State Kepler's Three Laws
- State the Law of Ellipses, Law of Equal Areas, and Law of Harmonies
- Explain the physical meaning of each law
- Apply these laws to planetary and satellite orbits
Derive Kepler's Third Law
- Derive $T^2 = (4\pi^2/GM)r^3$ from Newton's Law of Universal Gravitation
- State and justify all assumptions made in the derivation
- Explain why $T^2/r^3$ is constant for a given central mass
Apply Orbital Mechanics
- Apply orbital mechanics to planetary and satellite systems
- Calculate orbital periods, speeds, and transfer times
- Connect Kepler's laws to exoplanet detection methods
State Kepler's three laws in your own words. Don't look at any notes.
Kepler's Laws & Orbital Mechanics
State Kepler's three laws, derive the third law from Newton's law of gravitation, and apply orbital mechanics to real astronomical systems.
Kepler's First Law — Law of Ellipses
Planets orbit the Sun in elliptical paths with the Sun at one focus
Keplers Laws
Keplers Detailed
Kepler's First Law states that planets orbit the Sun in elliptical paths with the Sun located at one focus of the ellipse. An ellipse has two foci, and the sum of the distances from any point on the ellipse to the two foci is constant.
The shape of an ellipse is described by its eccentricity $e$, defined as:
Orbital Eccentricity
Eccentricity tells us how "stretched" the ellipse is:
- $e = 0$: the orbit is a perfect circle
- $e$ close to 0: the orbit is nearly circular
- $e$ approaching 1: the orbit is highly elongated
Most planetary orbits in our solar system are nearly circular. Here are some values:
| Planet | Eccentricity $e$ |
|---|---|
| Earth | 0.017 |
| Venus | 0.007 |
| Jupiter | 0.049 |
| Mercury | 0.206 |
| Mars | 0.094 |
For the purposes of HSC Physics, most orbits are approximated as circular. This means the Sun is treated as being at the centre of the orbit. This is a valid approximation because Earth's eccentricity ($e = 0.017$) is so small that the deviation from a circle is only about 1.7%.
Key terms for positions on an elliptical orbit:
- Perihelion: closest approach to the Sun ($r_\text{min} = a(1 - e)$)
- Aphelion: furthest distance from the Sun ($r_\text{max} = a(1 + e)$)
- Semi-major axis $a$: half the longest diameter of the ellipse (equals the orbital radius for circular orbits)
Kepler's Second Law — Law of Equal Areas
A line joining a planet to the Sun sweeps out equal areas in equal times
Kepler's Second Law states that a line joining a planet to the Sun sweeps out equal areas in equal times. This means planets move faster when closer to the Sun (perihelion) and slower when further from the Sun (aphelion).
This law is a direct consequence of the conservation of angular momentum. Since the gravitational force is always directed toward the Sun (a central force), there is no torque about the Sun, so angular momentum is conserved:
Conservation of Angular Momentum
For Earth's orbit, the effect is measurable but small due to the near-circular orbit:
- Perihelion distance: $r_p = 147.1 \times 10^6$ km (early January)
- Aphelion distance: $r_a = 152.1 \times 10^6$ km (early July)
- Speed ratio: $v_p / v_a = r_a / r_p = 152.1 / 147.1 = 1.034$
Earth moves about 3.4% faster at perihelion than at aphelion. This is why the period from autumn equinox to spring equinox (when Earth is near perihelion) is slightly shorter than the period from spring to autumn.
Worked Example — Earth's Orbital Speed at Perihelion and Aphelion
Calculate Earth's orbital speed at perihelion ($r_p = 147.1 \times 10^6$ km) and aphelion ($r_a = 152.1 \times 10^6$ km). The average orbital speed is $29.78$ km/s.
$r_p = 147.1 \times 10^9$ m, $r_a = 152.1 \times 10^9$ m, $v_\text{avg} = 29.78 \times 10^3$ m/s
Orbital speeds $v_p$ (perihelion) and $v_a$ (aphelion)
Use conservation of angular momentum: $v_p r_p = v_a r_a$. Also use the fact that the average speed corresponds to the semi-major axis $a = (r_p + r_a)/2$.
$a = \dfrac{147.1 \times 10^9 + 152.1 \times 10^9}{2} = 149.6 \times 10^9$ m
For circular orbit at $a$: $v = \sqrt{GM_\odot/a} = 29.78$ km/s.
From $v_p r_p = v_a r_a$ and conservation of angular momentum about the semi-major axis orbit speed, we use $v_p r_p = v_\text{circ} \cdot a$:
$v_p = \dfrac{v_\text{circ} \cdot a}{r_p} = \dfrac{(29.78 \times 10^3)(149.6 \times 10^9)}{147.1 \times 10^9}$
$$v_p = \frac{(29.78 \times 10^3)(149.6 \times 10^9)}{147.1 \times 10^9} = 3.029 \times 10^4 \text{ m/s} = 30.29 \text{ km/s}$$
$$v_a = \frac{v_p r_p}{r_a} = \frac{(30.29 \times 10^3)(147.1 \times 10^9)}{152.1 \times 10^9} = 2.930 \times 10^4 \text{ m/s} = 29.30 \text{ km/s}$$
$v_\text{perihelion} = 30.29$ km/s, $v_\text{aphelion} = 29.30$ km/s
Earth moves about 1 km/s faster at perihelion than at aphelion.
Kepler's Third Law — Law of Harmonies
The square of the orbital period is proportional to the cube of the semi-major axis
Kepler's Third Law states that the square of the orbital period $T$ is proportional to the cube of the semi-major axis $a$:
Kepler's Third Law
Derivation from Newton's Law of Universal Gravitation
For a circular orbit (valid approximation for low eccentricity), the gravitational force provides the centripetal force:
$$F_\text{grav} = F_\text{c}$$
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$
The orbital speed $v$ can be expressed in terms of the period $T$:
$$v = \frac{2\pi r}{T}$$
Substituting:
$$\frac{GMm}{r^2} = m \left(\frac{2\pi}{T}\right)^2 r$$
$$\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}$$
$$T^2 = \frac{4\pi^2}{GM} r^3$$
For elliptical orbits, $r$ is replaced by the semi-major axis $a$. This is the exact form of Kepler's Third Law.
The constant $T^2/a^3 = 4\pi^2/(GM)$ depends only on the central mass $M$, not on the orbiting body's mass $m$. This means all planets orbiting the Sun have the same value of $T^2/a^3$, all moons orbiting Jupiter have the same value of $T^2/a^3$, and all satellites orbiting Earth have the same value of $T^2/a^3$.
Worked Example — Jupiter's Orbital Period
Calculate Jupiter's orbital period given its average distance from the Sun is $r = 7.78 \times 10^{11}$ m.
$r = 7.78 \times 10^{11}$ m, $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, $M_\odot = 1.99 \times 10^{30}$ kg
Orbital period $T$
Use $T^2 = (4\pi^2/GM)r^3$ with $M = M_\odot$.
$$T^2 = \frac{4\pi^2 \times (7.78 \times 10^{11})^3}{(6.67 \times 10^{-11})(1.99 \times 10^{30})}$$
$$T^2 = \frac{4\pi^2 \times 4.707 \times 10^{35}}{1.327 \times 10^{20}}$$
$$T^2 = \frac{1.857 \times 10^{37}}{1.327 \times 10^{20}} = 1.399 \times 10^{17} \text{ s}^2$$
$$T = \sqrt{1.399 \times 10^{17}} = 3.74 \times 10^8 \text{ s}$$
$$T = \frac{3.74 \times 10^8}{3.156 \times 10^7 \text{ s/yr}} = \mathbf{11.9 \text{ years}}$$
This agrees well with the accepted value of 11.86 years.
Hohmann Transfer Orbits
The most energy-efficient way to transfer between circular orbits
A Hohmann transfer orbit is an elliptical orbit used to transfer between two circular orbits. It requires the minimum energy change of any transfer orbit and is the most efficient method for moving spacecraft between orbits.
The transfer orbit is an ellipse that is tangent to both the initial and final circular orbits:
- The periapsis (closest point) of the transfer ellipse touches the inner orbit
- The apoapsis (furthest point) of the transfer ellipse touches the outer orbit
Hohmann Transfer Orbit Properties
Hohmann transfers are used extensively in interplanetary missions. For example, missions to Mars typically use a Hohmann transfer, which takes about 8.5 months. The transfer requires two engine burns:
- First burn (at periapsis): accelerate to enter the transfer ellipse from the inner orbit
- Second burn (at apoapsis): accelerate again to circularise into the outer orbit
Worked Example — LEO to GEO Transfer
Calculate the transfer time from Low Earth Orbit ($r_1 = 6.8 \times 10^6$ m) to Geostationary Orbit ($r_2 = 4.22 \times 10^7$ m) using a Hohmann transfer orbit.
$r_1 = 6.8 \times 10^6$ m, $r_2 = 4.22 \times 10^7$ m, $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, $M_E = 5.97 \times 10^{24}$ kg
Transfer time $t_\text{transfer}$
Calculate semi-major axis of transfer ellipse, then use $t_\text{transfer} = \pi\sqrt{a^3/(GM)}$.
Step 1 — Semi-major axis:
$$a = \frac{r_1 + r_2}{2} = \frac{6.8 \times 10^6 + 4.22 \times 10^7}{2} = \frac{4.90 \times 10^7}{2} = 2.45 \times 10^7 \text{ m}$$
Step 2 — Transfer time:
$$t_\text{transfer} = \pi\sqrt{\frac{a^3}{GM}}$$
$$t_\text{transfer} = \pi\sqrt{\frac{(2.45 \times 10^7)^3}{(6.67 \times 10^{-11})(5.97 \times 10^{24})}}$$
$$t_\text{transfer} = \pi\sqrt{\frac{1.472 \times 10^{22}}{3.982 \times 10^{14}}} = \pi\sqrt{3.697 \times 10^7 \text{ s}^2}$$
$$t_\text{transfer} = \pi \times 6.080 \times 10^3 \text{ s} = 1.91 \times 10^4 \text{ s}$$
$t_\text{transfer} = 1.91 \times 10^4$ s $\approx$ 5.3 hours
This is the time to coast from LEO to GEO along the transfer ellipse. Two engine burns are required: one to enter the ellipse and one to circularise at GEO.
Essential Formulae — Kepler's Laws & Orbital Mechanics
Common Misconceptions
"Planetary orbits are perfect circles" — Right: Orbits are ellipses, though most planetary orbits have very low eccentricity and are nearly circular. Earth's eccentricity is only 0.017, but it is still an ellipse, not a circle.
"Kepler's laws only apply to planets" — Right: Kepler's laws apply to any orbiting system: moons around planets, artificial satellites around Earth, binary star systems, and exoplanets around their host stars. Any system where one body orbits another under gravity follows these laws.
Kepler's Third Law in the Search for Exoplanets
Kepler's Third Law is fundamental to the discovery and characterisation of exoplanets (planets orbiting stars other than our Sun). Two major detection methods both rely on this law:
1. The Radial Velocity Method detects the slight wobble of a star caused by an orbiting planet's gravity. As the star moves toward and away from us, its light is Doppler-shifted. The measured orbital period $T$ and the star's velocity amplitude, combined with the stellar mass $M_\star$ from spectral classification, allow astronomers to determine the planet's minimum mass using a form of Kepler's Third Law.
2. The Transit Method measures the slight, periodic dimming of a star when a planet passes in front of it. The transit gives the planet's radius directly, and the orbital period $T$ is measured from the timing between transits. With $M_\star$ known, Kepler's Third Law $T^2 = (4\pi^2/GM_\star)r^3$ gives the orbital radius.
Together, these methods allow astronomers to determine a planet's:
- Orbital period and semi-major axis (from $T$ and Kepler's Third Law)
- Planet radius (from transit depth)
- Planet mass (from radial velocity amplitude)
- Planet density (mass divided by volume)
- Habitability potential (from orbital distance and stellar luminosity)
As of 2024, over 5,500 exoplanets have been confirmed in the Milky Way, with thousands more candidates awaiting verification. Kepler's Third Law remains the cornerstone of every orbital characterisation.
Verify Kepler's Third Law
1 Verify Kepler's Third Law for Earth. Given $r = 1.50 \times 10^{11}$ m and $T = 1$ year $= 3.156 \times 10^7$ s, calculate $T^2/r^3$ and compare with $4\pi^2/(GM_\odot)$.
2 A hypothetical planet orbits the Sun with a period of 8 Earth years. Use Kepler's Third Law to find its orbital radius in metres and in AU (1 AU = $1.50 \times 10^{11}$ m).
3 Calculate the Hohmann transfer time from Earth's orbit ($r_E = 1.50 \times 10^{11}$ m) to Mars' orbit ($r_M = 2.28 \times 10^{11}$ m). Express your answer in days. ($M_\odot = 1.99 \times 10^{30}$ kg)
Kepler's Laws and Conservation Laws
Explain how each of Kepler's three laws is connected to a fundamental conservation law or principle from physics:
- Kepler's First Law and the nature of the gravitational force ($F \propto 1/r^2$)
- Kepler's Second Law and conservation of angular momentum
- Kepler's Third Law and Newton's Law of Universal Gravitation
📝 Copy into Books
▾Key Definitions
- Kepler's First Law: planets orbit in ellipses with Sun at one focus
- Kepler's Second Law: equal areas swept in equal times (conservation of angular momentum)
- Kepler's Third Law: $T^2 \propto r^3$ for all bodies orbiting the same central mass
- Eccentricity: $e = c/a$, measures how stretched an ellipse is ($0 = \text{circle}$)
- Hohmann transfer: most energy-efficient orbit between two circular orbits
Key Formulae
- $T^2 = (4\pi^2/GM)r^3$
- $v_p / v_a = r_a / r_p$ (from $mvr = \text{const}$)
- $e = c/a$
- $a_\text{transfer} = (r_1 + r_2)/2$
- $t_\text{transfer} = \pi\sqrt{a^3/(GM)}$
Important Points
- Most planetary orbits are nearly circular ($e_\text{Earth} = 0.017$)
- $T^2/r^3$ is constant for all bodies orbiting the same central mass
- Kepler's laws apply to all orbiting systems, not just planets
- Planets move fastest at perihelion, slowest at aphelion
- Hohmann transfers require minimum energy but take the longest time
Common Errors
- Treating orbits as perfect circles when the question specifies ellipses
- Forgetting the factor of $4\pi^2$ in Kepler's Third Law
- Confusing altitude $h$ with centre-to-centre distance $r = R + h$
- Thinking Kepler's laws only apply to planets
- Forgetting that planets move at variable speed
Revisit Your Answer
Now that you have studied Kepler's laws, revisit your earlier answers.
Kepler's First Law: Planets orbit the Sun in ellipses with the Sun at one focus. Eccentricity $e = c/a$ describes how elongated the ellipse is. Most planetary orbits have very low eccentricity.
Kepler's Second Law: A line joining a planet to the Sun sweeps out equal areas in equal times. This means planets move faster when closer to the Sun (perihelion) and slower when further (aphelion). This follows from conservation of angular momentum.
Kepler's Third Law: $T^2 \propto r^3$, or equivalently $T^2 = (4\pi^2/GM)r^3$. This can be derived from Newton's Law of Universal Gravitation combined with centripetal force. The constant $T^2/r^3$ is the same for all bodies orbiting the same central mass.
Has your understanding changed? Write a revised explanation:
1. Kepler's First Law states that planetary orbits are:
2. A planet moves fastest at:
3. Kepler's Third Law $T^2 \propto r^3$ applies to:
4. For a moon orbiting Earth, if its orbital radius were doubled, its period would:
5. The ratio $T^2/r^3$ for all planets orbiting the Sun:
State Kepler's three laws of planetary motion in your own words.
Derive Kepler's Third Law $T^2 = (4\pi^2/GM)r^3$ starting from Newton's Law of Universal Gravitation and the expression for centripetal force. State all assumptions.
Assess the importance of Kepler's Third Law in the discovery and characterisation of exoplanets. Describe how astronomers use this law to determine properties of planets orbiting distant stars.
Model Answers
▾Question 1 (2 marks)
First Law: Planets orbit the Sun in elliptical paths with the Sun at one focus. (0.5 marks)
Second Law: A line joining a planet to the Sun sweeps out equal areas in equal times. (0.5 marks)
Third Law: The square of the orbital period is proportional to the cube of the semi-major axis ($T^2 \propto r^3$). (0.5 marks)
Award an additional 0.5 marks for expressing any of the laws in the student's own words with correct physical meaning.
Question 2 (3 marks)
Equate gravitational force to centripetal force: $F_\text{grav} = F_\text{c}$ (0.5 marks)
$\dfrac{GMm}{r^2} = m\left(\dfrac{2\pi}{T}\right)^2 r$ (0.5 marks)
$\dfrac{GM}{r^2} = \dfrac{4\pi^2 r}{T^2}$ (0.5 marks)
$T^2 = \dfrac{4\pi^2 r^3}{GM} = \dfrac{4\pi^2}{GM}r^3$ (0.5 marks)
Assumptions: (1 mark for any two of)
- Circular orbit (valid for low eccentricity orbits)
- Central body is much more massive than the orbiting body ($M \gg m$)
- Negligible other forces (no perturbations from other planets, no atmospheric drag)
- Two-body problem (only the central mass and orbiting body considered)
Question 3 (4 marks)
Kepler's Third Law is fundamental to exoplanet science. (1 mark)
The transit method measures the orbital period $T$ directly from the timing between transits and the orbital radius $r$ (semi-major axis) from orbital geometry and stellar parameters. With $M_\star$ known from stellar classification, $T^2 = (4\pi^2/GM_\star)r^3$ confirms the mass of the central star and allows verification of the planet's orbital parameters. (1 mark)
The radial velocity method detects stellar wobble caused by an orbiting planet's gravity. The measured stellar velocity amplitude $K_\star$, combined with the period $T$, gives the planet's minimum mass via $m_p \sin i = (M_\star^2 P/2\pi G)^{1/3} \times K_\star \times \sqrt{1 - e^2}$. (1 mark)
Together, these methods characterise planet size (from transit depth), mass (from radial velocity), density (mass/volume), and habitability potential (from orbital distance relative to the habitable zone). Over 5,500 exoplanets have been confirmed using these techniques. (1 mark)
Mark Lesson 17 Complete
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