Kepler's Laws of Planetary Motion
This topic is part of the HSC Physics syllabus under the section Motion in Gravitational Fields.
HSC Physics Syllabus
- Investigate the relationship of Kepler’s Laws of Planetary Motion to the forces acting on, and the total energy of, planets in circular and non-circular orbits using: (ACSPH101)
Kepler's Laws of Planetary Motion
In this video, we discuss Kepler's first, second and third law of planetary motion including calculation example for the third law.
What are Kepler's Laws of Planetary Motion?
Johannes Kepler, a German astronomer and mathematician, formulated three laws that describe the motion of planets around the Sun. These laws were revolutionary as they described planetary motion in a precise and predictive manner, moving beyond the qualitative descriptions of earlier models.
Kepler's First Law: The Law of Ellipses
Kepler's first law states that each planet orbits the Sun in an elliptical path, with the Sun at one of the two foci of the ellipse.
An ellipse is a flattened circle. In the context of planetary motion, this means that the distance (orbital radius) between a planet and the Sun changes as the planet orbits. When a planet is closest to the Sun (at perihelion), the Sun is at one focus of the ellipse; when it is farthest away (at aphelion), the Sun is at the other focus.
A circle is a special case of an ellipse where both foci coincide at the centre. Therefore, in a circular orbit, the central body (like the Sun for planets or Earth for satellites) is at the centre of the orbit.
Kepler's first law applies to any pair of objects where one is orbiting the other due to gravitational attraction, like moons orbiting planets, stars orbiting galactic centres, or even binary star systems.
Eccentricity of Orbit
Eccentricity is a parameter that determines the amount by which an orbit deviates from a perfect circle. Eccentricity ranges from 0 to 1 for elliptical orbits. An eccentricity of 0 corresponds to a perfect circle, and as the eccentricity approaches 1, the ellipse becomes more elongated.
Earth's orbital eccentricity is approximately 0.0167. This value is very close to 0, indicating that Earth's orbit is nearly circular but slightly elliptical. Because Earth's eccentricity is so small, the difference in distance between Earth and the Sun at perihelion (closest approach) and aphelion (farthest point) is not very large. This small variation allows for simplifications in certain calculations or models where treating Earth's orbit as circular can yield reasonably accurate results.
Kepler's Second Law: The Law of Equal Areas
Kepler's second law states that an imaginary line joining a planet and the Sun sweeps out equal areas during equal intervals of time. This law implies that planets move faster when they are closer to the Sun and slower when they are farther away. When a planet is at the perihelion, it travels the most quickly, and when it is at the aphelion, it moves the slowest.
In a perfectly circular orbit, the speed of the orbiting object remains constant. However, Kepler's second law still holds, as the area swept per unit time remains constant (since the radius of the orbit is constant).
Kepler's Third Law: The Law of Harmonies
Kepler's third law states that the square of the orbital period of a planet (`T^2`) is directly proportional to the cube of the semi-major axis (`a^3`) of its orbit. The semi-major axis of the ellipse (half of the longest diameter of the elliptical orbit).
This law is expressed mathematically as:
$$T^2 \propto a^3$$
$$T^2 = ka^3 \text{ where k is the proportionality constant}$$
This law establishes a precise relationship between the distance of a planet from the Sun and the time it takes to complete its orbit. For example, a planet that is twice as far from the Sun as another planet will have an orbital period about `\sqrt{8}` times longer.
Kepler's third law applies to any two bodies orbiting each other under gravity. For instance, it can be used to determine the period of a moon orbiting a planet, or the orbital period of a binary star system, by knowing the distance between the two objects.
Kepler's Third Law Applies to Circular Orbits
For circular orbits, the semi-major axis of the ellipse is simply the radius of the circle. Kepler's third law can be used to relate the period of the orbit to this radius: the square of the period is directly proportional to the cube of the radius. This is particularly useful in calculating the circular orbits of satellites around the Earth.
We can derive an equation for Kepler's third law for circular orbits by considering the orbital velocity formula, and expressing this in terms of the circumference of the circular orbit:
$$v_{\text{orbital}} = \sqrt{\frac{GM}{r}} = \frac{2\pi r}{T}$$
$$\frac{GM}{r} = (\frac{2\pi r}{T})^2$$
$$\frac{r^3}{T^2} = \frac{GM}{4 {\pi}^2}$$
where:
- `r` is the orbital radius in metres (m)
- `T` is the orbital period in seconds (s)
- `G` is the universal gravitational constant = `6.67 xx 10^{-11}`
- `M` is mass an object is orbiting around in kg
Rearranging the above equation we get:
$$T^2 = \frac{4 {\pi}^2}{GM} \times r^3$$
This equation shows that the proportionality constant `k` between `T^2` and `r^3` or `a^3` is related to the central mass (`M`) around which objects are orbiting. In the case of planets in the Solar system, `M` is mass of the Sun.
$$k = \frac{4 {\pi}^2}{GM}$$
The directly proportional relationship between `r^3` and `T^2` can be represented by a straight line on the graph.
For example, the square of orbital period of moons of Jupiter are plotted against the cube of their orbital radius around Jupiter, we get a linear relationship.
The gradient of the line of best fit represents the proportionality constant because:
$$\text{gradient} = k = \frac{T^2}{r^3} = \frac{4 {\pi}^2}{GM}$$
Therefore, we can use the value of the gradient to determine the mass every object is orbiting around i.e. Jupiter.
$$M = \frac{4 {\pi}^2}{G \times \text{gradient}}$$
Example 1
Assuming the orbital period of Earth around the Sun is 365 days. Calculate the average distance between the Earth and the Sun.
Use the following information in your calculations:
- Mass of Sun = `2.0 xx 10^30` kg
- Mass of Earth = `6.0 xx 10^24` kg
Solution:
Since Earth's orbit around the Sun has a low eccentricity, we can model its orbit as circular.
$$\frac{r^3}{T^2} = \frac{GM}{4 {\pi}^2}$$
$$\frac{r^3}{(365 \times 24 \times 60 \times 60)^2} = \frac{(6.67 \times 10^{-11})(2.0 \times 10^{30})}{4 {\pi}^2}$$
$$r^3 = 3.36 \times 10^{33}$$
$$r = 1.50 \times 10^{11} \text{ m}$$
Example 2
A satellite orbits the planet Xerus in a circular path with a radius of `4.0 xx 10^6` m and orbital period of 4 hours.
Calculate the period for a satellite orbiting Xerus at a radius of `1.5 xx 10^7` m.
Solution:
Since both satellites orbit Xerus, the proportionality constant between their `T^2` and `r^3` is the same.
$$k = \frac{{r_1}^3}{{T_1}^2} = \frac{{r_2}^3}{{T_2}^2}$$
$$\frac{(4.0 \times 10^6)^3}{(4)^2} = \frac{(1.5 \times 10^7)^3}{{T_2}^2}$$
$$T^2 = 843.75$$
$$T = 29 \text{ hours}$$
Note that when equating `\frac{r^3}{T^2}` of orbiting objects in the same system, we do not need to substitute values with SI units. Here, we substituted time in hours instead of seconds, which will give a value for the orbital period of the second satellite also in hours.
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