

1. Law of orbits. Every planet moves in an ellipse with the Sun at one focus (not the centre). A circle is just the special case of zero eccentricity.
2. Law of areas. The line from the Sun to the planet sweeps out equal areas in equal times. This is nothing but conservation of angular momentum: with gravity always pointing at the Sun (a central force), there is no torque about the Sun, so L is constant — and the areal velocity dA/dt = L/2m is therefore constant too. It is why a comet whips around the Sun and dawdles out at the fringes.
3. Law of periods. The square of the period is proportional to the cube of the semi-major axis: T² ∝ a³.

The third law falls straight out of what we already have. For a circular orbit the period is the circumference over the speed, T = 2πr/v₀. Substitute v₀ = √(G M/r):

Deriving Kepler's third law T = 2π r√(G M/r) = 2π √(r³G M) ⟹ T² = 4π²G M r³

The bracket 4π²/G M is the same for every body orbiting the same central mass, so T²/r³ is a constant of the system — exactly Kepler's statement, now with the proportionality constant supplied by Newton.

◆ Activity — orbit sandbox

Fire a satellite sideways from the same point near a planet and vary only its launch speed (shown as a fraction of the circular-orbit speed, ≈ 7.9 km s⁻¹ here). Too slow and its orbit dips into the planet — it crashes. At just the right speed it settles into a stable orbit (a perfect circle when the speed equals the circular value). Push past √2 times that — about 11.2 km s⁻¹ — and it has enough energy to escape and never return.

Worked example

Find the orbital radius, altitude and speed of a geostationary satellite. Take G M = g R² = 4.0 × 10¹⁴ m³ s⁻², Earth's radius R = 6.4 × 10⁶ m, and period T = 24 h = 86 400 s.

Rearrange Kepler's third law for the radius: r³ = G M · T²4π².

Numerator: G M · T² = (4.0 × 10¹⁴)(86 400)² ≈ 2.99 × 10²⁴.
Divide by 4π² ≈ 39.5: r³ ≈ 7.6 × 10²² m³.
Cube root: r ≈ 4.2 × 10⁷ m = 42 000 km from the centre.
Altitude: h = r − R ≈ 42 000 − 6 400 ≈ 36 000 km.
Speed: v₀ = 2π r / T = 2π(4.2 × 10⁷)/86 400 ≈ 3.1 km s⁻¹.

So a communications satellite parked 36 000 km up circles once a day, keeping perfect pace with the ground below. (Cross-check with v₀ = √(G M/r) = √(4.0×10¹⁴/4.2×10⁷) ≈ 3.1 km s⁻¹ ✓.)

9Common confusions to clear up

G and g are completely different. G is a universal constant (6.67 × 10⁻¹¹), the same on every planet; g = G M/R² is a local acceleration that varies with the body and with position on it.
Astronauts are not beyond gravity. At a typical space-station altitude g is still about 90% of its surface value. They feel "weightless" because they and the station are in continuous free fall together around the Earth — falling and missing the ground, forever.
g is maximum at the surface, and decreases whether you go up or down — but by different laws: g(1 − 2h/R) going up versus g(1 − d/R) going down.
Escape speed is not orbital speed. Orbiting needs √(G M/r); escaping needs √2 times as much. And escape speed is independent of the mass thrown and of the launch direction.
A higher satellite is slower, not faster. Since v₀ ∝ 1/√r, raising the orbit lowers the speed while lengthening the period.
Potential energy is negative because we set it to zero at infinity and gravity binds. "More energy" means less negative — closer to escape, not closer to the planet.
Kepler's second law is just angular momentum conservation. The planet speeds up near the Sun and slows down far away so that equal areas are swept in equal times.

10Check yourself

Class 11 Physics · Gravitation · aligned to NCERT Ch 8 · SmartStudy.School