On a clear winter evening, look for a small, tight group of stars to the north-west of Orion. Indian astronomers called it Krittika; it is also known as the Pleiades. A telescope shows about a thousand stars in it, all moving together through space. Each star pulls every other star. How do we add up the energy that holds such a crowd together?
Building a system one mass at a time
The last lesson gave the energy of two masses. For more than two, build the system step by step. Start with every mass infinitely far from every other, where \(U = 0\). Then bring the masses closer, one at a time, and add the potential energy that each arrival creates.
The first mass arrives in empty space, so nothing pulls it, and the energy stays zero. The second mass is pulled only by the first, so the pair gains \(-Gm_1m_2/r_{12}\), where \(r_{12}\) is their separation. The third mass is pulled by both. Forces add as vectors, so the works add, and the third arrival contributes \(-Gm_1m_3/r_{13} - Gm_2m_3/r_{23}\). This rule, that the effects of several masses simply add, is called superposition.
Each new mass adds one term for every mass already in place. So when all have arrived, the system holds exactly one term for every pair. The order of arrival does not matter: a pair contributes the same term whichever of its two masses came first.
The gravitational potential energy of a system of point masses is the sum of \(-Gm_im_j/r_{ij}\) over every pair. Here \(r_{ij}\) is the distance between masses \(i\) and \(j\). Each pair is counted once, and \(U = 0\) when all the masses are infinitely far apart.
Counting the pairs
Counting pairs is the same problem as counting handshakes. In a room of \(n\) people who each shake hands with everyone else once, each person shakes \(n - 1\) hands. That counts every handshake twice, once from each end. So there are \(n(n - 1)/2\) handshakes: 1 for two people, 3 for three, 6 for four.
With all four masses in Fig 2.1, four pairs lie along the sides, at a distance \(a\). Two lie along the diagonals, at \(a\sqrt2\). So \(U = -4Gm^2/a - 2Gm^2/(a\sqrt2) = -(4 + \sqrt2)\,Gm^2/a\). That is \(-5.41\,Gm^2/a\). The diagonals add less than the sides, because they are longer.
Some books write the sum over all \(i\) and \(j\) with \(i \ne j\), and put a factor \(\tfrac12\) in front. That sum meets every pair twice, as \(ij\) and as \(ji\), and the half removes the double count. Summing over pairs, as here, needs no half. Do not use both at once.
Masses of \(2\,\text{kg}\), \(4\,\text{kg}\) and \(2\,\text{kg}\) sit on the \(x\) axis at \(x = 0\), \(2\,\text{m}\) and \(3\,\text{m}\). Find the potential energy of the system.
(a) The pairs are (1, 2) at \(2\,\text{m}\), (1, 3) at \(3\,\text{m}\) and (2, 3) at \(1\,\text{m}\): three pairs for three masses.
(b) \(U = -G\left(\dfrac{2 \times 4}{2} + \dfrac{2 \times 2}{3} + \dfrac{4 \times 2}{1}\right) = -G(4 + 1.33 + 8) = -13.3\,G\).
(c) With \(G = 6.67 \times 10^{-11}\), \(U = -8.9 \times 10^{-10}\,\text{J}\). The closest pair, 2 and 3, holds most of it.
Four masses are in place. A fifth mass is brought from very far away. How many new terms does its arrival add to the potential energy?
Four \(2\,\text{kg}\) masses sit at the corners of a square of side \(1\,\text{m}\). How much work is needed to separate them all to infinity, in units of \(10^{-10}\,\text{J}\)? Take \(G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\).
When the masses move
Mutual gravity is a conservative force, so a system of masses that move only under their own pull keeps its total energy. The kinetic energy of all the masses plus the sum over all pairs stays constant. As the masses fall together, the pair terms grow more negative, and the kinetic energy grows by exactly as much.
Three equal masses \(m\) start at rest at the corners of an equilateral triangle of side \(r\). They move under their mutual pull alone. Find the speed of each when the side has shrunk to \(r/2\).
System: all three masses. No outside force acts, so \(K + U\) is constant. There are three pairs, all of equal length.
\(U_i = -3Gm^2/r\) and \(U_f = -3Gm^2/(r/2) = -6Gm^2/r\). So \(K_f = U_i - U_f = 3Gm^2/r\).
By symmetry the three masses move alike, so each has \(\tfrac12 mv^2 = Gm^2/r\), and \(v = \sqrt{2Gm/r}\).
Each mass moves at \(\sqrt{2Gm/r}\), straight towards the centre of the triangle. Their momenta, equal in size and \(120°\) apart, add to zero, as they must.
Often some masses are held fixed while one moves, such as a spacecraft passing between planets that barely move. The pair terms among the fixed masses do not change. They appear on both sides of the energy account and cancel. So only the terms that involve the moving mass need to be written. The next lesson turns that idea into the potential of a point.
Two heavy masses \(M_1\) and \(M_2\) are held fixed. A small mass \(m\) moves from a point near them to another point. Which potential energy terms must enter the energy equation for \(m\)?
Masses of \(1\,\text{kg}\), \(2\,\text{kg}\) and \(3\,\text{kg}\) sit at the corners of an equilateral triangle of side \(1\,\text{m}\). How much work is needed to separate them to infinity, in units of \(10^{-10}\,\text{J}\)?
The energy inside one body
A planet or a star is itself a crowd of masses, every grain pulling every other. The sum over all these pairs is the body's self-energy. It is the energy released when the body is assembled from matter spread out at infinity. It is also minus the work needed to scatter the body again. For a continuous body, the sum over pairs becomes an integral.
Start with a thin spherical shell of mass \(M\) and radius \(R\). Build it in small pieces \(dm\) brought from far away. Suppose the shell already holds a mass \(\mu\). Every part of that mass is at the same distance \(R\) from the arriving piece, which lands on the shell. So the arrival adds \(-G\mu\,dm/R\). Adding \(\mu\) from \(0\) to \(M\) gives \(U = -GM^2/2R\).
A uniform solid sphere is built in shells, from the centre outward. The chapter on Gravitational Force and Field showed that a uniform sphere pulls an outside mass as if its mass sat at the centre. So a new shell lands on a sphere that acts as a point at its centre.
\[ m(r) = M\,\dfrac{r^3}{R^3} \]
Why. the density is uniform, so the mass already built grows as the volume, \(r^3\).
\[ dm = \dfrac{3Mr^2}{R^3}\,dr \]
Why. differentiate \(m(r)\). This is the mass of the new shell of thickness \(dr\).
\[ dU = -\dfrac{G\,m(r)\,dm}{r} = -\dfrac{3GM^2 r^4}{R^6}\,dr \]
Why. the sphere of radius \(r\) pulls the arriving shell as a point mass at the centre, a distance \(r\) away.
\[ U = -\dfrac{3GM^2}{R^6}\int_0^R r^4\,dr = -\dfrac{3GM^2}{5R} \]
Read it. add the shells from \(r = 0\) to \(r = R\). The integral of \(r^4\) is \(R^5/5\).
The solid sphere is more tightly bound than a shell of the same mass and radius, \(3/5\) against \(1/2\). In the sphere much of the mass sits closer together, and closer pairs have larger terms. For the Earth, with \(M = 6.0 \times 10^{24}\,\text{kg}\) and \(R = 6.4 \times 10^6\,\text{m}\), the formula gives about \(-2.3 \times 10^{32}\,\text{J}\), treating the Earth as uniform.
A thin shell and a uniform solid sphere have the same mass and the same radius. Which needs more energy to scatter completely to infinity?
The Sun has a mass of about \(2 \times 10^{30}\,\text{kg}\) and a radius of about \(7 \times 10^8\,\text{m}\). It radiates about \(3.8 \times 10^{26}\,\text{W}\). If its only supply were its gravitational self-energy, for how many years could it shine?
In 1854 the German physicist Hermann von Helmholtz asked what keeps the Sun hot. Burning coal would last only a few thousand years. He proposed that the Sun slowly shrinks, turning its gravitational self-energy into heat. In 1862 the British physicist Lord Kelvin used the same idea to limit the age of the Sun to some tens of millions of years. Geologists and Charles Darwin needed far longer for the rocks and for evolution. The quarrel lasted for decades. It ended in 1939, when Hans Bethe showed that nuclear fusion powers the Sun. The self-energy calculation was right; the Sun simply had a bigger store.
Return to the Krittika cluster. For about a thousand stars there are \(1000 \times 999/2\), roughly half a million, pairs. The energy that holds the cluster is the sum of half a million terms \(-Gm_im_j/r_{ij}\). Astronomers estimate such sums from the typical mass and spacing of the stars, and they find it negative. So the cluster is bound, and its stars move together.
- Counting each pair twice.Sum over pairs, \(n(n - 1)/2\) of them. A sum over all \(i \ne j\) needs a factor \(\tfrac12\).
- Leaving out the diagonal pairs of a square.List every pair. Four masses give six pairs: four sides and two diagonals.
- Writing the self-energy of a solid sphere as \(-GM^2/2R\).That is the thin shell. The solid sphere has \(-3GM^2/5R\).