Chapter 7 · Gravitation

7.1 Kepler's Laws and the Universal Law of Gravitation

Planets move on ellipses, sweep out equal areas in equal times and obey $T^2\propto a^3$. One law explains all three: every pair of masses attracts with $F=\dfrac{Gm_1m_2}{r^2}$.

Watch the night sky for a few months and most stars keep their places, wheeling round together as the Earth spins. A handful of bright points do not: Mars, Jupiter and Saturn drift slowly against the background stars, and every so often one of them stops, moves backwards for a few weeks, and then carries on. For two thousand years astronomers tried to explain these wanderings with circles upon circles. The answer, when it came, was far simpler — and it turned out to be the same rule that makes an apple fall.

In this section you will meet Kepler's three laws, which describe how planets move, and Newton's universal law of gravitation, which explains why. Newton showed that one attractive force, weakening as the inverse square of the distance, accounts for the shape of every orbit, the changing speed of a planet along it, and the exact link between the size of an orbit and the time taken to go round it. With the measured value of the gravitational constant $G$, the same law lets you "weigh" the Sun and the planets from nothing more than the sizes and periods of their orbits.

Kepler's three laws of planetary motion

Ancient astronomers, including Ptolemy, placed the Earth at the centre of the universe. In the fifth century Aryabhata argued that the daily turning of the sky is really the Earth spinning on its axis, and in 1543 Copernicus put the Sun at the centre with the planets going round it. At the end of the sixteenth century Tycho Brahe, working without a telescope, recorded the positions of the planets more accurately than anyone before him. His assistant Johannes Kepler spent years fitting these numbers and found that no combination of circles would do: the orbit of Mars is an ellipse.

📘 Definition 7.1 — Ellipse, semi-major axis and eccentricity

An ellipse is the set of points $P$ for which the sum of the distances to two fixed points $F_1$ and $F_2$, the foci, is constant: $PF_1+PF_2=2a$. Half the longest diameter, $a$, is the semi-major axis; half the shortest, $b$, is the semi-minor axis. If the foci are a distance $c$ from the centre,

$$b^2=a^2-c^2,\qquad e=\frac{c}{a}\quad(0\le e\lt 1),$$

where $e$ is the eccentricity: $e=0$ is a circle, and $e$ close to $1$ is a long, thin ellipse. For a planet with the Sun at a focus, the nearest point of the orbit (perihelion) and the farthest (aphelion) are at distances

$$r_p=a(1-e),\qquad r_a=a(1+e),\qquad a=\frac{r_p+r_a}{2}.$$
📐 Law 7.1 — Kepler's first law (law of orbits)

Every planet moves in an ellipse with the Sun at one focus. (The other focus is empty.)

The planets' ellipses are only slightly squashed — the Earth's eccentricity is $0.017$, so a scale drawing of its orbit looks like a circle — but the Sun is noticeably off-centre. Comets, by contrast, often move on very elongated ellipses.

📐 Law 7.2 — Kepler's second law (law of areas)

The line joining the Sun to a planet sweeps out equal areas in equal intervals of time. In other words, the areal velocity $\dfrac{\Delta A}{\Delta t}$ of the planet is constant.

Near perihelion the line from the Sun is short, so to sweep the same area each day the planet must move faster; near aphelion it moves slowest. Why should the area rate be constant? In a short time $\Delta t$ the planet moves $v\,\Delta t$, and the swept region is a thin triangle with base $r$ and height $v_\perp\Delta t$, where $v_\perp$ is the part of the velocity perpendicular to the Sun–planet line. So

$$\frac{\Delta A}{\Delta t}=\tfrac12 r\,v_\perp=\frac{L}{2m},$$

where $L=mrv_\perp$ is the planet's angular momentum about the Sun (Chapter 6). The Sun's pull always points along the Sun–planet line, so it exerts no torque about the Sun and $L$ is conserved. Kepler's second law is conservation of angular momentum, and it holds for any force directed towards a fixed centre (a central force). At perihelion and aphelion the velocity is perpendicular to the radius, which gives a very useful equation:

$$v_p\,r_p=v_a\,r_a .$$
📐 Law 7.3 — Kepler's third law (law of periods)

The square of a planet's period of revolution $T$ is proportional to the cube of the semi-major axis $a$ of its orbit:

$$T^2\propto a^3,\qquad\text{i.e.}\quad \frac{T^2}{a^3}\ \text{is the same for every planet of the Sun.}$$

Measuring $T$ in years and $a$ in astronomical units ($1\ \text{AU}=1.496\times10^{11}\ \text{m}$, the Earth's semi-major axis) makes the constant $1$ for the Sun's planets: $T^2=a^3$.

✏️ Example 1 — Periods and orbit sizes
(a) Mars has a semi-major axis of $1.524\ \text{AU}$. Find the length of its year. (b) A comet returns every $76$ years. Find the semi-major axis of its orbit. (c) The comet's perihelion distance is $0.59\ \text{AU}$. Find its aphelion distance and the eccentricity of its orbit.
(a) In years and AU, $T^2=a^3$: $T=a^{3/2}=1.524^{1.5}=1.88\ \text{yr}$ — about 687 days.
(b) $a=T^{2/3}=76^{2/3}=17.9\ \text{AU}$.
(c) $r_a=2a-r_p=2(17.94)-0.59=35.3\ \text{AU}$ (beyond Neptune). $e=\dfrac{r_a-r_p}{r_a+r_p}=\dfrac{34.7}{35.9}=0.97$.
Check $r_p=a(1-e)=17.94\times(1-0.967)=0.59\ \text{AU}$ ✓. Keep the unrounded $a$ in (c): subtracting nearly equal numbers magnifies rounding errors.
✏️ Example 2 — Fastest at perihelion
The Earth is $1.471\times10^{11}\ \text{m}$ from the Sun at perihelion (early January) and $1.521\times10^{11}\ \text{m}$ at aphelion (early July). Its speed at perihelion is $30.29\ \text{km/s}$. (a) Find its speed at aphelion. (b) Find the eccentricity of the Earth's orbit. (c) For the comet of Example 1, how many times faster does it move at perihelion than at aphelion?
Idea At both ends of the major axis the velocity is perpendicular to the radius, so the law of areas gives $v_pr_p=v_ar_a$.
(a) $v_a=v_p\dfrac{r_p}{r_a}=30.29\times\dfrac{1.471}{1.521}=29.29\ \text{km/s}$.
(b) $e=\dfrac{r_a-r_p}{r_a+r_p}=\dfrac{0.050}{2.992}=0.0167$ — very nearly a circle.
(c) $\dfrac{v_p}{v_a}=\dfrac{r_a}{r_p}=\dfrac{35.3}{0.59}\approx60$. The comet crawls through the outer Solar System and whips round the Sun.
Check In general $\dfrac{v_p}{v_a}=\dfrac{1+e}{1-e}$; for the Earth $\dfrac{1.0167}{0.9833}=1.034$ and $30.29/29.29=1.034$ ✓.
Interactive 1 — Kepler's Orbit Lab: Equal Areas in Equal Times

A planet moves on an ellipse with the Sun at the right-hand focus. Its position at each moment is computed from Kepler's laws. The orbit period is cut into 12 equal time intervals and the region swept in each is shaded: the slices near perihelion are short and fat, those near aphelion long and thin — yet every slice has the same area. Raise the eccentricity and watch the speed change along the orbit.

0.60
distance $r/a$ now—
speed ÷ aphelion speed—
area of each slice—
$v_p:v_a$—

Newton's universal law of gravitation

Kepler's laws describe the orbits; they do not say what holds the planets on them. Newton's insight was that the force which pulls an apple to the ground also reaches out to the Moon. In Section 3.4 you found that the Moon, $3.84\times10^8\ \text{m}$ away and going round once in $27.3$ days, accelerates towards the Earth at about $2.72\times10^{-3}\ \text{m/s}^2$. That is $\dfrac{9.8}{2.72\times10^{-3}}\approx3600=60^2$ times smaller than $g$ — and the Moon is about $60$ Earth radii away. A force that weakens as $\dfrac1{r^2}$ fits exactly. Newton then showed that an inverse-square attraction towards the Sun produces elliptical orbits obeying all three of Kepler's laws, and boldly proposed that every pair of bodies in the universe attracts.

📐 Law 7.4 — Newton's universal law of gravitation

Every particle of matter attracts every other particle. For two particles of masses $m_1$ and $m_2$ a distance $r$ apart, the force on each has magnitude

$$F=\frac{Gm_1m_2}{r^2},$$

and is directed along the line joining them, towards the other particle. $G$ is the universal gravitational constant, $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$. In vector form, if $\hat{\mathbf{r}}$ is the unit vector from $m_1$ to $m_2$, the force on $m_2$ is

$$\mathbf{F}_{21}=-\frac{Gm_1m_2}{r^2}\,\hat{\mathbf{r}},$$

and the force on $m_1$ is $\mathbf{F}_{12}=-\mathbf{F}_{21}$ — an action–reaction pair (Newton's third law).

Three features matter in every problem. The force is always attractive. It obeys the inverse-square law: double the separation and the force falls to a quarter. And it is universal: it does not depend on what the bodies are made of, or on anything between them. The law is stated for particles; for real bodies we need one more result, which Newton proved with the calculus he invented for the purpose.

📐 Result 7.5 — Gravitation of a uniform spherical shell and a sphere

(i) A uniform thin spherical shell of mass $M$ attracts a particle outside it as if all of $M$ were concentrated at the centre of the shell: $F=\dfrac{GMm}{r^2}$, with $r$ measured from the centre.

(ii) On a particle anywhere inside the shell the net gravitational force due to the shell is zero.

(iii) Hence a solid sphere whose density depends only on the distance from its centre (the Earth, to a good approximation) attracts any particle outside it as if all its mass were at its centre.

When several bodies act on one particle, each attracts it as though the others were not there, and the net force is the vector sum of the separate forces — the principle of superposition:

$$\mathbf{F}=\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3+\cdots$$
✏️ Example 3 — How strong is gravity?
(a) Two $60\ \text{kg}$ students sit with their centres $1.0\ \text{m}$ apart. Find the gravitational force between them and compare it with the weight of one student. (b) Find the gravitational force between the Earth ($M_E=5.97\times10^{24}\ \text{kg}$) and the Moon ($7.35\times10^{22}\ \text{kg}$), whose centres are $3.84\times10^8\ \text{m}$ apart.
(a) $F=\dfrac{(6.67\times10^{-11})(60)(60)}{1.0^2}=2.4\times10^{-7}\ \text{N}$. The weight is $60\times9.8=588\ \text{N}$, about $2.4\times10^{9}$ times larger.
(b) $F=\dfrac{(6.67\times10^{-11})(5.97\times10^{24})(7.35\times10^{22})}{(3.84\times10^8)^2}=\dfrac{2.927\times10^{37}}{1.475\times10^{17}}=1.98\times10^{20}\ \text{N}$.
Meaning Between everyday objects gravity is far too weak to notice; it dominates only when at least one mass is astronomically large. The Earth pulls the Moon — and the Moon pulls the Earth — with $2\times10^{20}\ \text{N}$.
Check Dividing the force in (b) by the Moon's mass gives $2.7\times10^{-3}\ \text{m/s}^2$, the Moon's centripetal acceleration quoted above ✓.
✏️ Example 4 — Adding gravitational forces
Two $5.0\ \text{kg}$ spheres A and B are fixed with their centres at $(0,\,0)$ and $(0.40\ \text{m},\,0)$. A $1.0\ \text{kg}$ sphere C is placed with its centre at $(0.20\ \text{m},\,0.15\ \text{m})$. (a) Find the net gravitational force on C due to A and B. (b) What is the net force on C if it is moved to $(0.20\ \text{m},\,0)$?
Distances C is $\sqrt{0.20^2+0.15^2}=0.25\ \text{m}$ from each of A and B (a 3–4–5 triangle).
Each force $F_A=F_B=\dfrac{(6.67\times10^{-11})(5.0)(1.0)}{0.25^2}=5.34\times10^{-9}\ \text{N}$, along CA and CB.
(a) By symmetry the $x$-components cancel. Each $y$-component is $F_A\times\dfrac{0.15}{0.25}=0.60F_A$, pointing towards the line AB. Net force $=2(0.60)(5.34\times10^{-9})=6.4\times10^{-9}\ \text{N}$, in the $-y$ direction.
(b) At the midpoint the two pulls are equal and opposite: net force zero.
Check Vector form: $\mathbf{F}_A\propto(-0.20,-0.15)$ and $\mathbf{F}_B\propto(+0.20,-0.15)$; their sum $\propto(0,-0.30)$, i.e. $(0,-0.30)\times\dfrac{5.34\times10^{-9}}{0.25}=(0,\,-6.4\times10^{-9})\ \text{N}$ ✓.

The gravitational constant $G$

Newton could not find $G$: he knew the product $GM$ for the Sun and the Earth from orbits, but not $G$ or the masses separately. Measuring $G$ means detecting the tiny pull between laboratory masses, which Henry Cavendish first did in 1798 with a torsion balance. A light rod carrying a small lead sphere at each end hangs from a very fine wire. Two large lead spheres are brought up close to the small ones, on opposite sides, so that their attractions form a couple and twist the rod. The wire twists until its restoring torque balances the gravitational torque. Calibrating the wire (from the rod's slow oscillations) gives the torque per radian of twist, so measuring the tiny angle gives the force $F$, and then $G=\dfrac{Fr^2}{Mm}$.

Modern measurements give $G=6.674\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$; in this course we use $6.67\times10^{-11}$. Its dimensional formula is $[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$, and its unit can also be written $\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$. Once $G$ was known, the mass of the Earth followed at once from $g$ and $R_E$ — you will do this in Section 7.2 — which is why Cavendish's experiment is often called "weighing the Earth".

Kepler's third law from Newton's law

Most planetary orbits are nearly circular, so take a planet of mass $m$ moving uniformly round a circle of radius $r$ about the Sun (mass $M\gg m$). Gravity provides the centripetal force (Section 3.4):

$$\frac{GMm}{r^2}=\frac{mv^2}{r}\quad\Longrightarrow\quad v=\sqrt{\frac{GM}{r}} .$$

One revolution takes $T=\dfrac{2\pi r}{v}$, so $T^2=\dfrac{4\pi^2r^2}{v^2}=\dfrac{4\pi^2r^2\cdot r}{GM}$.

📐 Result 7.6 — Kepler's third law from the law of gravitation

A body moving in a circular orbit of radius $r$ about a much more massive body $M$ has period

$$T^2=\frac{4\pi^2}{GM}\,r^3 .$$

The constant $\dfrac{4\pi^2}{GM}$ depends only on the central mass, not on the orbiting body. For an elliptical orbit the same formula holds with $r$ replaced by the semi-major axis $a$ (proved in higher courses).

Result 7.6 is how astronomers measure masses in space: time a moon, a planet or a star going round, measure the orbit's size, and solve for $M=\dfrac{4\pi^2r^3}{GT^2}$.

✏️ Example 5 — Weighing the Sun and the Earth
(a) The Earth orbits the Sun at an average distance of $1.50\times10^{11}\ \text{m}$ once every $365.25$ days. Find the mass of the Sun. (b) The Moon orbits the Earth at $3.84\times10^8\ \text{m}$ once every $27.3$ days. Estimate the mass of the Earth.
Setup $M=\dfrac{4\pi^2r^3}{GT^2}$, with $T$ in seconds.
(a) $T=365.25\times86\,400=3.156\times10^7\ \text{s}$. $M_S=\dfrac{4\pi^2(1.50\times10^{11})^3}{(6.67\times10^{-11})(3.156\times10^7)^2}=\dfrac{1.332\times10^{35}}{6.64\times10^{4}}=2.0\times10^{30}\ \text{kg}$.
(b) $T=27.3\times86\,400=2.359\times10^6\ \text{s}$. $M_E=\dfrac{4\pi^2(3.84\times10^8)^3}{(6.67\times10^{-11})(2.359\times10^6)^2}=6.0\times10^{24}\ \text{kg}$.
Check The accepted mass of the Earth is $5.97\times10^{24}\ \text{kg}$. The small difference arises mainly because the Earth itself is not fixed: Earth and Moon both circle their common centre of mass. Dimensions: $\dfrac{[\text{L}^3]}{[\text{M}^{-1}\text{L}^3\text{T}^{-2}][\text{T}^2]}=[\text{M}]$ ✓.
✏️ Example 6 — Same star, different planets
Two planets P and Q orbit the same star in circular orbits, with $r_Q=4r_P$. Find (a) the ratio of their periods $T_Q:T_P$, (b) the ratio of their orbital speeds $v_Q:v_P$. (c) If P takes $90$ days, how long does Q take?
(a) $T\propto r^{3/2}$: $\dfrac{T_Q}{T_P}=4^{3/2}=8$, so $8:1$.
(b) $v=\sqrt{\dfrac{GM}{r}}\propto r^{-1/2}$: $\dfrac{v_Q}{v_P}=4^{-1/2}=\dfrac12$, so $1:2$.
(c) $T_Q=8\times90=720$ days.
Check $v=\dfrac{2\pi r}{T}$: $\dfrac{v_Q}{v_P}=\dfrac{r_Q}{r_P}\cdot\dfrac{T_P}{T_Q}=4\times\dfrac18=\dfrac12$ ✓. Outer planets are slower and have farther to go.
Interactive 2 — Testing T² ∝ a³ on the Solar System

Left: six planets orbit the Sun, each taking its real period (radii are drawn on a compressed scale so that Saturn fits). Right: $\log T$ against $\log a$ for the same planets. If $T^2=a^3$ then $\log T=\tfrac32\log a$, a straight line of slope $1.5$ through $(0,0)$ — and every planet sits on it. Pick a planet to read its $T^2/a^3$, or choose "your planet" and slide its distance to predict its year.

2.8 AU
semi-major axis $a$—
period $T$—
$T^2/a^3$ (yr² AU⁻³)—
orbital speed—
⚠️ Common mistakes

1. Measuring $r$ from the surfaces. In $F=\dfrac{Gm_1m_2}{r^2}$, $r$ is the distance between the centres of spherical bodies. For a satellite $h$ above the Earth, $r=R_E+h$.

2. "The Sun is at the centre of the ellipse." It is at a focus. The centre of the ellipse is an empty point.

3. "Equal areas means equal distances." Equal areas are swept in equal times; the distances travelled along the orbit are unequal — longest near perihelion.

4. Squaring and cubing the wrong quantity. $T^2\propto a^3$, so $T\propto a^{3/2}$ and $a\propto T^{2/3}$. Doubling $a$ multiplies $T$ by $2\sqrt2\approx2.83$, not by $8$.

5. Forgetting that gravity acts on both bodies. The apple pulls the Earth up exactly as hard as the Earth pulls the apple down; the Earth's acceleration is negligible only because its mass is enormous.

Exercises

Take $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$, $M_E=5.97\times10^{24}\ \text{kg}$, $R_E=6.37\times10^6\ \text{m}$, mass of the Sun $1.99\times10^{30}\ \text{kg}$, mass of the Moon $7.35\times10^{22}\ \text{kg}$ and $1\ \text{AU}=1.50\times10^{11}\ \text{m}$ where needed.

I
Kepler's laws.
  • 1 Jupiter's semi-major axis is $5.20\ \text{AU}$. Find its period in years.
  • 2 A planet of the Sun has a period of $8.0$ years. Find its semi-major axis.
  • 3 An asteroid goes round the Sun every $4.6$ years. Find its semi-major axis.
  • 4 Saturn's period is $29.5$ years. How many times farther from the Sun than the Earth is it, on average?
  • 5 A body's distance from the Sun varies between $2.0\ \text{AU}$ and $6.0\ \text{AU}$. Find its semi-major axis, the eccentricity of its orbit and its period.
  • 6 For the body of part 5, find the ratio of its speed at perihelion to its speed at aphelion.
  • 7 A comet moves at $60\ \text{km/s}$ at perihelion, $0.50\ \text{AU}$ from the Sun. Its aphelion is $5.0\ \text{AU}$ from the Sun. Find its speed at aphelion.
  • 8 Explain, using angular momentum, why Kepler's second law holds for any force directed towards the Sun, whether or not it obeys the inverse-square law.
1 $T=5.20^{3/2}=11.9\ \text{yr}$.
2 $a=8.0^{2/3}=4.0\ \text{AU}$.
3 $a=4.6^{2/3}=2.8\ \text{AU}$ ($2.77$) — in the asteroid belt between Mars and Jupiter.
4 $a=29.5^{2/3}=9.55\ \text{AU}$, so about $9.5$ times farther.
5 $a=\tfrac12(2.0+6.0)=4.0\ \text{AU}$; $e=\dfrac{6.0-2.0}{6.0+2.0}=0.50$; $T=4.0^{3/2}=8.0\ \text{yr}$.
6 $\dfrac{v_p}{v_a}=\dfrac{r_a}{r_p}=\dfrac{6.0}{2.0}=3.0$ (also $\dfrac{1+e}{1-e}=\dfrac{1.5}{0.5}=3$).
7 $v_a=v_p\dfrac{r_p}{r_a}=60\times\dfrac{0.50}{5.0}=6.0\ \text{km/s}$.
8 A force along the Sun–planet line has zero moment (torque) about the Sun, so the planet's angular momentum $L=mrv_\perp$ about the Sun is constant. The areal velocity is $\tfrac12rv_\perp=\dfrac{L}{2m}$, hence also constant. The inverse-square law is needed for the first and third laws, not the second.
II
The law of gravitation and superposition.
  • 1 Find the gravitational force between two $1000\ \text{kg}$ cars whose centres of mass are $3.0\ \text{m}$ apart.
  • 2 Find the gravitational force exerted by the Earth on a $70\ \text{kg}$ person standing on its surface. Compare it with $mg$ for $g=9.8\ \text{m/s}^2$.
  • 3 The distance between two bodies is doubled and the mass of one of them is tripled. By what factor does the gravitational force change?
  • 4 Find the gravitational force between the Sun and the Earth ($1.50\times10^{11}\ \text{m}$ apart).
  • 5 Masses of $2.0\ \text{kg}$ and $8.0\ \text{kg}$ are fixed $0.60\ \text{m}$ apart. Where on the line joining them can a third mass be placed so that the net gravitational force on it is zero?
  • 6 In part 5, the third mass is nudged a little towards the $2.0\ \text{kg}$ mass, along the line. Does it return, or move further away? Explain.
  • 7 A uniform thin spherical shell has mass $100\ \text{kg}$ and radius $1.0\ \text{m}$. Find the gravitational force it exerts on a $1.0\ \text{kg}$ particle (a) $0.50\ \text{m}$ from its centre, (b) $2.0\ \text{m}$ from its centre.
  • 8 Four $2.0\ \text{kg}$ spheres sit at the corners of a square of side $1.0\ \text{m}$. Find the magnitude and direction of the net gravitational force on any one of them.
1 $F=\dfrac{(6.67\times10^{-11})(1000)^2}{3.0^2}=7.4\times10^{-6}\ \text{N}$.
2 $F=\dfrac{GM_Em}{R_E^2}=\dfrac{(6.67\times10^{-11})(5.97\times10^{24})(70)}{(6.37\times10^6)^2}=687\ \text{N}\approx6.9\times10^2\ \text{N}$; $mg=686\ \text{N}$ — the same within rounding, as it should be (Section 7.2).
3 $F'=\dfrac{G(3m_1)m_2}{(2r)^2}=\tfrac34F$.
4 $F=\dfrac{(6.67\times10^{-11})(1.99\times10^{30})(5.97\times10^{24})}{(1.50\times10^{11})^2}=3.5\times10^{22}\ \text{N}$.
5 Between the masses, at $x$ from the $2.0\ \text{kg}$ mass: $\dfrac{2.0}{x^2}=\dfrac{8.0}{(0.60-x)^2}$, so $\dfrac{0.60-x}{x}=2$, $x=0.20\ \text{m}$ from the $2.0\ \text{kg}$ mass ($0.40\ \text{m}$ from the $8.0\ \text{kg}$ mass). Outside the pair both pulls point the same way, so there is no other point.
6 Closer to the $2.0\ \text{kg}$ mass its pull grows and the other's shrinks, so the net force points towards the $2.0\ \text{kg}$ mass: the third mass moves further away. The equilibrium is unstable along the line.
7 (a) Inside the shell: zero. (b) Outside, as if all the mass were at the centre: $F=\dfrac{(6.67\times10^{-11})(100)(1.0)}{2.0^2}=1.7\times10^{-9}\ \text{N}$.
8 Two neighbours at $1.0\ \text{m}$ each pull with $F_1=\dfrac{G(2.0)^2}{1.0^2}=2.67\times10^{-10}\ \text{N}$, at right angles; their resultant is $\sqrt2F_1$ along the diagonal. The opposite corner, $\sqrt2\ \text{m}$ away, pulls with $\tfrac12F_1$ along the same diagonal. Net: $(\sqrt2+\tfrac12)F_1=1.91\times2.67\times10^{-10}=5.1\times10^{-10}\ \text{N}$, towards the centre of the square.
III
The constant $G$ and Kepler's third law.
  • 1 Use $F=\dfrac{Gm_1m_2}{r^2}$ to find the dimensional formula of $G$ and express its unit in SI base units.
  • 2 The Moon is $3.84\times10^8\ \text{m}$ from the Earth's centre and orbits once in $27.3$ days. Find its centripetal acceleration and show that $g$ divided by it is close to $(r/R_E)^2$.
  • 3 Jupiter's moon Io orbits at $4.22\times10^8\ \text{m}$ from Jupiter's centre with a period of $1.77$ days. Find the mass of Jupiter.
  • 4 Find the period, in minutes, of a satellite in a circular orbit of radius $7.0\times10^6\ \text{m}$ about the Earth.
  • 5 The radius of a satellite's circular orbit is increased by $1.0\%$. By about what percentage does its period increase?
  • 6 In a torsion-balance experiment, a $158\ \text{kg}$ lead sphere and a $0.73\ \text{kg}$ lead sphere have their centres $0.23\ \text{m}$ apart. Find the gravitational force between them.
  • 7 Show that the value of $\dfrac{T^2}{r^3}$ is the same for all satellites of the Earth, and calculate it.
  • 8 Venus orbits the Sun at $1.08\times10^{11}\ \text{m}$ with a period of $225$ days. Find the mass of the Sun from these data.
1 $G=\dfrac{Fr^2}{m_1m_2}$: $\dfrac{[\text{MLT}^{-2}][\text{L}^2]}{[\text{M}^2]}=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$; unit $\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$.
2 $T=2.359\times10^6\ \text{s}$, $a=\dfrac{4\pi^2r}{T^2}=2.72\times10^{-3}\ \text{m/s}^2$. $\dfrac{g}{a}=\dfrac{9.8}{2.72\times10^{-3}}=3.6\times10^3$; $\left(\dfrac{3.84\times10^8}{6.37\times10^6}\right)^2=60.3^2=3.6\times10^3$ ✓.
3 $T=1.77\times86\,400=1.529\times10^5\ \text{s}$; $M=\dfrac{4\pi^2r^3}{GT^2}=1.9\times10^{27}\ \text{kg}$ (about 318 Earth masses).
4 $T=2\pi\sqrt{\dfrac{r^3}{GM_E}}=2\pi\sqrt{\dfrac{3.43\times10^{20}}{3.98\times10^{14}}}=5.83\times10^3\ \text{s}=97\ \text{min}$.
5 $T\propto r^{3/2}$: $1.01^{1.5}=1.015$, an increase of about $1.5\%$ (small changes: $\dfrac{\Delta T}{T}\approx\tfrac32\dfrac{\Delta r}{r}$).
6 $F=\dfrac{(6.67\times10^{-11})(158)(0.73)}{0.23^2}=1.5\times10^{-7}\ \text{N}$ — roughly the weight of a grain of fine sand.
7 From Result 7.6, $\dfrac{T^2}{r^3}=\dfrac{4\pi^2}{GM_E}$, which contains only the Earth's mass: $\dfrac{39.48}{(6.67\times10^{-11})(5.97\times10^{24})}=9.9\times10^{-14}\ \text{s}^2\,\text{m}^{-3}$.
8 $T=225\times86\,400=1.944\times10^7\ \text{s}$; $M=\dfrac{4\pi^2(1.08\times10^{11})^3}{(6.67\times10^{-11})(1.944\times10^7)^2}=2.0\times10^{30}\ \text{kg}$ — the same Sun as in Example 5 ✓.
IV
Applications and harder problems.
  • 1 (Modelling) A planet is found orbiting a star of mass $0.80$ solar masses at $0.050\ \text{AU}$. Using $T^2\propto\dfrac{a^3}{M}$ and the Earth's orbit for comparison, find its period in days.
  • 2 At what distance from the centre of the Earth, on the line to the Moon, is the net gravitational force of the Earth and the Moon on a spacecraft zero? (Centre-to-centre distance $3.84\times10^8\ \text{m}$.)
  • 3 Ganymede orbits Jupiter at $1.07\times10^9\ \text{m}$. Using the data for Io in III 3, find Ganymede's period without using $G$.
  • 4 Show that for an orbit of eccentricity $e$, the ratio of greatest to least speed is $\dfrac{1+e}{1-e}$.
  • 5 A planet orbits a star of four times the Sun's mass at $2.0\ \text{AU}$. Find its period in years.
  • 6 Estimate the gravitational pull on a $60\ \text{kg}$ student from a building of mass $1.0\times10^7\ \text{kg}$ whose centre of mass is $50\ \text{m}$ away (treat both as particles). Why do you not notice it?
1 In years, AU and solar masses, $T^2=\dfrac{a^3}{M}$: $T=\sqrt{\dfrac{0.050^3}{0.80}}=0.0125\ \text{yr}=4.6$ days.
2 $\dfrac{M_E}{x^2}=\dfrac{M_M}{(d-x)^2}$: $\dfrac{x}{d-x}=\sqrt{\dfrac{5.97\times10^{24}}{7.35\times10^{22}}}=9.01$, so $x=\dfrac{9.01}{10.01}d=3.46\times10^8\ \text{m}$ — nine-tenths of the way to the Moon.
3 Same central mass: $T_G=T_I\left(\dfrac{r_G}{r_I}\right)^{3/2}=1.77\times\left(\dfrac{1.07\times10^9}{4.22\times10^8}\right)^{3/2}=1.77\times4.04=7.15$ days.
4 $v_pr_p=v_ar_a$ with $r_p=a(1-e)$, $r_a=a(1+e)$: $\dfrac{v_p}{v_a}=\dfrac{r_a}{r_p}=\dfrac{1+e}{1-e}$. Speed is greatest at perihelion and least at aphelion.
5 $T=\sqrt{\dfrac{a^3}{M}}=\sqrt{\dfrac{8.0}{4}}=1.4\ \text{yr}$.
6 $F=\dfrac{(6.67\times10^{-11})(1.0\times10^7)(60)}{50^2}=1.6\times10^{-5}\ \text{N}$, about $3\times10^{-8}$ of the student's weight ($588\ \text{N}$) — far smaller than the friction from the floor, so nothing happens.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State Kepler's three laws of planetary motion. Using the second law, explain why the Earth moves faster in January than in July.

Laws (1) Each planet moves in an ellipse with the Sun at one focus. (2) The line from the Sun to the planet sweeps equal areas in equal times. (3) $T^2\propto a^3$, where $a$ is the semi-major axis. [½ + ½ + ½]
Reason The Earth is at perihelion (nearest the Sun) in early January and at aphelion in early July. To sweep equal areas in equal times with a shorter radius, it must move faster: $v_pr_p=v_ar_a$, so $v_p\gt v_a$. [1½]
Exam tip Seasons are caused by the tilt of the Earth's axis, not by its distance from the Sun — January is winter in India despite perihelion.

E2. NEET The masses of two bodies are each doubled and the distance between them is halved. The gravitational force between them becomes (a) $4F$ (b) $8F$ (c) $16F$ (d) $F/4$

Idea $F'=\dfrac{G(2m_1)(2m_2)}{(r/2)^2}=\dfrac{4}{1/4}\cdot\dfrac{Gm_1m_2}{r^2}=16F$.
Answer (c) $16F$.
Exam tip (b) $8F$ comes from halving $r$ but forgetting to square; (a) from changing only the masses.

E3. JEE Main A planet moves round the Sun in an elliptical orbit of eccentricity $0.20$. The ratio of its greatest speed to its least speed is (a) $1.25$ (b) $1.5$ (c) $2.25$ (d) $0.67$

Idea Angular momentum: $v_pr_p=v_ar_a$, so $\dfrac{v_p}{v_a}=\dfrac{r_a}{r_p}=\dfrac{a(1+e)}{a(1-e)}$.
Answer $\dfrac{1.20}{0.80}=1.5$ — (b).
Exam tip (c) $2.25=1.5^2$ is the ratio of kinetic energies, not speeds; (d) is the inverted ratio.

E4. JEE Main NEET Two satellites A and B circle the Earth with periods in the ratio $T_A:T_B=1:8$. The ratio of their orbital speeds $v_A:v_B$ is (a) $2:1$ (b) $1:2$ (c) $4:1$ (d) $8:1$

Radii $r\propto T^{2/3}$: $r_A:r_B=1:8^{2/3}=1:4$.
Speeds $v=\sqrt{\dfrac{GM}{r}}\propto r^{-1/2}$: $v_A:v_B=\sqrt4:1=2:1$.
Answer (a) $2:1$.
Exam tip Shortcut: $v\propto T^{-1/3}$, and $8^{1/3}=2$. (c) $4:1$ is the ratio of radii inverted, a common slip.

E5. JEE Advanced (One or more correct) A planet moves in an elliptical orbit about the Sun, the only force on it being the Sun's gravity. Which statements are correct? (A) The planet's angular momentum about the Sun is constant. (B) The magnitude of its linear momentum is constant. (C) The net work done on it by gravity over one complete orbit is zero. (D) At perihelion the gravitational force on it is perpendicular to its velocity.

(A) The force is central (towards the Sun), so its torque about the Sun is zero: $\mathbf{L}$ is conserved ✓.
(B) The speed varies (fastest at perihelion), so $p=mv$ varies ✗.
(C) After one orbit the planet is back where it started with the same speed: $\Delta K=0$, so by the work–energy theorem the net work is zero ✓.
(D) At perihelion (and aphelion) the velocity is perpendicular to the Sun–planet line, along which the force acts ✓.
Answer (A), (C), (D).
Exam tip Between perihelion and aphelion the force has a component against the velocity (the planet slows); on the way back it has a component along it. Only at the two ends is it exactly perpendicular.

E6. Board · Case study Newton compared the fall of an apple with the "fall" of the Moon. The Moon orbits the Earth in a nearly circular orbit of radius $3.84\times10^8\ \text{m}$ with a period of $27.3$ days; the Earth's radius is $6.37\times10^6\ \text{m}$ and $g=9.8\ \text{m/s}^2$. (i) Calculate the Moon's centripetal acceleration. [2] (ii) Find the ratio $g/a_\text{Moon}$. [1] (iii) What does the comparison of this ratio with $(r/R_E)^2$ tell us about the force of gravity? [1]

(i) $T=27.3\times86\,400=2.36\times10^6\ \text{s}$; $a=\dfrac{4\pi^2r}{T^2}=\dfrac{39.48\times3.84\times10^8}{(2.359\times10^6)^2}=2.72\times10^{-3}\ \text{m/s}^2$. [2]
(ii) $\dfrac{9.8}{2.72\times10^{-3}}=3.6\times10^3$. [1]
(iii) $\dfrac{r}{R_E}=60.3$ and $60.3^2=3.6\times10^3$ — the same. The Earth's pull falls off as $\dfrac{1}{r^2}$, and the force that makes apples fall is the one that keeps the Moon in orbit. [1]
Exam tip Convert days to seconds before squaring — leaving $T$ in days gives an answer wrong by a factor of $86\,400^2$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How do astronomers find the size of a planet's orbit around another star — and decide whether it might be warm enough for liquid water?

Why it matters. Space telescopes have found thousands of exoplanets by the tiny dip in a star's brightness each time a planet passes in front of it. The dips repeat once per orbit, so they give the period $T$ directly. With the star's mass from its colour and brightness, Kepler's third law (Result 7.6) turns the period into an orbit size. The red dwarf TRAPPIST-1, about 40 light-years away, has seven Earth-sized planets found this way.

What to do.

  1. Show that, for a star of mass $M$ (in solar masses), Result 7.6 becomes $a^3=MT^2$ with $a$ in AU and $T$ in years.
  2. TRAPPIST-1 has a mass of about $0.090$ solar masses. One of its planets has a period of $6.10$ days. Find its orbital radius in AU and in metres, and compare it with Mercury's ($0.387\ \text{AU}$).
  3. The star gives out only about $5.5\times10^{-4}$ times the Sun's power. A planet receives the same light per square metre as the Earth if its distance (in AU) is about $\sqrt{L}$, where $L$ is the star's power in units of the Sun's. Find this distance and compare it with your answer to step 2.
  4. Find the planet's orbital speed and compare it with the Earth's $30\ \text{km/s}$.
  5. Look up two other exoplanets in a public exoplanet catalogue and repeat steps 2–3.

Tools: Calculator or spreadsheet; a public exoplanet archive run by a space agency or university.

Hand in: Your derivation from step 1, a table of period, star mass, orbit radius and "Earth-like light distance" for three planets, and a short paragraph on which of them might hold liquid water.

Going further. The transit method finds the period but not the planet's mass. How could the star's own small wobble about the common centre of mass reveal it?

Instructor note: $T=0.01670\ \text{yr}$, so $a=(0.090\times0.01670^2)^{1/3}$ $=0.0293\ \text{AU}$ $=4.38\times10^9\ \text{m}$, about $1/13$ of Mercury's distance. $\sqrt{5.5\times10^{-4}}=0.023\ \text{AU}$, so the planet is close to the "Earth-like light" distance (slightly farther, hence slightly cooler). Orbital speed $2\pi a/T=5.2\times10^4\ \text{m/s}\approx52\ \text{km/s}$. Students should see that small, dim stars have their temperate planets very close in, with periods of days.

🧪 Self-check quiz

0/5 correct

Q1. According to Kepler's first law, the Sun lies

Q2. Kepler's second law tells us that a planet moves

Q3. The distance between two point masses is tripled. The gravitational force between them becomes

Q4. The SI unit of the gravitational constant $G$ is

Q5. A planet's orbit has four times the semi-major axis of the Earth's. Its period is

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