Gravitation and Orbits

Beginner

Drop a pencil and it falls; look up and the Moon hangs in the sky. Newton's great idea was that one and the same pull explains both.

Step 1. Every object with mass attracts every other object, more strongly when the masses are bigger: you feel the Earth's pull, but not the tiny pull of the person sitting next to you.

Step 2. The pull weakens with distance, and fast. Distance is measured from centre to centre: twice as far, the pull drops to a quarter; three times as far, to a ninth. Quick check: if two objects are moved ten times farther apart, what happens to their pull? (Answer: it becomes a hundred times weaker.)

Step 3. Why doesn't the Moon fall onto the Earth? It does fall, all the time. Without the pull it would carry straight on; gravity bends its path toward the Earth. Picture a cannon on a very high mountain firing sideways. A slow ball lands nearby, a faster one farther away. Fired fast enough, the ball drops just as quickly as the curved ground falls away beneath it, so it never lands: it is in orbit. Astronauts float for the same reason: they and their spacecraft fall together, even though gravity up there is almost as strong as on the ground.

Step 4. Most orbits are slightly stretched circles, called ellipses, with the star off-centre. A planet speeds up as it falls in toward the star and slows down as it climbs away.

In the animation, the red arrow is the star's pull: always aimed at the star, longer when the planet is close. The blue dashed line shows where the planet would go if gravity switched off. The grey dots, left at equal time steps, bunch up far from the star and spread out near it.

Intermediate

The law of gravitation and the field strength

The pull that makes a pencil fall also bends the Moon's path. Every body with mass attracts every other body. For masses $m_1$ and $m_2$ whose centres are a distance $r$ apart, Newton's law of universal gravitation gives the size of the pull:
$F=\frac{G{m}_1{m}_2}{r^2}$

Here $G=6.674\times 10^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/k{\mathrm{g}}^2$ is the gravitational constant, so small that two 70 kg people 1 m apart attract each other with only $3.3\times 10^{-7}\mathrm{N}$. The force acts along the line joining the centres, and by Newton's third law each body pulls the other equally hard: you pull the Earth up with the same 690 N with which it pulls you down. For a planet or a star, $r$ is measured from its centre.

Dividing the force on a small mass $m$ by $m$ gives the field strength of a body of mass $M$, the pull per kilogram:
$g=\frac{GM}{r^2}$

Its unit, the newton per kilogram, is the same as m/s²: $g$ is the free-fall acceleration at that place. At the Earth's surface ($M=5.97\times 10^{24}\mathrm{kg}$, $R=6371\mathrm{km}$) it is 9.8 m/s². At $r=2R$, one Earth radius up, it is a quarter of that, about 2.45 m/s², and 400 km up it is still about 8.7 m/s².

Circular orbits

As in Circular Motion, a body moving at speed $v$ on a circle of radius $r$ needs a net force $m{v}^2/r$ toward the centre. For a satellite, gravity alone supplies it:
$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$
$v=\sqrt{\frac{GM}{r}}\text{(circular orbit)}$

The satellite's own mass $m$ cancels: a space station and a spanner dropped by its crew share the same orbit. One lap of length $2\pi r$ takes $T=2\pi r/v$, so $T^2=4{\pi }^2{r}^3/(GM)$. Higher orbits are slower, $v\propto 1/\sqrt{r}$, and their laps much longer, $T\propto r^{3/2}$: a satellite skimming the ground would circle in about 84 minutes, while the Moon, 60 Earth radii away, takes 27.3 days.

Ellipses and Kepler's three laws

A satellite launched sideways a little faster or slower than $\sqrt{GM/r}$ follows an ellipse instead. Johannes Kepler found three laws of planetary motion between 1609 and 1619, and Newton's law explains all three.

• Every planet moves on an ellipse with the Sun at one focus, not at the centre.
• The line from the Sun to the planet sweeps equal areas in equal times, so the planet is fastest at perihelion, its closest point, and slowest at aphelion, its farthest.
• The square of the period is proportional to the cube of the orbit's size.

The size is the semi-major axis $a$, half the longest diameter and the average of the closest and farthest distances. The eccentricity $e$ says how stretched the ellipse is: the closest distance is $a(1-e)$, the farthest $a(1+e)$, and a circle has $e=0$. In Newton's form the third law is
$T^2=\frac{4{\pi }^2{a}^3}{GM}$
the circular result with $r$ replaced by $a$, whatever the eccentricity. Around the Sun, with $T$ in years and $a$ in astronomical units ($1\mathrm{AU}=1.496\times 10^{11}\mathrm{m}$), it reads simply $T^2=a^3$.

Energy in a gravitational field

Pulling two masses apart takes work, so their potential energy grows with distance. Taking it as zero at infinite separation gives
$U=-\frac{GMm}{r}$

The minus sign means that a body sits in a well: lifting it from rest at $r$ to infinity takes $GMm/r$ of work. In orbit the total energy $E=\frac{1}{2}m{v}^2-GMm/r$ stays constant, so the body speeds up as it falls inward. With $E\ge 0$ the body escapes; from the Earth's surface that takes $\sqrt{2GM/R}\approx 11.2km/s$.

On an ellipse the energy depends only on its size, $E=-GMm/(2a)$. Equating the two expressions gives the speed at any distance, the vis-viva relation:
$v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)$
For a circle, $r=a$ and it reduces to $v^2=GM/r$.

Worked example: the planet in the animation

The star has the Sun's mass, $GM=1.327\times 10^{20}{\mathrm{m}}^3/{\mathrm{s}}^2$, and the ellipse has $a=1\mathrm{AU}$ and $e=0.5$, so the planet swings between 0.5 AU and 1.5 AU and, with the Earth's $a$, takes one year per lap. How strong is the pull per kilogram at the two ends? At perihelion, $7.48\times 10^{10}\mathrm{m}$ from the star,
$g=\frac{1.327\times 10^{20}}{(7.48\times 10^{10}{)}^2}\approx 0.0237m/{\mathrm{s}}^2=23.7mm/{\mathrm{s}}^2$
and three times farther out it is nine times smaller, 2.6 mm/s². These are the extremes of the $F/m$ readout, and the red $F$ arrow shrinks by the same factor.

For the speeds, the circular speed at 1 AU is $\sqrt{GM/a}\approx 29.8km/s$, the Earth's speed. With distances in AU, the vis-viva relation gives
$v=29.8\sqrt{2/0.5-1}=29.8\sqrt{3}\approx 51.6km/s$
$v=29.8\sqrt{2/1.5-1}=29.8/\sqrt{3}\approx 17.2km/s$
at perihelion and aphelion, the two ends of the $v$ readout.

The speed ratio, 3, is the inverse of the distance ratio, as equal areas demand: at both ends of the long axis the velocity is perpendicular to the radius, a short time $\Delta t$ sweeps a thin triangle of area $\frac{1}{2}rv\Delta t$, and $0.5\cdot 51.6=1.5\cdot 17.2$. The energy per kilogram, $\frac{1}{2}{v}^2-GM/r$, comes out as $-4.4\times 10^{8}J/kg$ at both points, which is $-GM/(2a)$.

Watch the angle between the arrows. On the way in, the purple $v$ arrow is less than 90° from the red $F$ arrow, so the pull has a component along the motion and speeds the planet up; on the way out the angle exceeds 90° and the pull brakes it. At perihelion and aphelion the arrows are perpendicular and the speed is momentarily steady.

Worked example: two Earth satellites

The International Space Station flies about 400 km up, so $r=6371+400=6771\mathrm{km}$ from the Earth's centre. With $GM=3.986\times 10^{14}{\mathrm{m}}^3/{\mathrm{s}}^2$ for the Earth,
$v=\sqrt{\frac{3.986\times 10^{14}}{6.771\times 10^6}}\approx 7672m/s\approx 7.67km/s$
$T=\frac{2\pi r}{v}=\frac{2\pi \cdot 6.771\times 10^6}{7672}\approx 5545\mathrm{s}$
about 92 minutes. Up there $g=GM/r^2\approx 8.7m/{\mathrm{s}}^2$, 89 % of its value on the ground.

A geostationary satellite must circle once per turn of the Earth relative to the stars, $T=86164\mathrm{s}$ (23 h 56 min), to hang above one point of the equator. The third law gives
$r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}\approx 4.216\times 10^7\mathrm{m}$
about 35 800 km above the equator, at $2\pi r/T\approx 3.07km/s$. As a check, the radius is $42164/6771\approx 6.23$ times the station's, so the period should be $6.23^{3/2}\approx 15.5$ times longer, and indeed $1436/92.4\approx 15.5$.

Common mistakes

• Believing there is no gravity in orbit. At 400 km $g$ is still 8.7 m/s²; astronauts float because they and the station fall together.
• Measuring $r$ from the surface. With 400 km instead of 6771 km, the station's speed would come out as 31.6 km/s, not 7.67 km/s.
• Thinking that heavier satellites or higher orbits need more speed. The formula $v=\sqrt{GM/r}$ contains no satellite mass, and the geostationary satellite is slower than the station.
• Putting the farthest distance into the third law. Halley's comet ranges from 0.59 AU to 35.1 AU, so $a=17.8\mathrm{AU}$ and $T=17.8^{3/2}\approx 75$ years, not 208 years.
• Assuming the planet speeds up because the pull is strong. At perihelion the pull peaks, at 23.7 mm/s² in the animation, yet it is perpendicular to the velocity, so the speed has just stopped rising.

Where you meet it

Navigation satellites 20 200 km up with 12-hour periods, geostationary weather and television satellites, the tides, and the masses of planets and stars, read off from the orbits around them with the third law: all rest on this lesson.

The equal-area law is angular momentum conservation (Angular Momentum and Its Conservation); $U=-GMm/r$ generalises the $mgh$ of Work, Energy and Power; the same inverse square governs charges in Coulomb's Law; and Gravity as Curved Spacetime starts from the fact that every body falls with the same acceleration.

Advanced

From the apple to the Moon

This section sets out the force law and the first evidence that one law rules both the ground and the sky. Two masses $M$ and $m$ a distance $r$ apart attract each other with a force of size $GMm/r^2$, where $G$ is the gravitational constant. In vector form, with $M$ at the origin and $m$ at the position $\vec{r}$,
$\vec{F}=-\frac{GMm}{r^2}\hat{r}$
where $\hat{r}$ is the unit vector pointing from $M$ towards $m$. The minus sign turns the force back towards $M$: gravity attracts.

Newton's first test compared the Moon with a falling apple. If the pull weakens as $1/r^2$, then 60 Earth radii away it should be $60^2=3600$ times weaker than at the surface. The Moon is $r=3.844\times 10^8\mathrm{m}$ away, about 60 Earth radii, and goes round the Earth once in 27.32 days ($T=2.36\times 10^6\mathrm{s}$). To stay on its circle it needs the centripetal acceleration
$a_c=\frac{4{\pi }^{2}r}{T^2}\approx 2.72\times 10^{-3}m/{\mathrm{s}}^2$
and the inverse-square law predicts $g/60^2=9.81/3600\approx 2.72\times 10^{-3}m/{\mathrm{s}}^2$. The two agree: the apple and the Moon obey the same law.

Why may the distance be measured from the Earth's centre? Because of the shell theorem: a uniform spherical shell attracts a mass outside it exactly as if all the shell's mass sat at its centre, and exerts no net force at all on a mass inside it. A planet is a nest of such shells, so from outside it acts as a point mass. Inside a uniform planet only the shells nearer the centre pull. Their mass grows as $r^3$ while their pull per unit mass falls as $1/r^2$, so $g$ falls in proportion to $r$, down to zero at the centre.

Once $G$ is known, the same law weighs the Earth. Cavendish measured $G$ with a torsion balance in 1798. At the surface $g=GM/R^2$, with $R$ the Earth's radius, so $M=g{R}^2/G\approx 5.97\times 10^{24}\mathrm{kg}$. That is a mean density of 5.5 g/cm³, twice that of surface rocks, so the deep interior, above all the core, must be far denser.

Equal areas from angular momentum

Kepler found that the line from the star to a planet sweeps out equal areas in equal times, so the planet moves fast near the star and slowly far from it. His rule is the conservation of angular momentum in another form, as the steps below show.

Legend

$M$	The star, of mass $M$.	$\mathrm{kg}$
$r$	Distance from the star to the planet.	$\mathrm{m}$
$\vec{F}$	Pull of the star on the planet, along the radius towards $M$.	$\mathrm{N}$
$\vec{v}dt$	The planet's step in the short time $dt$: velocity times time.	$\mathrm{m}$
$dr$	Part of the step along the radius: the change in $r$.	$\mathrm{m}$
$rd\theta$	Part of the step across the radius: an arc of radius $r$.	$\mathrm{m}$
$d\theta$	Angle the radius vector turns through during the step.	$\mathrm{rad}$
$dA$	Area of the thin triangle swept by the radius vector during the step.	${\mathrm{m}}^2$

Figure 1 enlarges one short step of a planet at the distance $r$ from its star $M$. The direction of the radius vector is given by the angle $\theta$. In a short time $dt$ the planet moves by $\vec{v}dt$, the blue arrow. Split this step into two parts. The part $dr$ along the radius only changes the distance; it sweeps no area, because it slides the planet along the line it already lies on. The part across the radius is the short arc $rd\theta$, traced as the radius vector turns through the small angle $d\theta$. Only this part sweeps new area: the gold triangle, of base $r$ and height $rd\theta$, so $dA=\frac{1}{2}r\cdot rd\theta$.

Dividing by $dt$ gives the areal speed, the area swept per unit time: $dA/dt=\frac{1}{2}{r}^2\dot{\theta }$, where $\dot{\theta }=d\theta /dt$ is the rate at which the radius vector turns. The planet's speed across the radius is $r\dot{\theta }$, so its angular momentum about the star has the size $L=mr\cdot r\dot{\theta }=m{r}^2\dot{\theta }$. Hence
$\frac{dA}{dt}=\frac{L}{2m}$

Now look at the pull $\vec{F}$ in Figure 1: it points along the radius, towards $M$. The torque $\vec{r}\times \vec{F}$ is therefore zero, because the cross product of two parallel vectors vanishes, and without a torque the angular momentum $\vec{L}$ cannot change. Its size $L$ fixes the areal speed, which is Kepler's second law. Its direction, perpendicular to both $\vec{r}$ and $\vec{v}$, never changes either, so the orbit stays in one plane.

Nothing here used the $1/r^2$. Equal areas hold for every central force, that is, every force that points along the line to a fixed centre. Kepler's second law therefore tests the direction of the force, not how its strength depends on distance.

The animation's planet makes this concrete. It orbits a star of one solar mass on an ellipse with semi-major axis $a=1\mathrm{AU}$ and eccentricity $e=0.5$. The semi-major axis is half the longest diameter. The eccentricity is a measure of how flattened the ellipse is: 0 for a circle, close to 1 for a long, thin one. The ellipse's semi-minor axis, half the shortest diameter, is $b=a\sqrt{1-e^2}\approx 0.866\mathrm{AU}$, and the planet's period, as the third law below shows, is one year.

In the expert view each gold sector is swept in an eighth of the period, about 45.7 days, so both have one eighth of the ellipse's area, $\pi ab/8\approx 0.34\mathrm{A}{\mathrm{U}}^2$. The sector at perihelion, the point nearest the star, is short and wide; the one at aphelion, the farthest point, is long and narrow.

At the two ends of the major axis the velocity is perpendicular to the radius, so there $L/m=rv$ exactly. With the speeds found from the vis-viva relation below, 51.6 km/s at 0.5 AU and 17.2 km/s at 1.5 AU, this gives $0.5\cdot 51.6=1.5\cdot 17.2\approx 25.8\mathrm{AU}\cdot km/s$ at both ends, as it must.

Energy and the radial motion

Angular momentum alone does not say how close to the star the planet comes or how far it strays. Energy does, and it reduces the orbit to a one-dimensional problem in the distance $r$.

Legend

$U_{eff}$	Effective potential energy of the radial motion: $L^2/(2m{r}^2)-GMm/r$.	$\mathrm{J}$
$E$	Total energy: constant along the orbit, negative for a bound orbit; drawn as a horizontal level.	$\mathrm{J}$
$r$	Distance from the star.	$\mathrm{m}$
$r_{\min }$	Perihelion distance $a(1-e)$: the inner turning point, where $\dot{r}=0$.	$\mathrm{m}$
$r_{\max }$	Aphelion distance $a(1+e)$: the outer turning point, where $\dot{r}=0$.	$\mathrm{m}$
$K_r$	Radial kinetic energy $\frac{1}{2}m{\dot{r}}^2$: the gap between $E$ and $U_{eff}$.	$\mathrm{J}$
$\frac{L^2}{2m{r}^2}$	Centrifugal term: the kinetic energy of the sideways motion at distance $r$.	$\mathrm{J}$
$U$	Gravitational potential energy, $-GMm/r$.	$\mathrm{J}$

Start with the potential energy. Choose $U=0$ when the two bodies are infinitely far apart. Then $U$ at the distance $r$ is the work gravity does on $m$ as it moves from $r$ out to infinity, ${\int }_r^{\infty }(-GMm/r^{\prime 2})d{r}^{\prime }$, which gives
$U=-\frac{GMm}{r}$
It is negative everywhere and rises towards zero far away: the lower dashed curve, $U$, in Figure 2.

Next the kinetic energy. Split the velocity into a radial part $\dot{r}$, the rate at which the distance changes, and a sideways part $r\dot{\theta }$. From $L=m{r}^2\dot{\theta }$ the sideways part equals $L/(mr)$, so its kinetic energy is $L^2/(2m{r}^2)$. The conserved total energy is then
$E=\frac{1}{2}m{\dot{r}}^2+\frac{L^2}{2m{r}^2}-\frac{GMm}{r}$

Because $L$ is constant, the last two terms depend on $r$ alone, so they can be treated together as a single potential energy for the radial motion: the effective potential energy $U_{eff}$, the purple curve of Figure 2. It is the sum of the two dashed curves. The upper one, the centrifugal term $L^2/(2m{r}^2)$, climbs steeply as $r$ shrinks; it acts as a centrifugal barrier that keeps the planet from falling into the star. The lower one is $U$. Close in the barrier wins, far out gravity wins, and in between the sum forms a well.

Now read the figure like a ball rolling in a valley. The energy $E$ is the horizontal dashed level, the same at every $r$. The gap between that level and the purple curve is the radial kinetic energy $K_r=\frac{1}{2}m{\dot{r}}^2$. It cannot be negative, so the planet can only be where the curve lies below the level.

For $E<0$, as drawn, the level cuts the curve at two turning points, $r_{\min }$ and $r_{\max }$. There $K_r=0$, so $\dot{r}=0$: the distance stops shrinking or growing and turns round. The distance therefore oscillates between $r_{\min }$, the perihelion, and $r_{\max }$, the aphelion. If the level just touches the bottom of the well, the two turning points merge and the orbit is a circle. For $E\ge 0$ the level meets the curve only once, at the barrier, and the planet escapes to infinity.

The shape of the orbit

Figure 2 gives the range of distances but not the curve the planet traces. To find the curve, rewrite the radial equation of motion in terms of the angle $\theta$ instead of the time.

Along the radius, Newton's second law reads $m(\ddot{r}-r{\dot{\theta }}^2)=-GMm/r^2$; the term $-r{\dot{\theta }}^2$ is the centripetal acceleration that comes from the turning of the radius vector. Two changes make the equation solvable. First, use $u=1/r$ as the variable. Second, trade time for angle with $\dot{\theta }=L/(m{r}^2)=(L/m)u^2$. Then $\dot{r}=-(L/m)du/d\theta$ and $\ddot{r}=-(L/m{)}^2{u}^2{d}^{2}u/d{\theta }^2$, and after dividing by $-m(L/m{)}^2{u}^2$ the law becomes linear:
$\frac{d^{2}u}{d{\theta }^2}+u=\frac{GM{m}^2}{L^2}$

This is the equation of a harmonic oscillator pushed by a constant force, so its solution is that constant plus a cosine: $u=(1+e\cos \theta )/p$, with $p=L^2/(GM{m}^2)$ and $\theta$ measured from perihelion. In terms of $r$,
$r=\frac{p}{1+e\cos \theta }$
which is a conic section with the star at a focus: an ellipse for $e<1$ ($E<0$), a parabola for $e=1$ ($E=0$) and a hyperbola for $e>1$ ($E>0$).

For the ellipse, perihelion at $\theta =0$ and aphelion at $\theta =\pi$ give $r_{\min }=p/(1+e)$ and $r_{\max }=p/(1-e)$. Their sum is the major axis $2a$, so $p=a(1-e^2)=b^2/a$, $r_{\min }=a(1-e)$ and $r_{\max }=a(1+e)$.

The inverse square is special. By Bertrand's theorem only two forces give bound orbits that always close: the inverse square and the linear spring force. Under any other power law a bound orbit does not close, and its perihelion drifts round a little on every lap.

The energy fixes the size of the ellipse. At either turning point $\dot{r}=0$, so $E=L^2/(2m{r}^2)-GMm/r$. Multiplied by $r^2$ this is a quadratic, $E{r}^2+GMmr-L^2/(2m)=0$, whose two roots are $r_{\min }$ and $r_{\max }$. The roots of a quadratic add up to minus the middle coefficient over the first one, so $r_{\min }+r_{\max }=-GMm/E$. That sum is $2a$, hence
$E=-\frac{GMm}{2a}$
The energy sets the size $a$; for a given $a$, the angular momentum sets $p$ and with it the shape $e$.

Setting this equal to the energy at any point of the orbit, $\frac{1}{2}m{v}^2-GMm/r$, and dividing by $m/2$ gives the vis-viva relation, which the animation uses for its speed readout:
$v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)$
On a circle, $r=a$ and $v^2=GM/r$; at 1 AU from the Sun this is 29.8 km/s, the Earth's orbital speed. For the animation's planet, with $a=1\mathrm{AU}$, the speed is $29.8\sqrt{2/r-1}km/s$ with $r$ in AU: 51.6 km/s at perihelion and 17.2 km/s at aphelion.

Kepler's third law

The third law links the period to the size of the orbit and, read backwards, weighs the central body.

Over one period the radius vector sweeps the whole ellipse, of area $\pi ab$, at the constant rate $L/(2m)$, so $T=2\pi abm/L$. Squaring gives $T^2=4{\pi }^2{a}^2{b}^2{m}^2/L^2$. From the orbit equation $L^2=GM{m}^{2}p$, and $p=b^2/a$, so $L^2=GM{m}^2{b}^2/a$. Substituting, $b^2$ and $m^2$ cancel:
$T^2=\frac{4{\pi }^2{a}^3}{GM}$
The period depends on $a$ alone, not on the shape: a circle of radius 1 AU and the animation's ellipse with $a=1\mathrm{AU}$ and $e=0.5$ both take one year.

Read backwards, the law weighs the Sun. The Earth's orbit, $a=1.496\times 10^{11}\mathrm{m}$ and $T=3.156\times 10^7\mathrm{s}$, gives $GM=4{\pi }^2{a}^3/T^2\approx 1.327\times 10^{20}{\mathrm{m}}^3/{\mathrm{s}}^2$, and so a solar mass of $1.99\times 10^{30}\mathrm{kg}$.

Time along the orbit: Kepler's equation

The laws so far give the shape and the period, but not where the planet is at a given moment. That needs one more angle, the eccentric anomaly, and one equation for it.

Legend

$O$	Centre of the ellipse and of the dashed circle.
$M$	The star, at one focus of the ellipse; its mass is $M$.	$\mathrm{kg}$
$a$	Semi-major axis: from the centre to the end of the long axis; also the radius of the dashed circle.	$\mathrm{m}$
$b$	Semi-minor axis: from the centre to the end of the short axis.	$\mathrm{m}$
$ae$	Distance from the centre to the star: the eccentricity $e$ times $a$.	$\mathrm{m}$
$r$	Distance from the star to the planet.	$\mathrm{m}$
$\psi$	Eccentric anomaly: the angle at $O$, from the axis towards perihelion to the point of the circle straight above the planet.	$\mathrm{rad}$

Figure 3 shows the ellipse with its centre $O$, semi-major axis $a$ and semi-minor axis $b$. The star $M$ sits at a focus, a distance $ae$ from $O$; perihelion is the end of the long axis on the star's side. Around $O$ draw the dashed circle of radius $a$, which touches the ellipse at both ends of the long axis. From the planet draw a line perpendicular to the long axis, the thin dashed line, until it meets the circle. The line from $O$ to that point of the circle makes the angle $\psi$ with the axis towards perihelion: this angle is the eccentric anomaly.

Squeezing the circle towards the axis by the factor $b/a$ turns it into the ellipse, so measured from $O$ the planet is at $(a\cos \psi ,b\sin \psi )$, while the star is at $(ae,0)$. Pythagoras gives the distance between them, and with $b^2=a^2(1-e^2)$ the terms collect into a perfect square:
$r^2=a^2(\cos \psi -e{)}^2+b^2{\sin }^2\psi =a^2(1-e\cos \psi {)}^2$
that is, $r=a(1-e\cos \psi )$: smallest, $a(1-e)$, at $\psi =0$ and largest, $a(1+e)$, at $\psi =\pi$.

Now the time. By the law of areas, the time $t$ since perihelion is proportional to the area swept since then, and $t/T$ is that area divided by the whole ellipse, $\pi ab$. The area is easiest to find on the circle and then squeeze.

On the circle, the region swept from perihelion to the point at $\psi$, as seen from the star, is the sector at $O$ with angle $\psi$, of area $\frac{1}{2}{a}^2\psi$, minus the triangle with corners $O$, the star and that point. This triangle has base $ae$ and height $a\sin \psi$, so its area is $\frac{1}{2}{a}^{2}e\sin \psi$. Squeezing by $b/a$ shrinks every area by the same factor, so the area swept on the ellipse is $\frac{1}{2}ab(\psi -e\sin \psi )$. Dividing by $\pi ab$ and multiplying by $2\pi$ gives Kepler's equation:
$\frac{2\pi t}{T}=\psi -e\sin \psi$

Given $\psi$, the time follows at once; given the time, there is no closed-form solution for $\psi$. The animation finds $\psi$ by Newton's iteration at each of its time steps, then the position. The $r(t)$ graph of the expert view shows the result: a broad, flat crest at 1.5 AU and a narrow, sharp trough at 0.5 AU. The planet crosses the minor axis at $\psi =\pi /2$ and $\psi =3\pi /2$; putting these into Kepler's equation shows that it spends a fraction $\frac{1}{2}+e/\pi \approx 0.66$ of its year on the far side of the minor axis, about 241 days, against 124 days on the near side.

Worked problem: a transfer orbit

Kepler orbits are also how spacecraft travel. The cheapest two-burn route between two circular orbits around the same star is a Hohmann transfer: half of an ellipse that touches both circles.

Legend

$M$	The star, at the centre of both circles; its mass is $M$.	$\mathrm{kg}$
$r_1$	Radius of the inner circular orbit, where the craft starts.	$\mathrm{m}$
$r_2$	Radius of the outer circular orbit, the destination.	$\mathrm{m}$
$\Delta v_1$	First burn, forward at the inner circle: from the circular speed up to the perihelion speed of the ellipse.	$m/s$
$\Delta v_2$	Second burn, forward at aphelion, the far end of the ellipse: from the aphelion speed up to the circular speed.	$m/s$

Figure 4 draws it to scale for $r_1=0.5\mathrm{AU}$ and $r_2=1.5\mathrm{AU}$, so that the transfer ellipse is the animation's ellipse. The craft starts on the inner circle. A forward burn $\Delta v_1$ puts it on the ellipse whose perihelion touches the inner circle and whose aphelion touches the outer one. Its major axis is therefore $r_1+r_2$, so $a=(r_1+r_2)/2=1\mathrm{AU}$. Only the solid half of the ellipse is flown; at its far end a second forward burn $\Delta v_2$ makes the orbit circular.

The first burn. On a circle the vis-viva relation gives $v=\sqrt{GM/r}$, so the circular speed at $r_1$ is $29.8\sqrt{2}\approx 42.1km/s$. The transfer ellipse needs the perihelion speed of 51.6 km/s, so $\Delta v_1=9.5km/s$.

The second burn. Half an orbit later, after $T/2\approx 183$ days, the craft reaches aphelion at 17.2 km/s. That is short of the circular speed there, $29.8/\sqrt{1.5}\approx 24.3km/s$, so $\Delta v_2=7.1km/s$.

Both burns pushed forward, 16.6 km/s in all, yet the craft ends 17.8 km/s slower than it started. Energy resolves the paradox. Per kilogram, a circular orbit has the total energy $E=-GM/(2r)$, the kinetic energy $GM/(2r)=-E$ and the potential energy $-GM/r=2E$. From $r_1$ to $r_2$ the total rises from $-8.87\times 10^8$ to $-2.96\times 10^{8}J/kg$, so the engine supplies $5.91\times 10^{8}J/kg$. The potential energy rises by twice that, $1.18\times 10^{9}J/kg$, and the kinetic energy falls by $5.91\times 10^{8}J/kg$ to make up the difference. Higher orbits are slower, but they hold more energy.

The same route from the Earth to Mars, from 1 AU to 1.52 AU, has $a\approx 1.26\mathrm{AU}$. With $T$ in years and $a$ in AU the third law reads $T=a^{3/2}$, so the trip takes $\frac{1}{2}\cdot 1.26^{3/2}\approx 0.707$ years, about 258 days.

Beyond two point masses

Real systems differ from one point mass circling a fixed one in three ways: both bodies move, bodies have size, and there are usually more than two.

Two bodies orbit their common centre of mass, and their relative orbit obeys the same laws with $M$ replaced by $M+m$: $T^2=4{\pi }^2{a}^3/(G(M+m))$. The Earth–Moon centre of mass lies 4670 km from the Earth's centre, inside the Earth; the Sun–Jupiter one lies 1.07 solar radii from the Sun's centre, just above its surface. The small wobble of a star around such a point reveals planets around other stars.

A body of finite size also feels how the field changes across it. The field $GM/d^2$ of a body of mass $M$ at the distance $d$ changes over the Earth's radius $R$ by about $R$ times its rate of change with distance, $2GMR/d^3$. For the Moon this is $1.1\times 10^{-6}m/{\mathrm{s}}^2$: the tidal acceleration, which stretches the oceans into two bulges, one facing the Moon and one on the far side. The Sun's field at the Earth is about 180 times the Moon's, yet its tidal effect is only about half the Moon's, because the difference falls as $1/d^3$ and the Sun is about 390 times farther away.

Other planets perturb every orbit; they advance Mercury's perihelion by about 530″ per century. The Earth's equatorial bulge turns the planes of satellite orbits, by about 1° a day for the near-polar orbits chosen to keep pace with the Sun. For three or more bodies there is no general formula, and planetary tables come from numerical integration.

Limits of validity

The inverse-square law has been confirmed from separations below a tenth of a millimetre up to planetary distances, and the equality of gravitational and inertial mass to about one part in $10^{15}$. Newtonian gravity is the weak-field, slow-motion limit of general relativity. The size of the corrections is set by the pure number $GM/(r{c}^2)$, with $c$ the speed of light: about $7\times 10^{-10}$ at the Earth's surface and $1\times 10^{-8}$ at 1 AU from the Sun.

Small as they are, the corrections are measured. Mercury's perihelion advances by an extra 43″ per century; the animation's planet, with $a=1\mathrm{AU}$ and $e=0.5$, would turn by about 5″ per century. Navigation-satellite clocks 20 200 km up gain about 46 microseconds a day on ground clocks because gravity is weaker there; their orbital speed takes back about 7, leaving a net gain of about 38. Near a neutron star, where $GM/(r{c}^2)\approx 0.2$, and at a black-hole horizon, where it reaches 0.5, Newton's law fails outright.

Gravity is also extremely weak: two electrons repel each other electrically about $4\times 10^{42}$ times more strongly than they attract each other gravitationally. It rules the sky only because mass comes in one sign, so it always adds up, while the positive and negative charges of large bodies cancel. At the Planck length, about $1.6\times 10^{-35}\mathrm{m}$, gravity and quantum mechanics would both matter, and no tested theory covers that regime.

Where this leads

The equal-area law is the conservation of angular momentum met in Angular Momentum and Its Conservation, and $U=-GMm/r$ is the general form of the $mgh$ of Work, Energy and Power. The same inverse square governs charges in Coulomb's Law, its field picture continues in Electric Field and Potential, and the electron moves on a Kepler orbit with quantised angular momentum in Atomic Models: the Bohr Atom.

Inside a uniform Earth $g\propto r$, a restoring pull that grows with distance from the centre, which is the condition for Simple Harmonic Motion: a train in a frictionless tunnel through the centre would cross in about 42 minutes, half the 84-minute period of an orbit grazing the surface. Finally, the equality of gravitational and inertial mass is the starting point of Gravity as Curved Spacetime, where orbits become the straightest paths through curved spacetime.