Gravitation

For most of history the heavens and the Earth were thought to run on different rules. Newton’s boldest move was to insist they don’t: the force that drops an apple also bends the Moon’s path.

Before Newton#

Astronomers had tracked the planets for centuries. Aryabhata, in fifth-century India, argued that the Earth spins on its axis. A thousand years later Tycho Brahe’s precise naked-eye measurements let Johannes Kepler distil planetary motion into three rules:

1. Each planet moves on an ellipse with the Sun at one focus.
2. The line from the Sun to a planet sweeps out equal areas in equal times.
3. The square of a planet’s period is proportional to the cube of the semi-major axis of its ellipse.

These described the motion beautifully but didn’t say why. That needed a force law.

The Moon is falling#

The Moon circles the Earth at about $R = 3.85\times10^8$ m once every 27.3 days, so it is accelerating toward the Earth at

An apple falls with 9.8 m/s², 3600 times more. The apple is about 6400 km from the Earth’s centre and the Moon about 60 times further, and 60² = 3600. So the acceleration produced by the Earth falls off as the inverse square of distance.

The law#

Acceleration ∝ 1/r² means the force on a body of mass m is ∝ m/r². By the third law the body pulls the Earth equally hard, so the force must also be proportional to the Earth’s mass. Newton then made the leap that every pair of bodies in the universe attracts this way:

G is so small that two people standing a metre apart attract each other with well under a millionth of a newton. Henry Cavendish measured it in 1798 with little more than a stick, a wire and four lead balls.

The torsion balance#

Two small balls of mass m are fixed to the ends of a light rod of length l, which hangs horizontally from a very fine quartz fibre. Two large balls of mass M are placed beside the small ones, on opposite sides, all four centres on one horizontal circle. Each small ball feels a sideways pull

where r is the centre-to-centre distance (a uniform sphere acts like a point at its centre). The two pulls are perpendicular to the rod and turn it the same way, giving a torque $2F\cdot\tfrac l2 = Fl$. As the rod turns, the fibre twists and pushes back with a torque proportional to the twist, kθ. The rod settles where the two balance:

Making a tiny angle visible#

θ is far too small to read off a protractor. So a small mirror is fixed to the fibre and a light beam is bounced off it onto a distant scale. A mirror turned through θ turns the reflected beam through 2θ.

To double the effect again, the big balls are moved to the mirror-image position on the other side. The rod swings from +θ to −θ, a total of 2θ, so the beam sweeps through 4θ. If the spot moves a distance d on a scale a distance D away, $4\theta = d/D$, and

Every quantity on the right can be measured. (k itself comes from timing the rod’s slow twisting oscillations.)

Weighing the Earth#

Once G is known, the Earth’s mass follows from the acceleration of a falling apple: $g = GM/R^2$, so

Near the ground, lifting something by h costs mgh. But g weakens with distance, so for satellites and rockets we need the exact potential energy of two bodies at any separation.

Moving one particle away from another#

Recall from the chapter on energy that for a conservative force, $U_f - U_i = -\int_i^f \vec F\cdot d\vec r$. Hold particle m₁ fixed and move m₂ outward so the separation grows from r to r + dr. Gravity pulls m₂ back toward m₁ with $Gm_1m_2/r^2$, so it does work $dW = -\dfrac{Gm_1m_2}{r^2}dr$, and the potential energy rises by $dU = -dW = \dfrac{Gm_1m_2}{r^2}dr$. From r₁ to r₂:

Where to put zero#

Only differences matter, so we may choose where U = 0. The natural choice is infinite separation, where the particles no longer interact. Put r₁ → ∞ and r₂ = r:

U depends only on the separation, not on which particle moved, so the formula holds generally. For three or more particles, add the potential energy of every pair: N particles form N(N − 1)/2 pairs.

Where mgh comes from#

For a body of mass m lifted from the surface (r = R) to a height h, the exact formula gives

Potential energy belongs to a pair of bodies. Divide out the test mass and you get a property of the place itself: the gravitational potential, a scalar map of how deep each point lies in the well.

Definition#

Equivalently, V at a point is the work per kilogram an external agent must do to bring a mass slowly from the reference point to that point. Because V is a scalar, contributions from different masses simply add, with no directions to worry about.

Calculating V#

A point mass

Using U = −GMm/r with zero at infinity:

A ring, on its axis

Every bit dm of a ring of radius a is the same distance $z = \sqrt{a^2 + r^2}$ from a point P on the axis a distance r from the centre. So each contributes $-G\,dm/z$ and the sum is easy:

A thin spherical shell

Slice a shell of mass M, radius a, into rings by angle θ from the line OP (O the centre, OP = r). The ring between θ and θ + dθ has mass $dm = \tfrac M2\sin\theta\,d\theta$, and all of it is at distance z from P, where $z^2 = a^2 + r^2 - 2ar\cos\theta$. Differentiating, $z\,dz = ar\sin\theta\,d\theta$, so $dm = \dfrac{M}{2ar}z\,dz$, and the ring’s potential at P is

Now integrate over z. The limits are where the cases differ:

• P outside (r > a): z runs from r − a to r + a, a range of 2a. $V = -\dfrac{GM}{2ar}(2a) = -\dfrac{GM}{r}$ (11.9), exactly as if all the mass were at the centre.
• P inside (r < a): z runs from a − r to a + r, a range of 2r. $V = -\dfrac{GM}{2ar}(2r) = -\dfrac{GM}{a}$ (11.10): the same at every point inside.

A solid sphere

Outside, treat it as a stack of concentric shells, each acting as a point at the centre: $V = -GM/r$ (11.11). Inside, at distance r from the centre, split the sphere in two. The inner ball (mass $Mr^3/a^3$) acts like a point at the centre; the outer shells each give their own constant interior potential $-G\,dm/x$, with $dm = 3Mx^2dx/a^3$. Adding up:

At the centre this is $-\tfrac32 GM/a$, one and a half times as deep as at the surface.

How does the Earth pull the Moon across 384 000 km of empty space? The modern answer: it doesn’t reach across. It sets up a gravitational field around itself, and the field acts on the Moon where the Moon is.

A two-step picture#

“Action at a distance” has a problem: if the Sun moved, would distant planets feel it instantly? The field picture splits the interaction in two. A mass creates a field throughout the space around it, a physical thing with energy and momentum of its own. Any other mass placed in that field feels a force from the field at its own location.

Fields from several sources add as vectors. Near the Earth’s surface the field is just $\vec g$: the same size and direction as the acceleration due to gravity, though the two are different quantities.

Field from potential, and back#

Moving a mass m by $d\vec r$, the field does work $m\vec E\cdot d\vec r$, so the potential energy changes by $-m\vec E\cdot d\vec r$ and the potential by

(∂/∂x is a partial derivative: differentiate with respect to x holding y and z fixed.) Know the field, integrate for the potential; know the potential, differentiate for the field.

Fields of common shapes#

Point mass

Ring, on its axis

Each bit dm pulls along the line to itself, at distance $z = \sqrt{a^2+r^2}$. Sideways components cancel round the ring; the components along the axis add, each a fraction $\cos\alpha = r/z$ of the whole:

Disc, on its axis

Build the disc from rings of radius x and mass $2Mx\,dx/a^2$, add their (parallel) fields and substitute $z^2 = r^2 + x^2$:

where θ is the angle a radius of the disc subtends at P.

Thin shell

Repeating the ring-slicing argument for the field gives $GM/r^2$ outside (as if a point) and exactly zero inside. Inside a shell, the near part of the wall pulls harder per kilogram but there is less of it; the far part pulls more weakly but there is more of it. With a 1/r² law the two balance exactly, in every direction.

Solid sphere

Outside: $GM/r^2$ (11.23). Inside at distance r, shells outside r contribute nothing, and the inner ball of mass $Mr^3/a^3$ acts as a point at the centre:

We usually treat g as a constant 9.8 m/s². It isn’t. It changes with height, depth, latitude and even what lies under your feet.

With height#

At a height h above the surface of the Earth (mass M, radius R):

where $g = GM/R^2$ is the surface value. For h much smaller than R, the binomial approximation gives

With depth#

In a mine a depth d below the surface you are inside the Earth. Treating the Earth as uniform, only the ball beneath you pulls (the shell above gives zero field), so from the solid-sphere result:

Other effects#

The Earth spins

In the rotating frame of the Earth, a body at colatitude θ (angle from the north pole) also feels a centrifugal force $m\omega^2R\sin\theta$ outward from the axis. At the poles this vanishes; at the equator it points straight up, opposing gravity:

We met this in the chapter on circular motion: apparent g is least at the equator, greatest at the poles.

The Earth isn’t a sphere

The spin has also flattened the Earth: the equatorial radius is about 21 km larger than the polar radius. Being further from the centre at the equator lowers g there too, reinforcing the rotation effect.

The Earth isn’t uniform

Mountains, ocean basins, dense ore bodies and oil-bearing rock all nudge g locally. Geologists map these tiny variations (gravimetry) to find what lies underground.

Weighing the Earth#

“Weight of the Earth” is meaningless (weight is the Earth’s pull on something), but its mass follows from g:

Throw a stone sideways and it curves to the ground. Throw it fast enough and the ground curves away beneath it as fast as it falls. That endless fall is an orbit.

Circular orbits#

Planetary orbits are ellipses, but most are nearly circular, so circles give excellent estimates. Take a body of mass m in a circle of radius a around a much heavier body M (the Sun for planets, the Earth for satellites). Gravity supplies the centripetal force:

The kinetic energy is $\tfrac12mv^2 = GMm/2a$, the potential energy $-GMm/a$, so the total is

Geostationary satellites

A satellite above the equator, moving the same way as the Earth turns, with a period of one day, stays above the same spot. Dishes can point at it without tracking. Solving $T = 2\pi\sqrt{a^3/GM}$ for a with T = 24 h:

That is about 36 000 km above the surface. Communications and weather satellites live there.

Kepler’s laws, explained#

1. Ellipses. An inverse-square attraction produces conic-section paths. A circle needs exactly the right speed, √(GM/a), aimed exactly sideways; anything else gives an ellipse with the attracting body at a focus (or, fast enough, an escape).
2. Equal areas in equal times. Gravity points straight at the Sun, so it exerts no torque about the Sun, and the planet’s angular momentum L is conserved. In a short time dt the radius sweeps a thin triangle of area $\tfrac12 r\,(v_\perp dt)$, so $dA/dt = L/2m$: constant. Planets speed up near the Sun and slow down far from it.
3. T² ∝ a³. For circles we just derived $T^2 = (4\pi^2/GM)a^3$; the same holds for ellipses with a the semi-major axis. The constant is the same for every planet of the Sun.

Weightlessness#

Gravity at the ISS is still about 89% of its surface value. So why do astronauts float? Consider a body on a surface inside an orbiting satellite. The forces on it are gravity $GMm/r^2$ and the push N from the surface. The body moves with the satellite, accelerating at $GM/r^2$ toward the Earth, so

Throw a ball up and it comes back. Throw it harder and it goes higher. Is there a speed beyond which it never returns? Energy says yes, and it says the speed is the same whatever you throw.

The escape condition#

Launch a body of mass m from the surface of a planet (mass M, radius R) at speed u. Ignoring air, energy is conserved. At height h its speed v satisfies

The last term is positive but shrinks toward zero as h grows. If the bracket is zero or positive, v² stays positive forever: the body never stops, and so never turns back. The threshold is

Binding energy#

A body at rest on the surface has total energy $-GMm/R$. The minimum energy that frees it completely (brings its total to zero) is called the binding energy of the planet–body system:

For a satellite already in a circular orbit of radius a, the binding energy is smaller: $GMm/2a$.

Black holes#

Imagine squeezing a star while keeping its mass fixed. $\sqrt{2GM/R}$ rises as R falls. When it reaches the speed of light c, nothing launched from the surface can escape, and since nothing travels faster than light, nothing escapes at all, not even light. Setting $\sqrt{2GM/R} = c$:

Astronomers have found many black holes: collapsed remnants of massive stars, and giants of millions to billions of solar masses at the centres of galaxies.

Inertial and gravitational mass#

There are two independent ways to compare the masses of two bodies:

• Inertial mass: apply equal forces and compare accelerations, $m_A/m_B = a_B/a_A$. This is the m in F = ma.
• Gravitational mass: compare the gravitational pulls the Earth exerts on them, $m_A/m_B = F_A/F_B$. This is what a spring balance measures.

Nothing in Newton’s theory requires these to agree, yet every experiment finds them equal to extraordinary precision. That is why all bodies fall with the same acceleration. Einstein made this equivalence the foundation of general relativity.