Third Law of Kepler

Did you know that Halley's Comet has a periodic orbit that makes it pass nearby the Earth once every 75 years? By directly applying Kepler's Third Law we can estimate how close it will pass by the Earth. Every satellite and planet in orbit follows the Third Law of Kepler, which allows us to describe the period of the orbit in an accurate manner. In this article, we will discuss and prove Kepler's Third Law.

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    Third Law of Kepler An image of Halley's Comet StudySmarterFig. 1 - This is an image of Halley's Comet. Kepler's Third Law applies to the comet.

    Kepler's Laws of Planetary Motion

    In the 1500s astronomer, Nicolaus Copernicus hypothesized that the Sun was stationary at the center of our universe, with the other planets, such as Earth, orbiting around it with circular revolutions. One of the major improvements to this model was originated by Johannes Kepler, who used astronomical observations to show that instead of having circular orbits, the planets actually followed an elliptical motion around the Sun.

    These orbital trajectories were described by Kepler's Laws, which we will be discussing and expanding upon in this article. Even though these were originally formulated for the planets in the Solar System orbiting the Sun, they can be applied universally. They can be applied to exoplanets around other stars, as well as describing the motion of a satellite orbiting a planet. However, if we want to apply Kepler's Laws we must make sure that the center of mass of the satellite-planet system is inside the central body and that the satellite is much less massive than the planet's center of mass.

    Kepler's First Law states that planets move in elliptical orbits with the sun at its focus.

    Third Law of Kepler Diagram of an ellipse StudySmarterFig. 2 - Elipses have two focus points as well as a semi-major axis and a semi-minor axis.

    Kepler's Second Law states that the distance vector between the planet and the Sun sweeps equal areas of the ellipse at equal times.

    In a circular orbit, the satellite will move at the same speed throughout the orbit. However, according to Kepler's Second Law, in an elliptical orbit, a satellite travels faster when it is nearer the planet and moves slower when farthest away from the planet. Third Law of Kepler A graphic showcasing the area swept out by elliptical orbits StudySmarterFig. 3 - A visual representation of Kepler's Second Law. Notice how equal areas of space are swept across equal periods of time.

    Third Law of Kepler: Definition

    Kepler was so amazed by his finding that he stated the following in his book Harmonies of the World (1619):

    I first believed I was dreaming... But it is absolutely certain and exact that the ratio which exists between the period times of any two planets is precisely the ratio of the 3/2th power of the mean distance."

    Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of the ellipse.


    The planets in the solar system exhibit different orbital periods. For example, let's compare the orbital periods of Venus and Mercury. Venus' orbital period is 224 Earth days and Mercury's orbital period is 88 Earth days. We are familiar with describing orbital periods in Earth days, but we have to remember that days last 24 hours only on planet Earth. Some planets even have days longer than a year. An example of this is the planet Venus. It takes 224 Earth days for Venus to orbit the Sun. In comparison, it takes 243 Earth days for Venus to complete a rotation about its axis.

    Why is it that different planets have different orbital periods? If we look at the distances of the respective planets to the Sun, we see that Mercury is the closest planet to the Sun. It, therefore, has the shortest orbital period of the planets.

    Third Law of Kepler: Formula

    Kepler's Third Law explains that the period for a satellite to orbit Earth increases rapidly with the radius of its orbit. For an elliptical orbit, Kepler's Third Law can be expressed as follows,

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

    where \(T\) is the orbit's period in seconds \(\mathrm s\), \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(M\) is the planet's mass in \(\text{kg}\), and \(a\) is the semi-major axis in \(\text{m}\). The semi-major axis is equal in length to half of the longest diameter of an ellipse. In the case of a circular orbit, the semi-major axis can be replaced with the radius of the orbit.

    Third Law of Kepler A diagram visualizing Kepler's third law StudySmarterFig. 4 - This is a diagram visualizing Kepler's Third Law. According to Kepler's Third Law, the ratio of the square of the orbital period with the cube of the semi-major axis of its orbit is a constant.

    Proving the Third Law of Kepler

    This relationship may be obtained in the case of a circular orbit using Newton's Law of Gravitation. To find it, we can equal the centripetal force of the satellite to the gravitational force between the planet and the orbiting object.

    Newton's Law of Gravitation is given by

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

    \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(M\) is the planet's mass in kilograms \(\mathrm{kg}\), \(m\) is the orbiting object's mass in kilograms \(\mathrm{kg}\), and \(r\) is the distance between them in meters \(\mathrm m\).

    The centripetal force is expressed below:

    $$F=m\omega^2r,$$

    where we define \(m\) as the orbiting object's mass in kilograms \(\mathrm{kg}\), while \(\omega\) is the angular velocity in radians per second \(\frac{\mathrm{rad}}{\mathrm s}\), and \(r\) is the distance between the planet and the orbiting object in meters \(\mathrm m\).

    Now we equal the centripetal force of the satellite to the gravitational force between the planet and the Sun,

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

    We can express the angular velocity in terms of the orbital period and then rearrange the equation:

    $$\begin{align*}\frac{GMm}{r^2}&=mr\left(\frac{2\pi}T\right)^2,\\mr\frac{4\pi^2}{T^2}&=\frac{GMm}{r^2}.\end{align*}$$

    If we solve for \(T^2\), we find the expression for the Third Law of Kepler:

    $$T^2=\left(\frac{4\pi^2}{GM}\right)r^3.$$

    Since the value in the parenthesis is a constant, we can see that \(T^2\) is proportional to \(r^3\).

    Instead of circles, a more comprehensive derivation may be made using generic elliptical orbits. As a result, a circle's radius \(r\) is replaced by the semi-major axis \(a\) of an ellipse.

    As a result, the formula of Kepler's Third Law for elliptical orbits is given by

    $$T^2=\left(\frac{4\pi^2}{GM}\right)a^3.$$

    We can also derive the Third Law of Kepler by examining the Second Law of Kepler and using calculus. The Second Law of Kepler is given by

    $$\frac{{\text{d}}A}{{\text{d}}t}=\frac{L}{2m},$$

    where \(L\) is the angular momentum and \(m\) is the mass of the orbiting planet. Since the whole point of the Second Law of Kepler is that the orbit sweeps equal areas of the ellipse at equal times, the right-hand side of the equation above is constant. The area swept out in an orbit can be expressed in terms of the period of the orbit \(T\):

    $$A={\frac{L}{2m}}T.$$

    We know the total area of an ellipse is given by:

    $$A=\pi{a}{b},$$

    where \(a\) is the semi-major axis and \(b\) is the semi-minor axis of the ellipse. These quantities are related to the eccentricity \(e\) that determines the orbit shape:

    $$b^2={a^2}\left(1-e^2\right).$$

    Now we express the area swept out in an orbit as:

    $$\frac{\pi{a}{b}}{T}=\frac{L}{2m}.$$

    Next, we will examine the orbit equation for the case of a mass \(m\) orbiting a central mass \(M\):

    $$r\left(\theta\right)=\frac{L^2}{GMm^2\left(1+e\cos{\theta}\right)}.$$

    The orbit equation accurately predicts the position of mass \(m\) at any point in the orbit and angle \(\theta\) with respect to the semi-major axis. Let's examine the case of aphelion, which the largest distance between the two masses. At aphelion, \(\theta=0\) and \(r=a\left(1-e\right)\), so the orbit equation becomes:

    $$a\left(1-e\right)=\frac{L^2}{GMm^2\left(1+e\right)}.$$

    We can solve this equation to find an expression for the square of the angular momentum \(L\) that allows us to substitute into the equation for the area swept out in an orbit:

    $$\begin{align*}\frac{L^2}{m^2}&=a\left(1-e\right)GM\left(1+e\right),\\\frac{L^2}{m^2}&=a\left(1-e^2\right)GM.\end{align*}$$

    Now we square both sides of the equation for the area swept out in an orbit and substitute the obtained equations for \(\frac{L^2}{m^2}\) and \(b^2\):

    $$\begin{align*}\frac{{\pi}^2{a}^2{b}^2}{T^2}&=\frac{L^2}{4m^2},\\\frac{{\pi}^2{a}^2\bcancel{\left(1-e^2\right)}}{T^2}&=\frac{a\bcancel{\left(1-e^2\right)}GM}{4},\\\frac{4{\pi}^2{a^3}}{GM}&=T^2\end{align*}$$

    We have proved the Third Law of Kepler.

    Satellite \(A\) has an orbital period of four days \(\left(T_{\text{A}}=4\,\text{d}\right)\). Satellite \(B\) is also in orbit, but all we know is that its radius is twice of satellite \(A\). Determine the orbital period of Satellite \(B\).

    Third Law of Kepler Satellites A and B around a planet StudySmarterFig. 5 - Satellites \(A\) and \(B\) are around the planet. Satellite \(B\) has twice the radius of satellite \(A\).

    Satellite \(A\) has a radius \(r\), while Satellite \(B\) has a radius \(2r\). If we know the period of Satellite \(A\), then we can find the period of Satellite \(B\) using Kepler's Third Law equation.

    According to Kepler's third law, \(\frac{r^3}{T^2}\) is equal to a constant, as both satellites are orbiting the same planet. So, the proportionality will be equal for both satellites.

    For Satellite \(A\):

    $$\frac{r^3}{\left(4\;\mathrm{d}\right)^2}$$

    For Satellite \(B\):

    $$\frac{{(2r)}^3}{T_{\text{B}}}$$

    Now we can determine the orbital period for Satellite \(B\).

    $$\begin{align*}\frac{r^3}{\left(4\;\mathrm{d}\right)^2}&=\frac{\left(2r\right)^3}{\left(T_{\text{B}}\right)^2},\\\frac{r^3}{16\;\mathrm{d}^2}&=\frac{8r^3}{\left(T_{\text{B}}\right)^2},\\\frac{\cancel{r^3}}{16\;\mathrm{d}^2}&=\frac{8\cancel{r^3}}{\left(T_{\text{B}}\right)^2},\\\left(T_{\text{B}}\right)^2&=128\;\mathrm{d}^2,\\\sqrt{\left(T_{\text{B}}\right)^2}&=\sqrt{128\;\mathrm{d}^2},\\T_{\text{B}}&=11.3\;\mathrm{d}.\end{align*}$$

    Third Law of Kepler - Key takeaways

    • Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of the ellipse, \(T^2=\left(\frac{4\pi^2}{GM}\right)r^3\).
    • This relationship may be obtained in the case of a circular orbit using Newton's Law of Gravitation. To find it, we can equal the centripetal force of the satellite to the gravitational force between the planet and the orbiting object.
    • It applies to all objects orbiting around a central mass.

    References

    1. Fig. 1 - This is an image of Halley's Comet. Kepler's Third Law applies to the comet (https://commons.wikimedia.org/wiki/File:Comet_Halley_close_up-cropped.jpg), by ESA (https://www.esa.int/spaceinimages/Images/2012/11/Comet_Halley_close_up), licensed by CC BY-SA 3.0 IGO (https://creativecommons.org/licenses/by-sa/3.0/igo/deed.en)
    2. Fig. 2 - Elipses have two focus points as well as a semi-major axis and a semi-minor axis, StudySmarter Originals
    3. Fig. 3 - A visual representation of Kepler's Second Law. Notice how equal areas of space are swept across equal periods of time, StudySmarter Originals
    4. Fig. 4 - This is a diagram visualizing Kepler's Third Law. According to Kepler's Third Law, the ratio of the square of the orbital period with the cube of the semi-major axis of its orbit is a constant, StudySmarter Originals
    5. Fig. 5 - Satellites \(A\) and \(B\) are around the planet. Satellite \(B\) has twice the radius of satellite \(A\), StudySmarter Originals
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    Third Law of Kepler
    Frequently Asked Questions about Third Law of Kepler

    What is the 3rd law of Kepler?

    Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of the ellipse.

    How to proof Kepler's third law?

    To prove it, we can equal the centripetal force of the satellite to the gravitational force between the planet and the orbiting object.

    What is Kepler's third law of planetary motion?

    Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of the ellipse.

    What are examples of Kepler's third law?

    Examples of Kepler's third law are a satellite orbiting a planet and a planet orbiting the Sun.

    What is the formula of Kepler's third law?

    T^2=(r^3)(4pi^2)/(GM)

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