Gravitation

A spherical cavity is hollowed out of a uniform solid sphere, its centre offset from the sphere's centre. What is the gravitational field inside the cavity?

The expected answer is "it varies". It does not. The field is uniform — same magnitude and same direction at every point of the cavity.

Treat the hollowed sphere as a full sphere plus a negative sphere filling the cavity. Inside a uniform solid sphere the field is

${\vec{g}}_{inside}=-\frac{4}{3}\pi G\rho \vec{r}$

measured from that sphere's own centre. Adding the two contributions at a point $P$, with $\vec{r}$ from the big centre $O$ and $\vec{r}{}^{\prime }$ from the cavity centre $C$:

$\vec{g}=-\frac{4}{3}\pi G\rho \vec{r}+\frac{4}{3}\pi G\rho \vec{r}{}^{\prime }=-\frac{4}{3}\pi G\rho \left(\vec{r}-\vec{r}{}^{\prime }\right)=-\frac{4}{3}\pi G\rho \vec{d}$

where $\vec{d}=\overrightarrow{OC}$ is fixed. The position of $P$ has cancelled out entirely.

The whole result came from one idea: superposition, with a negative density standing in for removed matter. That idea, plus the fact that an orbit is completely determined by two conserved quantities, carries most of this chapter.

1. Field and potential of extended bodies

Point masses and spheres are handled by the shell theorem. Everything else needs integration:

$\vec{g}=-G\int \frac{dm}{r^2}\hat{r},V=-G\int \frac{dm}{r}$

Do the potential first whenever you can. $V$ is a scalar, so the integral is ordinary; $\vec{g}$ is a vector needing components. Then recover the field by differentiating, $g_r=-\frac{dV}{dr}$.

On the axis of a ring of mass $M$ and radius $R$, at distance $x$ from the centre, every element is the same distance $\sqrt{x^2+R^2}$ away, so the potential integral is trivial:

$V=-\frac{GM}{\sqrt{x^2+R^2}},g=\frac{GMx}{{\left(x^2+R^2\right)}^{3/2}}$

The field is zero at the centre by symmetry and zero at infinity, so it must peak somewhere between. Differentiating gives the maximum at

$x=\frac{R}{\sqrt{2}}$

Illustration 1

Find the field and potential on the axis of a uniform ring at $x=R$, and compare the field with that of a point mass $M$ at the same distance.

$V=-\frac{GM}{\sqrt{2R^2}}=-\frac{GM}{R\sqrt{2}}=-0.707\frac{GM}{R}$

$g=\frac{GMR}{{\left(2R^2\right)}^{3/2}}=\frac{GMR}{2\sqrt{2}{R}^3}=\frac{GM}{2\sqrt{2}{R}^2}=0.354\frac{GM}{R^2}$

A point mass at distance $R$ would give $\frac{GM}{R^2}$, so the ring produces only about 35% as much.

Why so much weaker: most of the ring's mass is not along the axis, so each element's pull has a sideways component that cancels against the element diametrically opposite. Only the axial components survive, and each is reduced by the factor $\cos \theta =x/\sqrt{x^2+R^2}$.

From a ring to a disc to an infinite sheet

Build a disc out of concentric rings and integrate. With surface density $\sigma$,

$g_{disc}=2\pi G\sigma \left(1-\frac{x}{\sqrt{x^2+R^2}}\right)$

Two limits are worth extracting, because both are standard Advanced answers.

Very close to the disc, or equivalently $R\to \infty$: the bracket tends to 1 and

$g_{sheet}=2\pi G\sigma$

The field of an infinite sheet does not depend on distance at all. Move twice as far away and it is unchanged — the same structure as the electric field of a charged plane, and for the same geometric reason.

Very far away, $x\gg R$: expanding the square root gives $g\to \frac{GM}{x^2}$ with $M=\sigma \pi R^2$, the point-mass result. Any finite body looks like a point from far enough away.

Illustration 2

Find the gravitational field on the axis of a uniform disc of radius $R$ at $x=R$, as a fraction of its value just above the surface.

At $x=R$: $\frac{x}{\sqrt{x^2+R^2}}=\frac{R}{R\sqrt{2}}=0.707$

$g=2\pi G\sigma \left(1-0.707\right)=0.293\left(2\pi G\sigma \right)$

Just above the surface ($x\to 0$) the field is $2\pi G\sigma$, so at one radius out it has fallen to about $29\%$.

Note how fast that is. For a point mass, moving from the surface to one radius above cuts the field to a quarter. The disc is not far off — which is the general lesson that a body's exact shape stops mattering surprisingly quickly with distance.

Illustration 3

A sphere of radius $R$ and density $\rho$ has a spherical cavity of radius $R/2$ carved out, the cavity touching the centre and the surface. Find the field at the cavity's centre.

The cavity's centre is at $d=R/2$ from the sphere's centre.

$g=\frac{4}{3}\pi G\rho d=\frac{4}{3}\pi G\rho \left(\frac{R}{2}\right)=\frac{2\pi G\rho R}{3}$

directed from the cavity's centre toward the sphere's centre.

Express it through the original mass $M=\frac{4}{3}\pi R^3\rho$:

$g=\frac{GM}{2R^2}$

The striking part is what the answer does not contain. Move the test point anywhere else inside the cavity and you get the same vector — same size, same direction. A body released anywhere in the cavity accelerates uniformly, exactly as in a uniform gravitational field.

2. Self-energy: what it costs to assemble a body

The gravitational potential energy of a body with itself is the work needed to bring its matter in from infinity. Build a sphere shell by shell: when a shell of radius $r$ and thickness $dr$ is added onto the mass already present,

$dU=-\frac{G\left(\frac{4}{3}\pi r^3\rho \right)\left(4\pi r^2\rho dr\right)}{r}$

Integrating from $0$ to $R$ and substituting $\rho =\frac{3M}{4\pi R^3}$:

$U_{self}=-\frac{3G{M}^2}{5R}$

The factor $\frac{3}{5}$ is worth memorising, and it is specific to a uniform sphere. A hollow shell gives $-\frac{G{M}^2}{2R}$ instead, because none of its mass had to be pushed to the centre.

Illustration 4

Find the energy needed to disperse the Earth's mass to infinity. Take $M=6\times 10^{24}$ kg, $R=6.4\times 10^6$ m, $G=6.67\times 10^{-11}$.

$U_{self}=-\frac{3G{M}^2}{5R}=-\frac{3(6.67\times 10^{-11})(3.6\times 10^{49})}{5(6.4\times 10^6)}$

$=-\frac{7.20\times 10^{39}}{3.2\times 10^7}=-2.25\times 10^{32}\text{J}$

The energy required is $+2.25\times 10^{32}$ J.

For scale, world energy consumption is around $6\times 10^{20}$ J per year, so this is about 400 billion years of it. This number is also what a self-gravitating gas cloud releases as it collapses into a star, which is why protostars glow before fusion ever begins.

3. Tunnels through a sphere

Inside a uniform sphere only the enclosed mass pulls, so the field grows linearly from the centre:

$g_{inside}=\frac{GMr}{R^3}$

Now bore a straight tunnel along any chord, at perpendicular distance $d$ from the centre. At a point $x$ along the tunnel from its midpoint, the distance from the centre is $r=\sqrt{x^2+d^2}$, and only the component of $\vec{g}$ along the tunnel drives the motion:

$a_{\parallel }=-\frac{GMr}{R^3}\cdot \frac{x}{r}=-\frac{GM}{R^3}x$

The $r$ cancels completely. The acceleration is proportional to the displacement and directed back to the midpoint, which is exactly simple harmonic motion:

$\omega =\sqrt{\frac{GM}{R^3}}=\sqrt{\frac{g}{R}},T=2\pi \sqrt{\frac{R}{g}}\approx 84.6\text{minutes}$

The period does not depend on $d$. A tunnel straight through the centre and a shallow chord near the surface give the same 84.6 minutes — and so does a satellite skimming the surface, since $2\pi \sqrt{R^3/GM}$ is the same expression. A dropped stone keeps pace with an orbiting spacecraft.

Illustration 5

A tunnel is bored along a chord whose perpendicular distance from the Earth's centre is $R/2$. A stone is released at one end. Find the period and the maximum speed. Take $R=6.4\times 10^6$ m, $g=9.8$.

Period, unchanged by the chord's position:

$T=2\pi \sqrt{\frac{6.4\times 10^6}{9.8}}=5077\text{s}=84.6\text{min}$

Amplitude is the half-length of the chord:

$A=\sqrt{R^2-{\left(\frac{R}{2}\right)}^2}=R\frac{\sqrt{3}}{2}=5.54\times 10^6\text{m}$

$\omega =\frac{2\pi }{T}=1.238\times 10^{-3}\text{rad/s}$

$v_{max}=\omega A=1.238\times 10^{-3}\left(5.54\times 10^6\right)=6.86\text{km/s}$

Compare a tunnel through the centre, where $A=R$ and $v_{max}=7.92$ km/s. The shorter chord gives a lower top speed but takes exactly as long, because SHM's period is independent of amplitude.

4. Energy of a system, and escaping from it

Gravitational potential energy belongs to pairs, so a system of $n$ masses has $\frac{n(n-1)}{2}$ terms:

$U=-G{\sum }_{\text{pairs}}\frac{m_i{m}_j}{r_{ij}}$

The work needed to disperse the whole system to infinity is $-U$, and the escape condition for a body from a system is that its total energy reach zero — with $U$ computed against every other mass, not just the nearest.

Two bodies released from rest is the standard Advanced setting, and it needs both conservation laws. Momentum gives $m_1{v}_1=m_2{v}_2$, and energy gives their sum, which combine into a clean statement about the relative speed:

$v_{rel}=\sqrt{2G\left(m_1+m_2\right)\left(\frac{1}{r}-\frac{1}{d}\right)}$

Illustration 6

Two particles of masses $m$ and $2m$ are released from rest a distance $d$ apart. Find their relative speed when the separation has halved, and the speed of each.

$v_{rel}=\sqrt{2G(3m)\left(\frac{2}{d}-\frac{1}{d}\right)}=\sqrt{\frac{6Gm}{d}}$

Momentum conservation (the centre of mass never moves) gives $m{v}_1=2m{v}_2$, so $v_1=2v_2$ and $v_{rel}=v_1+v_2=3v_2$:

$v_2=\frac{1}{3}\sqrt{\frac{6Gm}{d}},v_1=\frac{2}{3}\sqrt{\frac{6Gm}{d}}$

The lighter particle moves twice as fast, as momentum conservation demands, and it therefore carries twice the kinetic energy — since $K=p^2/2m$ and their momenta are equal.

Illustration 7

Three particles, each of mass $m$, are held at the corners of an equilateral triangle of side $a$. Find the speed each must be given, directed radially outward, to just escape the system.

Potential energy of three pairs, all at separation $a$:

$U=-\frac{3G{m}^2}{a}$

By symmetry each particle gets the same speed $v$, and "just escapes" means total energy zero:

$3\left(\frac{1}{2}m{v}^2\right)-\frac{3G{m}^2}{a}=0$

$v=\sqrt{\frac{2Gm}{a}}$

The trap to avoid: using only the nearest neighbour, which would give $\sqrt{Gm/a}$ and is out by $\sqrt{2}$. Escape is from the whole system, so every pair must appear in $U$.

5. Elliptical orbits: two numbers fix everything

An orbit is settled by two conserved quantities: energy and angular momentum. Everything else follows.

$E=-\frac{GMm}{2a},L=m\sqrt{GMa\left(1-e^2\right)}$

The energy depends only on the semi-major axis — not on the eccentricity at all. A circular orbit of radius $a$ and a wildly elongated ellipse of the same $a$ have identical energies and identical periods.

From $E=\frac{1}{2}m{v}^2-\frac{GMm}{r}$ and the expression above comes the single most useful orbital formula:

$\boxed{v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)}$

the vis-viva equation. Setting $r=a$ recovers the circular speed; setting $a\to \infty$ recovers the escape speed.

At the two apses the velocity is perpendicular to the radius, so $L=mvr$ directly:

$v_p{r}_p=v_a{r}_a,r_p=a(1-e),r_a=a(1+e)$

Illustration 8

A satellite's orbit has perigee $7000\text{km}$ and apogee $21000\text{km}$ from the Earth's centre. Take $GM=4\times 10^{14}$ SI. Find the eccentricity and both apsidal speeds.

$a=\frac{7000+21000}{2}=14000\text{km}=1.4\times 10^7\text{m}$

$e=\frac{r_a-r_p}{r_a+r_p}=\frac{14000}{28000}=0.5$

At perigee, using vis-viva with $r=7\times 10^6$:

$v_p^2=4\times 10^{14}\left(\frac{2}{7\times 10^6}-\frac{1}{1.4\times 10^7}\right)=4\times 10^{14}\left(2.857-0.714\right)\times 10^{-7}$

$v_p^2=8.57\times 10^7\Rightarrow v_p=9.26\text{km/s}$

At apogee, from angular momentum rather than repeating the algebra:

$v_a=v_p\frac{r_p}{r_a}=9.26\left(\frac{1}{3}\right)=3.09\text{km/s}$

Check with vis-viva at $r=2.1\times 10^7$: $v_a^2=4\times 10^{14}(0.952-0.714)\times 10^{-7}=9.52\times 10^6$, giving $3.09$ km/s. The two routes agree.

Illustration 9

A satellite in a circular orbit of radius $r$ is given a sudden tangential boost that raises its speed by a factor $k$ (with $1<k<\sqrt{2}$). Find the apogee of the new orbit.

The boost happens at $r$, which becomes the perigee of the new ellipse.

$v_{new}^2=k^2\frac{GM}{r}$

From vis-viva at $r$:

$k^2\frac{GM}{r}=GM\left(\frac{2}{r}-\frac{1}{a}\right)\Rightarrow \frac{1}{a}=\frac{2-k^2}{r}$

$a=\frac{r}{2-k^2},r_a=2a-r=\frac{k^{2}r}{2-k^2}$

Read the limits. At $k=1$ this gives $r_a=r$, a circle. As $k\to \sqrt{2}$ the denominator vanishes and $r_a\to \infty$ — the satellite escapes, which is exactly the condition $v_e=\sqrt{2}{v}_o$.

A $10\%$ boost, $k=1.1$: $r_a=\frac{1.21r}{0.79}=1.53r$. A tenth more speed buys half again the altitude at the far side, which is why orbital manoeuvres are so fuel-efficient when made at perigee.

6. Kepler's second law is angular momentum

Gravity is a central force — always along the line to the centre — so it exerts no torque about that point and $\vec{L}$ is conserved. The areal velocity follows immediately:

$\frac{dA}{dt}=\frac{L}{2m}=\text{constant}$

That is Kepler's second law, and the derivation is two lines rather than an empirical observation.

A useful consequence: since $L$ is fixed, a planet's speed and its distance are inversely related only at the apses, where $\vec{v}\perp \vec{r}$. Elsewhere $L=mvr\sin \theta$ and the angle matters.

Illustration 10

A comet moves in an orbit of eccentricity $0.8$. Find the ratio of its speed at perihelion to that at aphelion, and the ratio of the times it spends in the half-orbit nearer the Sun to the half further away.

$\frac{r_a}{r_p}=\frac{1+e}{1-e}=\frac{1.8}{0.2}=9$

$\frac{v_p}{v_a}=\frac{r_a}{r_p}=9$

For the times, use equal areas in equal times. The chord through the focus perpendicular to the major axis divides the ellipse into two unequal areas, and the comet spends time in proportion to the area swept.

Qualitatively but decisively: the far half is much the larger area, so the comet spends the overwhelming majority of its period out there and races through perihelion in a small fraction of it. This is why a comet is visible for weeks out of an orbit lasting decades.

7. Binary systems

Two comparable masses orbit their common centre of mass, which stays fixed.

$r_1=\frac{m_{2}d}{m_1+m_2},r_2=\frac{m_{1}d}{m_1+m_2}$

Both stars share one period, and they are always diametrically opposite the centre of mass. Applying gravity as the centripetal force to either one gives

$T=2\pi \sqrt{\frac{d^3}{G\left(m_1+m_2\right)}}$

This is Kepler's third law with the total mass in it. For a planet orbiting the Sun, $m_1+m_2\approx M_{Sun}$ and the familiar form is recovered. For two comparable stars the correction is large, and it is precisely how the masses of binary stars are measured.

Illustration 11

Two stars of masses $2M_s$ and $6M_s$ are separated by $d$. Find the ratio of their orbital radii and of their speeds, and where the more massive one sits.

$\frac{r_1}{r_2}=\frac{m_2}{m_1}=\frac{6}{2}=3$

The lighter star, $2M_s$, moves on the larger circle — three times the radius of the heavier one's.

Since both complete an orbit in the same time, $v=\frac{2\pi r}{T}$ gives

$\frac{v_1}{v_2}=\frac{r_1}{r_2}=3$

The observational consequence. A massive star with a light companion barely moves, tracing a tiny circle. Astronomers detect exoplanets exactly this way — by the small periodic wobble of the star about the common centre of mass, which is the only visible sign of an unseen partner.

8. Manoeuvres: raising an orbit and escaping from one

The energy of a circular orbit is $E=-\frac{GMm}{2r}$, so moving to a higher orbit requires energy even though the satellite ends up slower.

$\Delta E=\frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

Escaping from an orbit rather than from the surface is much cheaper, because half the work is already done:

$\Delta v=v_o\left(\sqrt{2}-1\right)=0.414v_o$

Illustration 12

A satellite of mass $500\text{kg}$ orbits at $r=2R$ where $R=6400$ km, with $GM=4\times 10^{14}$. Find the extra energy needed to escape, and the extra speed.

$r=1.28\times 10^7$ m.

$v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{4\times 10^{14}}{1.28\times 10^7}}=5590\text{m/s}$

$E=-\frac{GMm}{2r}=-\frac{4\times 10^{14}(500)}{2(1.28\times 10^7)}=-7.81\times 10^9\text{J}$

To reach $E=0$ the energy required is $7.81\times 10^9\text{J}$, which is exactly the satellite's own kinetic energy.

$\Delta v=0.414(5590)=2315\text{m/s}$

The general result worth keeping: for any circular orbit, the energy needed to escape equals the kinetic energy the satellite already has, and the extra speed is $41.4\%$ of the orbital speed.

Summary

• Superposition with negative density handles any cavity. The field inside a spherical cavity is uniform, $\frac{4}{3}\pi G\rho d$, independent of position.
• Compute $V$ before $\vec{g}$: the potential is a scalar integral, and $g_r=-dV/dr$ recovers the field.
• Ring on axis: $V=-\frac{GM}{\sqrt{x^2+R^2}}$, $g=\frac{GMx}{(x^2+R^2{)}^{3/2}}$, peaking at $x=R/\sqrt{2}$.
• Self-energy of a uniform sphere: $-\frac{3G{M}^2}{5R}$; of a shell, $-\frac{G{M}^2}{2R}$.
• Disc on axis: $2\pi G\sigma \left(1-\frac{x}{\sqrt{x^2+R^2}}\right)$, tending to $2\pi G\sigma$ for an infinite sheet — independent of distance.
• A straight tunnel along any chord gives SHM with $T=2\pi \sqrt{R/g}=84.6$ min, independent of the chord.
• System energy counts pairs: $n(n-1)/2$ terms. Escape is from the whole system, not the nearest neighbour.
• Two bodies released from rest: $v_{rel}=\sqrt{2G(m_1+m_2)\left(\frac{1}{r}-\frac{1}{d}\right)}$.
• $E=-\frac{GMm}{2a}$ depends only on the semi-major axis, not on eccentricity. So does the period.
• Vis-viva: $v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)$ — the one formula that covers circular, elliptical and escape cases.
• $r_p=a(1-e)$, $r_a=a(1+e)$, $e=\frac{r_a-r_p}{r_a+r_p}$, and $v_p{r}_p=v_a{r}_a$.
• Kepler's second law is angular momentum conservation: $dA/dt=L/2m$.
• Binary: both stars share one period, $T=2\pi \sqrt{\frac{d^3}{G(m_1+m_2)}}$, and the heavier star moves on the smaller circle.
• Raising an orbit costs energy while lowering the speed; escaping from a circular orbit needs $\Delta v=0.414v_o$ and an energy equal to the orbital kinetic energy.