Gravitation

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 7.

Why doesn't the Moon fall on us?

It does. The Moon is falling toward the Earth every second of every day. It just has so much sideways speed that, as it falls, the Earth's curved surface drops away beneath it, and it keeps missing. Newton saw that the force that drops an apple and the force that holds the Moon are one and the same. That single insight, plus a few lines of algebra, explains the tides, the orbit of every satellite, and why rockets need 11 km per second to leave home.

From epicycles to ellipses

The ancient geocentric model of Ptolemy put the Earth at the centre, with each planet riding on a small circle (an epicycle) that itself travelled on a bigger one. Aryabhata had already suggested a rotating Earth, and Copernicus revived the Sun-centred picture. Galileo's telescope, which showed Jupiter's moons and the phases of Venus, gave it direct support. The decisive step came from Johannes Kepler, who spent years analysing the naked-eye observations of Tycho Brahe and distilled them into three laws.

Why the second law holds. The area swept per second is dA/dt = ½ r² ω = L/(2m), where L is the planet's angular momentum about the Sun. The Sun's pull always points along the line joining the two bodies, so it exerts no torque about the Sun. With no torque, L stays constant, and so does the rate of sweeping area. This is true for any force directed along the line to a fixed centre, not just gravity.

Worked example: a comet's speeds. A comet passes the Sun at its closest point (1 AU) with a speed of 40 km s⁻¹. At its farthest point it is 4 AU from the Sun. What is its speed there?

Common wrong step. Equating kinetic energies, which are not equal here because the potential energy changes along the orbit. Speed and distance trade through angular momentum, not through kinetic energy.

Worked example: Kepler's third law. A planet orbits the Sun at a mean distance 4 times that of the Earth. What is its period?

Common wrong step. Writing T ∝ a, which gives 4 years. The period grows faster than the distance: T ∝ a^(3/2).

Newton's insight: one law for apples and moons

The Moon is about 60 times farther from the Earth's centre than an apple. Its centripetal acceleration is 4π²R/T², with R = 3.84 × 10⁸ m and T = 27.3 days, which comes to 0.0027 m s⁻². That is almost exactly g/3600, and 3600 = 60². So the pull of the Earth falls off as the inverse square of the distance. Newton's universal law of gravitation states that every body attracts every other with a force along the line joining them, proportional to the product of the masses and inversely proportional to the square of the distance:

$$F = \dfrac{G\,m_1 m_2}{r^2}, \qquad G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$$

The force is always attractive, directed along the line joining the bodies and toward the other one. The constant G is universal, with dimensions [M⁻¹L³T⁻²]. The forces on the two bodies are equal and opposite (Newton's third law), however unequal the masses. For several bodies, add the forces as vectors (superposition).

How we know G. In 1798 Henry Cavendish measured it using a torsion balance. Two small lead balls sit at the ends of a light rod hung from a thin fibre. Large lead spheres are brought close to the small ones, and their gravitational pull twists the fibre by a tiny angle. Comparing this with the fibre's known twisting resistance gives G. Once G is known, g = GM/R² gives the mass of the Earth: M = gR²/G = 9.8 × (6.4 × 10⁶)² / 6.67 × 10⁻¹¹ ≈ 6 × 10²⁴ kg. This is why Cavendish's experiment is described as "weighing the Earth".

Shell theorems. For uniform spherical shells, adding up all the pulls gives two clean results. Outside the shell, it attracts a point mass exactly as if all its mass were at its centre. Inside the shell, the net force on a point mass is zero everywhere. Together they let us treat the Earth, the Sun and the planets as point masses when we are outside them.

Worked example: the neutral point. Two bodies of mass M and 4M are 6 m apart. At what point on the line joining them is the net gravitational force on a small test mass zero?

Common wrong step. Choosing the midpoint (3 m). The point must be closer to the lighter body so that its smaller mass makes up for the shorter distance. Also, gravity is always attractive, so there is no neutral point outside the segment between the bodies.

Beyond the textbook. For the exam: the force inside a uniform shell is zero, but gravitational shielding is impossible. The shell does not stop the pull of outside bodies on anything inside. In reality: nothing blocks gravity because there is no repulsive gravitational charge to cancel it, unlike electricity where positive and negative charges cancel. Why it matters: this is why gravity, the weakest force by far between everyday objects, still shapes the whole universe: it only adds up.

Acceleration due to gravity

A body of mass m at the surface feels F = GMm/R², where M and R are the mass and radius of the Earth. Newton's second law gives its acceleration:

$$g = \dfrac{GM}{R^2} \approx 9.8\ \text{m s}^{-2}$$

It does not depend on m, which is why all bodies fall alike in vacuum. Note the two similar-looking symbols: G is a universal constant; g depends on the planet and the location.

Above the surface. At height h, g(h) = GM/(R + h)² = g · R²/(R + h)². For h much smaller than R, a binomial expansion gives

g(h) ≈ g (1 − 2h/R)

Below the surface. Inside a uniform Earth, the shell outside radius (R − d) pulls with zero net force, so only the inner sphere counts. Since mass ∝ r³ and the pull goes as 1/r², g is proportional to r:

$$g(d) = g\left(1 - \dfrac{d}{R}\right), \quad g = 0 \text{ at the centre}$$

So g is largest at the surface and falls both upward and downward. For small heights and depths the two formulas show that height lowers g twice as fast as depth does: g(h) = g(d) when d = 2h.

Worked example: when the approximation fails. At what height above the Earth's surface does g fall to one quarter of its surface value?

Common wrong step. Using g(1 − 2h/R) = g/4, which gives h = 3R/8. That formula needs h ≪ R. At h = 3R/8 it is already badly off. The approximation is good to about 1% only for heights up to a few tens of kilometres.

Pause and predict: at which depth below the surface of a uniform Earth is the weight of a body half its surface weight?

Check your answer: at d = R/2. g(d) = g(1 − d/R) = g/2 when d/R = ½, so the depth is about 3200 km.

Beyond the textbook. For the exam: treat the Earth as uniform, so g falls steadily from the surface to zero at the centre. In reality: the Earth's density rises steeply toward the core, so g actually stays about the same or slightly increases as you go down, reaching roughly 10.7 m s⁻² near the boundary between the mantle and the core, before dropping to zero at the centre. Why it matters: the linear fall of g with depth is a property of a uniform sphere, which the real Earth is not. Use the linear formula in the exam, and know that it is an idealisation.

Gravitational potential energy

Near the surface the potential energy is mgh, because g is almost constant. Over larger distances we must integrate the changing force. Taking the zero of potential energy at infinity, the energy of two point masses a distance r apart is

$$U = -\dfrac{G\,m_1 m_2}{r}$$

It is negative because gravity does positive work as the masses approach, so the system has less energy when they are closer than when they are infinitely far apart. The gravitational potential at a point is the potential energy per unit mass there, V = −GM/r. For several masses, add the potential energies of all pairs.

The familiar mgh is the same thing in disguise. The change from the surface (R) to height h is GMmh/(R(R + h)), and for h ≪ R this is mgh, since GM/R² = g.

Escape speed

What is the least speed that lets a body leave the Earth for good? "For good" means arriving at infinity with zero kinetic energy. Energy conservation then needs total energy at the start to be at least zero:

½ m v² − G M m / R = 0

$$v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR} \approx 11.2\ \text{km s}^{-1}$$

Two points matter for the exam. It does not depend on the mass of the projectile, and it does not depend on the direction of projection (as long as the path does not hit the ground). For the Moon, with its small mass and radius, v(escape) is only about 2.3 km s⁻¹.

Worked example: a bigger planet. A planet has 8 times the mass and 2 times the radius of the Earth. Find the escape speed from its surface and its value of g.

Common wrong step. Treating v(escape) as proportional to M/R, which gives 4 × 11.2. The escape speed involves a square root.

Beyond the textbook. For the exam: the Moon cannot hold an atmosphere because its escape speed is small. In reality: the same logic explains why the Earth has almost no free hydrogen or helium. Gas molecules move at a range of speeds, and the lightest molecules, being fastest at a given temperature, are the ones that reach escape speed from the top of the atmosphere over geological time. Heavier gases like nitrogen and oxygen stay.

Satellites

A satellite in a circular orbit of radius r = R + h needs a centripetal force, and gravity provides it:

m v² / r = G M m / r²

$$v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{gR^2}{R+h}}, \qquad T = 2\pi\sqrt{\dfrac{r^3}{GM}}$$

The speed does not depend on the satellite's own mass. For a satellite skimming the surface (h ≈ 0), v = √(gR) ≈ 7.9 km s⁻¹ and T₀ = 2π√(R/g) ≈ 85 minutes. The relation T² ∝ r³ is Kepler's third law for satellites. Comparing with the escape speed, v(escape) = √2 × v(orbit) at the same radius.

Worked example: a satellite at height R. A satellite circles the Earth at a height equal to the Earth's radius. Find its orbital speed and its period in terms of the near-surface values.

Common wrong step. Keeping the speed at 7.9 km s⁻¹ because "the satellite is in orbit". A farther orbit is a slower orbit, with a longer period.

Two special orbits are worth knowing. A geostationary satellite has a period of exactly 24 hours, orbits in the equatorial plane in the same direction as the Earth's rotation, and so seems to hang fixed over one point. Its height is about 36 000 km above the surface. Communication and broadcast satellites use it. Polar satellites orbit at a low height, typically 500 to 800 km, in a plane through the poles, with a period of about 100 minutes. As the Earth rotates beneath them, they scan the entire surface, which is useful for mapping and remote sensing.

Weightlessness. An astronaut in orbit seems to float, but not because gravity is absent. At the height of the International Space Station gravity is still about 88% of its surface value. The astronaut and the station are both in free fall with the same acceleration, so the floor does not push on the astronaut. The sensation of weight comes from that push.

Beyond the textbook. For the exam: weightlessness is free fall, not zero gravity. In reality: at about 400 km, the height of the space station, g is roughly 8.7 m s⁻², from g(R/(R + h))² with h = 400 km. If you could stand on a very tall tower of that height, you would weigh nearly 90% of your normal weight. The station does not float because it is "outside" gravity. It is falling around the Earth continuously.

Energy of an orbiting satellite

For a circular orbit, K = ½mv² = GMm/(2r) and U = −GMm/r. The total is

$$E = K + U = -\dfrac{GMm}{2r} = -K$$

The total energy is negative, which says the satellite is bound. The magnitude of the total equals the kinetic energy and half the magnitude of the potential energy. The binding energy, the energy needed to free the satellite, is GMm/(2r). For an elliptical orbit, the same formula holds with r replaced by the semi-major axis a.

Pause and predict: a satellite in a circular orbit slowly loses energy to the thin upper atmosphere. Does its speed go up or down?

Check your answer: it goes up. E = −GMm/(2r) becomes more negative, so r decreases. A smaller orbit needs a larger speed, v = √(GM/r). Drag takes energy away, yet the satellite speeds up, because the kinetic energy equals −E. This surprising behaviour is why a decaying satellite spirals in faster and faster.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: System of Particles and Rotational Motion gives the angular momentum conservation behind Kepler's second law.
• Next: Mechanical Properties of Solids drops the rigid-body idealisation and asks how real solids stretch, bend and break.
• Earlier idea reused: Work, Energy and Power introduced conservative forces and mgh, which become −GMm/r here.
• Class XII link: Electric Charges and Fields repeats the inverse square law with Coulomb's law, and Atoms compares electron orbits with planetary orbits.