Motion in Orbits

Lessons

Senior 5 · Physics

Motion in Orbits

4 lessons

Lesson 3 of 4

Kepler's Laws of Planetary Motion

25 min

Rules hidden in a notebook

For more than twenty years the Danish astronomer Tycho Brahe measured the positions of the planets night after night, without a telescope, more accurately than anyone before him. After his death in 1601 his assistant Johannes Kepler spent years trying to fit the measurements of Mars to circles, and failed by a few minutes of arc. Rather than ignore the small error, he tried other shapes, and found three simple laws (published in 1609 and 1619) that every planet obeys. Kepler did not know why they were true. Newton's law of gravitation, from Lesson 1, explains all three.

Kepler's first law: elliptical orbits

Each planet moves in an ellipse with the Sun at one focus. An ellipse is the set of points whose distances from two fixed points, the foci, add up to a constant $2 a$ ; $a$ is the semi-major axis, which plays the role of the orbit's radius. The planets' ellipses are very nearly circles: the Earth is 147.1 million km from the Sun at its closest (perihelion, early January) and 152.1 million km at its furthest (aphelion, early July).

Kepler's second law: equal areas

The line joining a planet to the Sun sweeps out equal areas in equal times. So a planet moves fastest when it is closest to the Sun and slowest when it is furthest away. This follows from the fact that gravity points towards the Sun, so it exerts no torque about the Sun and the planet's angular momentum $m v r$ (at perihelion and aphelion, where $v$ is perpendicular to $r$ ) stays constant: $v_{p} r_{p} = v_{a} r_{a}$ .

Kepler's third law: the law of periods

The square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit: $T^2 \propto a^3$ .

Deriving the third law for a circular orbit

For a planet of mass $m$ in a circular orbit of radius $r$ about the Sun (mass $M$ ), gravity provides the centripetal force:

\frac{GMm}{r^2} = m\omega^2r = m\frac{4\pi^2}{T^2}r \quad\Longrightarrow\quad T^2 = \frac{4\pi^2}{GM}\,r^3

The constant $\frac{4\pi^2}{GM}$ depends only on the central body, so it is the same for every planet of the Sun, and (with a different value) for every moon of Jupiter or every satellite of the Earth. The planet's own mass cancels.

Definition

Kepler's laws: (1) each planet moves in an ellipse with the Sun at one focus; (2) the line from the Sun to a planet sweeps out equal areas in equal times; (3) the square of the orbital period is proportional to the cube of the semi-major axis, $T^2 = \frac{4\pi^2}{GM}a^3$ . Example: Mars, at 1.524 times the Earth's distance, has a period of $1.524^{1.5} = 1.88$ years.

Activity 5.1: verifying Kepler's third law

The table gives NASA's values for the mean distance $a$ from the Sun and the orbital period $T$ of six planets. The last three columns convert them to astronomical units (AU, the Earth's mean distance) and years, and test whether $\frac{T^2}{a^3}$ is constant.

Planet	$a$ / million km	$T$ / days	$a$ / AU	$T$ / years	$\frac{T^2}{a^3}$
Mercury	57.9	88	0.387	0.241	1.002
Venus	108.2	224.7	0.723	0.615	1.001
Earth	149.6	365.2	1.000	1.000	1.000
Mars	228.0	687	1.524	1.881	1.000
Jupiter	778.5	4,331	5.204	11.859	0.998
Saturn	1432.0	10,747	9.572	29.428	0.987

Within 2%, $\frac{T^2}{a^3} = 1$ in these units for every planet, from Mercury to Saturn, although their periods range over a factor of more than a hundred. A second test is to plot $\lg T$ against $\lg a$ : if $T = ka^{n}$ , then $\lg T = n\lg a + \lg k$ , a straight line of gradient $n$ .

The gradient is 1.5, so $T \propto a^{1.5}$ , that is $T^2 \propto a^3$ . Your task: plot the graph yourself on graph paper, measure its gradient, and use the Earth's values with $T^2 = \frac{4\pi^2}{GM}a^3$ to find the mass of the Sun.

Worked example 1: the mass of the Sun and of Jupiter

(a) The Earth orbits the Sun at $1.496 \times 10^{11}\,\mathrm{m}$ in $3.156 \times 10^7\,\mathrm{s}$ . Find the mass of the Sun. (b) Jupiter's moon Io orbits Jupiter at $4.22 \times 10^8\,\mathrm{m}$ every 1.769 days. Find the mass of Jupiter.

Solution

(a) $M = \frac{4\pi^2r^3}{GT^2} = \frac{4\pi^2 \times (1.496 \times 10^{11})^3}{6.67 \times 10^{-11} \times (3.156 \times 10^7)^2} = 1.99 \times 10^{30}\,\mathrm{kg}$ .

(b) $T = 1.769 \times 86\,400 = 1.528 \times 10^5\,\mathrm{s}$ ; $M = \frac{4\pi^2 \times (4.22 \times 10^8)^3}{6.67 \times 10^{-11} \times (1.528 \times 10^5)^2} = 1.90 \times 10^{27}\,\mathrm{kg}$ , about a thousandth of the Sun's mass. Any planet with a moon can be "weighed" this way.

Worked example 2: speeds at perihelion and aphelion

Using $v_{p} r_{p} = v_{a} r_{a}$ and NASA's distances (147.1 and 152.1 million km), by what percentage is the Earth faster at perihelion than at aphelion?

Solution

$\frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{152.1}{147.1} = 1.034$ : about 3.4% faster in early January than in early July. This is also why the northern winter half of the year is a few days shorter than the summer half.

Work and energy in orbital motion

For a circular orbit, $\frac{mv^2}{r} = \frac{GMm}{r^2}$ gives $v = \sqrt{\frac{GM}{r}}$ : satellites in higher orbits move more slowly. The energies are

E_k = \tfrac{1}{2}mv^2 = \frac{GMm}{2r}, \qquad U = -\frac{GMm}{r}, \qquad E = E_k + U = -\frac{GMm}{2r}

The total energy is negative (the satellite is bound) and equal to minus the kinetic energy. To move a satellite to a higher orbit, work must be done on it, $\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$ , yet it ends up moving more slowly: the extra work, and more besides, goes into potential energy. Conversely, a satellite that loses energy to air drag in a low orbit sinks lower and speeds up.

Worked example 3: raising an orbit

A 1000 kg satellite is moved from a circular orbit 400 km up ( $r_1 = 6.77 \times 10^6\,\mathrm{m}$ ) to a geostationary orbit ( $r_2 = 4.22 \times 10^7\,\mathrm{m}$ ). Find the energy needed and the change in speed. ( $GM = 3.99 \times 10^{14}\,\mathrm{m^3\,s^{-2}}$ .)

Solution

$\Delta E = \frac{3.99 \times 10^{14} \times 1000}{2}\left(\frac{1}{6.77 \times 10^6} - \frac{1}{4.22 \times 10^7}\right) = 2.0 \times 10^{17} \times 1.24 \times 10^{-7} = 2.5 \times 10^{10}\,\mathrm{J}$ .

$v_1 = \sqrt{\frac{3.99 \times 10^{14}}{6.77 \times 10^6}} = 7.67\,\mathrm{km\,s^{-1}}$ ; $v_2 = \sqrt{\frac{3.99 \times 10^{14}}{4.22 \times 10^7}} = 3.07\,\mathrm{km\,s^{-1}}$ . The satellite gains 25 GJ of energy but ends up moving 4.6 km/s slower.

Warning

Common mistakes. (1) Using the planet's or satellite's mass as $M$ in $T^2 = \frac{4\pi^2}{GM}r^3$ : $M$ is the mass of the body at the centre. (2) Forgetting to convert days or years to seconds and km to m. (3) Thinking a higher orbit means a faster satellite: $v = \sqrt{\frac{GM}{r}}$ falls with $r$ . (4) Saying the Earth is closest to the Sun in the northern summer: perihelion is in early January.

Tip

For comparisons around the same central body, skip $G$ and $M$ entirely: $\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$ . In AU and years, $T^{2} = a^{3}$ for anything orbiting the Sun.

Explained in 5 questions: Kepler's laws of planetary motion (Encyclopaedia Britannica). Watch for how Tycho Brahe's data led Kepler to ellipses, and what the equal-area law says about a planet's speed.

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Practice

State Kepler's three laws.
Derive $T^2 = \frac{4\pi^2}{GM}r^3$ for a circular orbit, and show that the orbiting body's mass does not matter.
An asteroid orbits the Sun at a mean distance of 2.77 AU. Find its period in years.
Use NASA's values for Mercury (perihelion 46.0, aphelion 69.8 million km) to find the ratio of its fastest to slowest orbital speed.
The Moon's orbit has radius $3.84 \times 10^8\,\mathrm{m}$ . Use $GM_E = 3.99 \times 10^{14}\,\mathrm{m^3\,s^{-2}}$ to find its period in days.
Find the speed and the total energy of a 500 kg satellite in a circular orbit of radius $7.0 \times 10^6\,\mathrm{m}$ around the Earth.
Plot $\lg T$ against $\lg a$ for the six planets in the table, measure the gradient, and explain what it shows.
A satellite in a low orbit gradually loses energy to friction with the thin upper atmosphere. Evaluate the claim that "it therefore slows down".

Show answersHide answers

Elliptical orbits with the Sun at a focus; equal areas in equal times; $T^2 \propto a^3$ .
$\frac{GMm}{r^2} = m\frac{4\pi^2r}{T^2}$ ; $m$ cancels, giving $T^2 = \frac{4\pi^2r^3}{GM}$ .
$T = 2.77^{1.5} = 4.61$ years.
$\frac{v_p}{v_a} = \frac{69.8}{46.0} = 1.52$ .
$T = 2\pi\sqrt{\frac{(3.84 \times 10^8)^3}{3.99 \times 10^{14}}} = 2.37 \times 10^6\,\mathrm{s} = 27.4$ days.
$v = \sqrt{\frac{3.99 \times 10^{14}}{7.0 \times 10^6}} = 7.55\,\mathrm{km\,s^{-1}}$ ; $E = -\frac{3.99 \times 10^{14} \times 500}{2 \times 7.0 \times 10^6} = -1.43 \times 10^{10}\,\mathrm{J}$ .
A straight line of gradient 1.5: $T \propto a^{1.5}$ , so $T^2 \propto a^3$ , Kepler's third law.
The claim is wrong. Losing energy makes $E = -\frac{GMm}{2r}$ more negative, so $r$ decreases; and $v = \sqrt{\frac{GM}{r}}$ then increases. The satellite spirals down and speeds up, because it gains more kinetic energy from falling than it loses to friction.

Key points

Kepler: ellipses with the Sun at a focus; equal areas in equal times ( $v_{p} r_{p} = v_{a} r_{a}$ ); $T^2 \propto a^3$ .
Circular orbit: $T^2 = \frac{4\pi^2}{GM}r^3$ , so orbital data give the central mass.
$v = \sqrt{\frac{GM}{r}}$ ; $E_k = \frac{GMm}{2r}$ , $U = -\frac{GMm}{r}$ , $E = -\frac{GMm}{2r}$ .
A $\lg T$ – $\lg a$ graph of the planets has gradient 1.5.

Sources and credits

13.5 Kepler's Laws of Planetary Motion and 13.4 Satellite Orbits and Energy, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Planetary data: Planetary Fact Sheet, NASA Space Science Data Coordinated Archive (US government work, public domain).
Video: Explained In 5 Questions: Kepler's Law of Planetary Motion, Encyclopaedia Britannica, embedded from YouTube.

The activity, the three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Rules hidden in a notebook

2Kepler's first law: elliptical orbits

3Kepler's second law: equal areas

4Kepler's third law: the law of periods

Deriving the third law for a circular orbit

5Activity 5.1: verifying Kepler's third law

6Work and energy in orbital motion

Rules hidden in a notebook

Kepler's first law: elliptical orbits

Kepler's second law: equal areas

Kepler's third law: the law of periods

Activity 5.1: verifying Kepler's third law

Work and energy in orbital motion