Kepler's Laws

An orbit is not a circle, and that matters

In the last lesson you set gravity equal to the centripetal force and got a clean result: a satellite in a circle of radius $r$ around a mass $M$ moves at $v = \sqrt{GM/r}$. Every orbit in that lesson was a circle, and for a circle speed and distance never change, so there was nothing to track.

Real orbits are not circles. Halley's comet comes closer to the Sun than Venus and then goes out past Neptune. Somewhere in between it must speed up and slow down enormously, and the radius that appeared in $v=\sqrt{GM/r}$ is no longer a single number. So you need three things a circle never asked for: a shape, a rule for how fast the body moves at each point of that shape, and a period that still makes sense when there is no one radius.

Kepler found all three by staring at Tycho Brahe's naked-eye positions of Mars, decades before Newton could explain any of them. They are the three laws, and between them they answer almost every orbit question you will be asked.

The shape: ellipses, and the two numbers that fix them

Kepler's first law: each planet moves on an ellipse with the Sun at one focus. Not at the centre: at a focus, which is off to one side.

An ellipse has two special interior points, the foci. The defining property is that the sum of the distances from the two foci to any point on the curve is the same everywhere on the curve. Two numbers pin the ellipse down completely:

• $a$, the semi-major axis: half the long diameter. This is the size.
• $e$, the eccentricity, a pure number between 0 and 1. This is the shape.

The centre of the ellipse and either focus are separated by a distance $c = ae$. So $e$ measures how far off-centre the Sun sits, as a fraction of $a$. If $e=0$ the two foci merge at the centre and you have a circle. As $e$ climbs toward 1 the ellipse stretches into a sliver.

Two distances get used constantly. The closest point to the Sun is perihelion, on the near end of the long axis, at distance $a - c$. The farthest is aphelion, at $a + c$. So

$r_p = a(1-e), \qquad r_a = a(1+e)$

(For Earth satellites the words are perigee and apogee; the geometry is identical.) Add those two and the $e$ cancels:

$r_p + r_a = 2a$

That is the single most useful line on this page. Whenever a problem hands you the closest and the farthest distance, you already have $a$: it is their average. Subtract them instead and you get $r_a - r_p = 2ae$, so

$e = \frac{r_a - r_p}{r_a + r_p}$

The second law: equal areas, which is angular momentum

Kepler's second law: the line from the Sun to the planet sweeps out equal areas in equal times. Near perihelion the line is short, so to sweep the same area the planet must race around; near aphelion the line is long and the planet crawls.

Here is why it is true. In a short time $dt$ the planet moves by $\vec v\,dt$. The region swept is very nearly a triangle with one vertex at the Sun and two sides $\vec r$ and $\vec v\,dt$. The area of a triangle built on two vectors is half the magnitude of their cross product, so

$\frac{dA}{dt} = \tfrac12\,|\vec r \times \vec v| = \frac{L}{2m}$

where $L = m|\vec r \times \vec v|$ is the angular momentum about the Sun and $m$ is the planet's mass. Gravity pulls the planet straight back along $\vec r$ toward the Sun. A force parallel to $\vec r$ has zero torque about that point, so $L$ never changes, and therefore neither does $dA/dt$. Equal areas in equal times is conservation of angular momentum wearing a geometric disguise.

Because the argument only used "the force points at the Sun", the second law holds for any central force. The first and third laws need the force to be exactly inverse-square; the second does not.

The practical payoff comes at the two ends of the orbit. At perihelion and at aphelion the velocity is perpendicular to $\vec r$ (that is what it means to be at the closest or farthest point: the distance is momentarily not changing, so there is no radial velocity). With $\vec r \perp \vec v$, the cross product is just $rv$, and equating $L$ at the two ends:

$v_p r_p = v_a r_a \qquad\Longrightarrow\qquad \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{1+e}{1-e}$

The third law: period from size alone

Kepler's third law: $T^2 \propto a^3$. The period depends on the size of the orbit and on nothing else, not on the eccentricity, not on the orbiting body's mass.

For a circle you can derive it in two lines, and the result turns out to be exactly right for ellipses too if you read $r$ as $a$. Gravity supplies the centripetal force on a body of mass $m$ in a circle of radius $r$:

$\frac{GMm}{r^2} = m\omega^2 r$

The $m$ cancels, which is already the reason the law knows nothing about the planet. Put in $\omega = 2\pi/T$:

$\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2} \qquad\Longrightarrow\qquad T^2 = \frac{4\pi^2}{GM}\,a^3$

Read that in words. $T^2/a^3$ equals $4\pi^2/GM$, a constant that depends only on the central mass. Every object orbiting the same central body shares it. That is what makes the law a measuring instrument: watch a moon go round, time it, measure its distance, and you have weighed the planet without going there.

For solar-system work there is a shortcut that removes $G$ and $M$ entirely. Measure $a$ in astronomical units (1 AU is Earth's semi-major axis) and $T$ in years. Earth has $a=1$ and $T=1$, so the constant is 1, and

$T^2 = a^3 \quad\text{(AU and years, Sun at the focus)}$

so $T = a^{3/2}$. Most of the easy orbit problems you will meet are one press of this on a calculator.

Worked example 1: the period of Ceres

Ceres orbits the Sun with $a = 2.77\ \mathrm{AU}$. Find $T$.

Everything is already in AU and the central body is the Sun, so use $T=a^{3/2}$ directly. Take it in two steps so you can see the arithmetic:

$\sqrt{2.77} = 1.6643, \qquad T = 1.6643 \times 2.77 = 4.61\ \text{years}$

Going backwards is just as easy. If a comet has $T = 27.0$ years, then $a = T^{2/3} = 27^{2/3} = (27^{1/3})^2 = 3^2 = 9.00\ \mathrm{AU}$.

Worked example 2: how many orbits a day

A satellite orbits Earth at an altitude of $700\ \mathrm{km}$. Using $GM_E = 3.983\times10^{14}\ \mathrm{m^3/s^2}$ and Earth's radius $6.37\times10^6\ \mathrm{m}$, how many complete orbits does it make in a day?

First the trap: orbital radius is measured from the centre of the Earth, not from the ground. Altitude is not radius.

$r = 6.37\times10^6 + 0.700\times10^6 = 7.07\times10^6\ \mathrm{m}$

Now the SI form of the third law, solved for $T$:

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

$r^3 = (7.07\times10^6)^3 = 3.534\times10^{20}\ \mathrm{m^3}$

$\frac{r^3}{GM} = \frac{3.534\times10^{20}}{3.983\times10^{14}} = 8.873\times10^{5}\ \mathrm{s^2}$

$T = 2\pi\sqrt{8.873\times10^5} = 2\pi(941.9) = 5919\ \mathrm{s}$

That is about 99 minutes, which should feel right for low Earth orbit. In one day:

$\frac{86400}{5919} = 14.6 \text{ orbits}$

Speed at any point: vis-viva

The second law relates the two ends of an orbit to each other, and the third law gives the period. Neither one tells you the speed at a general point. For that you need energy.

The total energy of a body of mass $m$ in orbit is kinetic plus gravitational potential, $E = \tfrac12 mv^2 - GMm/r$, and it is conserved. What is not obvious, but is true for every ellipse, is that this energy depends only on the semi-major axis:

$E = -\frac{GMm}{2a}$

You can check it fast on a circle, where $r=a$ and $v^2 = GM/r$: $E = \tfrac12 m(GM/a) - GMm/a = -GMm/2a$. Setting the two expressions for $E$ equal and cancelling $m$ gives the vis-viva equation:

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

This is the workhorse. Given the central mass, your current distance $r$, and the size $a$ of the orbit you are on, it hands you your speed. Two special cases are worth noticing. On a circle $r=a$ and it collapses to $v=\sqrt{GM/r}$, the result from the last lesson. And $a\to\infty$ gives $v = \sqrt{2GM/r}$, escape speed: an infinitely large orbit is one that never comes back.

Turned around, vis-viva also finds the orbit from the motion. If you know $r$ and $v$ at one instant, solve for $a$, and then $r_a = 2a - r$ gives the far end.

Worked example 3: apogee of a fast launch

A projectile is launched horizontally from Earth's surface at $9000\ \mathrm{m/s}$ (ignore the atmosphere). How far from Earth's centre does it get? Use $GM_E = 3.983\times10^{14}$, $R_E = 6.37\times10^6\ \mathrm{m}$.

Horizontal launch means the velocity is perpendicular to $\vec r$, so the launch point is a turning point of the orbit: here, the perigee. Find $a$ from vis-viva at that point.

$\frac{1}{a} = \frac{2}{r} - \frac{v^2}{GM}$

$\frac{2}{r} = \frac{2}{6.37\times10^6} = 3.140\times10^{-7}\ \mathrm{m^{-1}}$

$\frac{v^2}{GM} = \frac{(9000)^2}{3.983\times10^{14}} = \frac{8.10\times10^7}{3.983\times10^{14}} = 2.034\times10^{-7}\ \mathrm{m^{-1}}$

$\frac{1}{a} = 3.140\times10^{-7} - 2.034\times10^{-7} = 1.106\times10^{-7}$

$a = 9.04\times10^{6}\ \mathrm{m}$

The apogee is the other end of the long axis, and the two ends add to $2a$:

$r_a = 2a - r_p = 1.808\times10^{7} - 6.37\times10^{6} = 1.17\times10^{7}\ \mathrm{m}$

Check the sign of the story: $9000\ \mathrm{m/s}$ is faster than the circular speed at the surface ($\sqrt{GM/R} = 7906\ \mathrm{m/s}$) and slower than escape ($11180\ \mathrm{m/s}$), so it should rise to an apogee and come back. It does.

Worked example 4: the burn that raises an apogee

A satellite is in a circular orbit of radius $7.00\times10^6\ \mathrm{m}$ and fires its engine to enter an ellipse with apogee $4.20\times10^7\ \mathrm{m}$. How big a speed change does it need? Use $GM_E = 3.983\times10^{14}$.

The engine burns briefly, so the satellite stays at $r = 7.00\times10^6$ during the burn; only its speed changes. Get the speed before and the speed after at that same radius, and subtract.

Before, on the circle:

$v_1 = \sqrt{\frac{3.983\times10^{14}}{7.00\times10^6}} = \sqrt{5.690\times10^{7}} = 7543\ \mathrm{m/s}$

After, on the ellipse. The burn point becomes the perigee, and the apogee is given, so

$a = \tfrac12(7.00\times10^6 + 4.20\times10^7) = 2.45\times10^{7}\ \mathrm{m}$

$v_2^2 = GM\left(\frac{2}{7.00\times10^6} - \frac{1}{2.45\times10^7}\right) = 3.983\times10^{14}\,(2.857\times10^{-7} - 0.408\times10^{-7})$

$v_2^2 = 3.983\times10^{14} \times 2.449\times10^{-7} = 9.754\times10^{7}$

$v_2 = 9877\ \mathrm{m/s}$

$\Delta v = 9877 - 7543 = 2333\ \mathrm{m/s}$

Both velocities point the same way (along the track) so the speeds subtract directly. If the burn had been at an angle you would have to subtract the vectors instead.

Transfer orbits and degenerate ones

Two classic setups are just the three laws pointed at an unusual question.

A Hohmann transfer moves a spacecraft between two circular orbits using an ellipse that is tangent to both: its perihelion sits on the inner circle and its aphelion on the outer one. So $2a = r_\text{inner} + r_\text{outer}$, and because you travel from one end of the long axis to the other, the trip takes exactly half a period.

Worked example 5: how long to Mars

A spacecraft leaves Earth's orbit ($1.00\ \mathrm{AU}$) on an ellipse whose aphelion just touches Mars's orbit ($1.524\ \mathrm{AU}$). How long is the crossing?

$a = \frac{1.000 + 1.524}{2} = 1.262\ \mathrm{AU}$

$T = a^{3/2}: \quad \sqrt{1.262} = 1.1234, \quad T = 1.1234 \times 1.262 = 1.4177\ \text{years}$

Half of that is the trip:

$t = \tfrac12(1.4177) = 0.709\ \text{years} \approx 259\ \text{days}$

The other setup is the degenerate ellipse. Stop a planet dead in its orbit and let it fall straight into the Sun. That radial line looks like nothing at all, but it is the limit $e \to 1$ of an ellipse: a path that goes out to a maximum distance $r_\text{max} = 2a$ and comes back. So the fall starts at "aphelion" with $r_a = 2a$, giving $a = r_\text{max}/2$, and the third law still applies. The fall to the centre is half of that orbit, so it takes $T/2$.

Earth stopped at $1.00\ \mathrm{AU}$ and released has $a = 0.500\ \mathrm{AU}$, so $T = 0.5^{3/2} = 0.3536$ years, and the fall takes $0.1768$ years, about 65 days. No integration required.

Weighing things with the third law

Rearrange $T^2 = 4\pi^2 a^3/GM$ for the mass:

$M = \frac{4\pi^2 a^3}{G T^2}$

In SI that needs $G$ and careful powers of ten. But there is a version with no constants at all. Since the Sun with $a$ in AU and $T$ in years gives $T^2 = a^3$, a central mass of $M$ solar masses gives $T^2 = a^3/M$, so

$M = \frac{a^3}{T^2} \quad \text{(solar masses, AU, years)}$

A star orbiting a galactic centre at $a = 1000\ \mathrm{AU}$ with $T = 15.0$ years gives $M = 10^9/225 = 4.44\times10^6$ solar masses. That is how the black hole at the centre of the Milky Way was weighed.

One caveat for binaries. When the two masses are comparable, both orbit their common centre of mass, and what the law returns is the sum $M_1 + M_2$, with $a$ the separation between them. Two stars with $T=50.0$ years and a separation of $20.0\ \mathrm{AU}$ have a combined mass of $20^3/50^2 = 8000/2500 = 3.2$ solar masses.

There is one more disguise worth spotting. For an orbit skimming the surface of a uniform sphere of density $\rho$, put $M = \tfrac43\pi R^3 \rho$ and $a = R$ into the third law:

$T^2 = \frac{4\pi^2 R^3}{G \cdot \tfrac43 \pi R^3 \rho} = \frac{3\pi}{G\rho}$

The radius cancels completely. A surface-skimming orbit has a period that depends only on density. For rock at $5500\ \mathrm{kg/m^3}$, $T = \sqrt{3\pi/(6.67\times10^{-11}\times5500)} = \sqrt{2.569\times10^{7}} = 5070\ \mathrm{s}$, about 84 minutes, which is why every low orbit around every rocky body takes roughly an hour and a half.

How to recognise which law you need

Read the problem for what it gives you and what it asks.

Two distances and a period, or a period and a distance. Third law. Convert to AU and years if the Sun is at the centre and skip $G$ entirely; use $T = 2\pi\sqrt{a^3/GM}$ if it is a planet. If the two distances are the ends of an orbit, average them for $a$ first.

Two speeds, or a speed at one end and a question about the other end. Second law: $v_p r_p = v_a r_a$. The word "perihelion", "aphelion", "perigee", "apogee", "closest", or "farthest" is the signal, because only at those points is $\vec v$ perpendicular to $\vec r$ and the relation that simple.

A speed at a general point, or a $\Delta v$, or "how far out does it get". Vis-viva. Any problem with a rocket burn, a launch speed, or a maximum distance wants $v^2 = GM(2/r - 1/a)$, usually run once to find $a$ and once to find a speed.

A mass you are asked to find. Third law rearranged. Check whether the units let you use $M = a^3/T^2$ in solar masses.

"Eccentricity" in the question. Convert immediately to $r_p = a(1-e)$ and $r_a = a(1+e)$, then treat it as a two-distance problem. Eccentricity almost never appears in a final formula; it is a way of stating the two distances.

One structural warning. The third law never involves speed and the second law never involves the period. If your chain of reasoning needs both a speed and a time, you almost certainly need vis-viva or energy in the middle.

The thread running through all of it is that the orbit is fixed by its energy and its angular momentum: energy sets $a$, and the pair together set $e$. In the next lesson you will look properly at the quantity that has been quietly doing the work here, in Work and the work-energy theorem.