"Gravitation" — Practice

Question 1

A spherical cavity is carved out of a uniform solid sphere, off-centre. Describe the gravitational field inside the cavity and justify the answer.

Accepted Answer

The field is **uniform** — identical in magnitude and direction at every point of the cavity.

Treat the object as a full sphere of density $+ho$ plus a sphere of density $-ho$ filling the cavity. Inside a uniform sphere, $\vec{g}=-	frac43\pi Gho\,\vec{r}$ from its own centre, so at a point $P$:

$\vec{g}=-	frac43\pi Gho\,\vec{r}+	frac43\pi Gho\,\vec{r}\,'=-	frac43\pi Gho\left(\vec{r}-\vec{r}\,'ight)$

$$\vec{g}=-	frac43\pi Gho\,\vec{d}$$

where $\vec{d}$ joins the two centres and is fixed. **The position of $P$ has cancelled entirely.**

*The essential ingredient* is that the interior field of a uniform sphere is linear in $\vec{r}$. Linearity is what allows the two position vectors to subtract into a constant.

Question 2

On the axis of a uniform ring, at what distance from the centre is the gravitational field greatest?

Accepted Answer

$g=\dfrac{GMx}{\left(x^{2}+R^{2}ight)^{3/2}}$

At the centre $x=0$ so $g=0$, by symmetry — every element's pull is cancelled by the one opposite. At infinity $g	o0$ as well. So there is a maximum in between.

Differentiating and setting $dg/dx=0$:

$\left(x^{2}+R^{2}ight)^{3/2}-x\cdot	frac32\left(x^{2}+R^{2}ight)^{1/2}(2x)=0$

$\left(x^{2}+R^{2}ight)-3x^{2}=0 \Rightarrow 2x^{2}=R^{2}$

$$x=\frac{R}{\sqrt2}$$

*Worth noting:* the same geometry and the same answer appear for the electric field on the axis of a charged ring, because both are inverse-square central fields.

Question 3

Two satellites orbit the Earth with the same semi-major axis, one on a circle and one on an ellipse of eccentricity $0.6$. Compare their total energies and their periods.

Accepted Answer

$E=-\dfrac{GMm}{2a}$ and $T=2\pi\sqrt{\dfrac{a^{3}}{GM}}$

**Neither expression contains $e$.** So for equal semi-major axes and equal masses:

$$E_{circle}=E_{ellipse}, \qquad T_{circle}=T_{ellipse}$$

*What does differ* is everything instantaneous. The elliptical satellite ranges between $r_p=0.4a$ and $r_a=1.6a$, and its speed varies by the factor $r_a/r_p=4$ over the orbit. The circular one holds a fixed radius and a fixed speed.

*The physical reading:* eccentricity redistributes energy between kinetic and potential around the orbit, but changes neither the total nor the time to go round.

Question 4

A sphere of radius $R$ and mass $M$ (before hollowing) has a cavity of radius $R/4$ whose centre lies at $3R/4$ from the sphere's centre. Find the field at the cavity's centre.

Accepted Answer

$g=	frac43\pi Gho\,d$ with $d=\dfrac{3R}{4}$

Substituting $ho=\dfrac{3M}{4\pi R^{3}}$:

$g=	frac43\pi G\left(\dfrac{3M}{4\pi R^{3}}ight)\left(\dfrac{3R}{4}ight)=\dfrac{GM}{R^{3}}\cdot\dfrac{3R}{4}$

$$g=\frac{3GM}{4R^{2}}$$

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

*Compare with the surface field* $GM/R^{2}$: the cavity field is $75\%$ of it, which is large. And crucially it is the same at every point of the cavity, not just at its centre.

*Note that the cavity's own radius never entered the answer* — only the offset $d$ matters. A larger cavity at the same offset gives the identical field, though of course it removes more mass.

Question 5

A uniform sphere of mass $M$ contracts under its own gravity from radius $2R$ to radius $R$. Find the energy released.

Accepted Answer

$U=-\dfrac{3GM^{2}}{5R_{sphere}}$

**Initially:** $U_i=-\dfrac{3GM^{2}}{5(2R)}=-\dfrac{3GM^{2}}{10R}$

**Finally:** $U_f=-\dfrac{3GM^{2}}{5R}=-\dfrac{6GM^{2}}{10R}$

$\Delta U=U_f-U_i=-\dfrac{3GM^{2}}{10R}$

$$	ext{Energy released}=\frac{3GM^{2}}{10R}$$

*This is how a protostar shines before fusion.* Gravitational contraction converts potential energy into heat and radiation, and the calculation above is the total budget available from halving the radius. Roughly half becomes internal heat and half is radiated, which is the content of the virial theorem.

Question 6

A satellite at distance $r=8	imes10^{6}\ 	ext{m}$ from the Earth's centre has speed $6\ 	ext{km/s}$. Take $GM=4	imes10^{14}$ SI. Find the semi-major axis of its orbit and say whether it is bound.

Accepted Answer

$v^{2}=GM\left(\dfrac{2}{r}-\dfrac{1}{a}ight)$

$3.6	imes10^{7}=4	imes10^{14}\left(\dfrac{2}{8	imes10^{6}}-\dfrac{1}{a}ight)$

$\dfrac{3.6	imes10^{7}}{4	imes10^{14}}=9.0	imes10^{-8}=2.5	imes10^{-7}-\dfrac1a$

$\dfrac1a=1.6	imes10^{-7} \Rightarrow a=\mathbf{6.25	imes10^{6}\ 	ext{m}}$

**Bound**, since $a$ is positive and finite. Its energy is $E=-\dfrac{GMm}{2a}<0$.

*Note the geometry:* $a<r$, so the satellite is currently **beyond** the semi-major axis and must be near apogee. Indeed $r_a=2a-r_p$ and with $a=6.25	imes10^{6}$, $r_p=2a-r=4.5	imes10^{6}$ m — which is below the Earth's surface, so this orbit would actually intersect the planet.

Question 7

A tunnel is bored straight through the centre of the Earth. Find the period of oscillation of a stone dropped in, and its maximum speed. Take $R=6.4	imes10^{6}$ m and $g=9.8$.

Accepted Answer

Inside a uniform sphere $g=\dfrac{GMr}{R^{3}}=\dfrac{g}{R}r$, directed to the centre, so

$a=-\dfrac{g}{R}x$ — **simple harmonic**, with $\omega=\sqrt{\dfrac{g}{R}}$

$T=2\pi\sqrt{\dfrac{R}{g}}=2\pi\sqrt{\dfrac{6.4	imes10^{6}}{9.8}}=\mathbf{5077\ 	ext{s}}=84.6\ 	ext{min}$

The amplitude is the full radius, so

$v_{max}=\omega R=\sqrt{gR}=\sqrt{9.8\left(6.4	imes10^{6}ight)}=\mathbf{7.92\ 	ext{km/s}}$

*Two things worth noticing.* The maximum speed equals the **orbital speed** of a surface-grazing satellite, and the period equals that satellite's period. A stone dropped through the Earth and a spacecraft skimming the surface arrive at the far side together.

*And the period is the same for any chord*, since the perpendicular offset cancels out of the along-tunnel acceleration.

Question 8

Two stars orbit their common centre of mass with period $T$ and constant separation $d$. Derive the expression for their total mass, and find the individual masses if one star's orbital radius is three times the other's.

Accepted Answer

For star 1, orbiting at radius $r_1=\dfrac{m_2d}{m_1+m_2}$:

$\dfrac{Gm_1m_2}{d^{2}}=m_1\omega^{2}r_1=m_1\omega^{2}\dfrac{m_2d}{m_1+m_2}$

$\dfrac{G}{d^{2}}=\dfrac{\omega^{2}d}{m_1+m_2} \Rightarrow \omega^{2}=\dfrac{G(m_1+m_2)}{d^{3}}$

$$T=2\pi\sqrt{\frac{d^{3}}{G\left(m_1+m_2\right)}} \qquad\Longrightarrow\qquad m_1+m_2=\frac{4\pi^{2}d^{3}}{GT^{2}}$$

**Individual masses.** $\dfrac{r_1}{r_2}=\dfrac{m_2}{m_1}=3$, so $m_2=3m_1$:

$$m_1=\frac{m_1+m_2}{4}=\frac{\pi^{2}d^{3}}{GT^{2}}, \qquad m_2=\frac{3\pi^{2}d^{3}}{GT^{2}}$$

*This is the actual method.* Observing a binary's period and separation gives the total mass; observing the ratio of the two wobble amplitudes splits it. It is the only direct way stellar masses are measured.

Question 9

A satellite at an altitude of $300\ 	ext{km}$ is moving tangentially at $8.5\ 	ext{km/s}$. Take $R=6400\ 	ext{km}$ and $GM=4	imes10^{14}$ SI. Find the apogee altitude and the eccentricity.

Accepted Answer

$r_p=6.7	imes10^{6}\ 	ext{m}$, $v=8500\ 	ext{m/s}$

$v^{2}=7.225	imes10^{7}=4	imes10^{14}\left(\dfrac{2}{6.7	imes10^{6}}-\dfrac1aight)$

$1.806	imes10^{-7}=2.985	imes10^{-7}-\dfrac1a \Rightarrow \dfrac1a=1.179	imes10^{-7}$

$a=8.48	imes10^{6}\ 	ext{m}$

$r_a=2a-r_p=1.696	imes10^{7}-6.7	imes10^{6}=1.026	imes10^{7}\ 	ext{m}$

$$	ext{Apogee altitude}=10260-6400=\mathbf{3860\ 	ext{km}}$$

$e=\dfrac{r_a-r_p}{r_a+r_p}=\dfrac{3560}{16960}=\mathbf{0.21}$

*Sanity check on the speed.* A circular orbit at $6.7	imes10^{6}$ m needs $\sqrt{GM/r}=7.73$ km/s. The satellite has $8.5$ km/s, more than circular but less than the escape value $10.93$ km/s — so an ellipse with perigee here, exactly as assumed.

Question 10

Four particles, each of mass $m$, are held at the corners of a square of side $a$. Find the speed each must be given radially outward so that the system just disperses to infinity.

Accepted Answer

Four particles give $\dfrac{4(3)}{2}=6$ pairs, of two kinds:

| Pairs | Number | Separation |
|---|---|---|
| Sides | 4 | $a$ |
| Diagonals | 2 | $a\sqrt2$ |

$U=-Gm^{2}\left[\dfrac{4}{a}+\dfrac{2}{a\sqrt2}ight]=-\dfrac{Gm^{2}}{a}\left(4+\sqrt2ight)=-\dfrac{5.414\,Gm^{2}}{a}$

'Just disperses' means total energy zero, with all four given the same speed by symmetry:

$4\left(	frac12mv^{2}ight)=\dfrac{5.414\,Gm^{2}}{a}$

$2v^{2}=\dfrac{5.414\,Gm}{a} \Rightarrow v=\mathbf{1.645\sqrt{\dfrac{Gm}{a}}}$

*The two standard errors:* counting 12 pairs by treating AB and BA as distinct, and forgetting that the diagonals are longer by $\sqrt2$ and so contribute less.

Question 11

A body is dropped from rest at a height equal to the Earth's radius above the surface. Find its impact speed, and compare with what $mgh$ would predict. Take $g=9.8$ and $R=6.4	imes10^{6}$ m.

Accepted Answer

It starts at $r=2R$ and lands at $r=R$, with $GM=gR^{2}$.

$-\dfrac{GMm}{2R}=-\dfrac{GMm}{R}+	frac12mv^{2}$

$	frac12v^{2}=\dfrac{GM}{R}-\dfrac{GM}{2R}=\dfrac{GM}{2R}$

$v=\sqrt{\dfrac{GM}{R}}=\sqrt{gR}=\sqrt{9.8\left(6.4	imes10^{6}ight)}=\mathbf{7.92\ 	ext{km/s}}$

**The $mgh$ estimate** would give $v=\sqrt{2gR}=11.2\ 	ext{km/s}$ — too high by $41\%$.

*Why it fails so badly.* Over a fall of one whole radius, $g$ rises from $g/4$ at the start to $g$ at the surface, so assuming the surface value throughout hugely overestimates the work done.

*Note the coincidence worth remembering:* $\sqrt{gR}$ is also the orbital speed of a surface-grazing satellite, and the escape speed $\sqrt{2gR}$ is exactly what $mgh$ wrongly predicted here.

Question 12

A satellite in a circular orbit of radius $r$ is given a tangential boost multiplying its speed by $k$, with $1<k<\sqrt2$. Find the apogee of the new orbit, and evaluate it for a $20\%$ boost.

Accepted Answer

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

$v_{new}^{2}=k^{2}\dfrac{GM}{r}=GM\left(\dfrac{2}{r}-\dfrac1aight)$

$\dfrac{k^{2}}{r}=\dfrac{2}{r}-\dfrac1a \Rightarrow \dfrac1a=\dfrac{2-k^{2}}{r} \Rightarrow a=\dfrac{r}{2-k^{2}}$

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

**For $k=1.2$:** $r_a=\dfrac{1.44r}{2-1.44}=\dfrac{1.44r}{0.56}=\mathbf{2.57\,r}$

*A $20\%$ speed increase raises the far side of the orbit by a factor of $2.57$.* That leverage is why orbital transfers are performed at perigee, where the existing speed is greatest and a given $\Delta v$ buys the most.

*Check the limits.* $k=1$ gives $r_a=r$, a circle. As $k	o\sqrt2$ the denominator vanishes and $r_a	o\infty$ — escape, which is exactly $v_e=\sqrt2\,v_o$.

Practice — "Gravitation"

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