Order in Planetary Orbits

Elliptical orbits

An ellipse is a shape formed by taking a diagonal slice through a cone. It is essentially the shape of a circle viewed at an angle.

An ellipse can be drawn by taking a piece of paper, two push-pins, a loop of string, and a pencil. The two pins are pushed through the paper into a suitable surface, providing the two foci for the ellipse. They should be closer together than the loop is long. The loop of string is placed around the base of these pins, leaving some slack. The pencil is now placed so that the pins and the loop form a triangle with a slight tension on the string.

Now try to draw a shape by moving the pencil about the pins while keeping the string taunt. The result should be an ellipse. The shape of the ellipse can be varied either by moving the pins closer together or further apart. This shape, according to Kepler, defines the path that a planet takes when it orbits the Sun.

Kepler's First Law - A planet orbits the Sun on an ellipse with the Sun at one focus.

Kepler's Second Law says in brief that an object speeds up as it gets closer to the Sun and slows down as it moves further away. Where the distance from the Sun to the orbital path is longer, only a smaller arc needs to be traversed to sweep out an area that requires a wider arc near the Sun.

As the planet moves closer to the Sun along its orbit the gravitational force works to increase the velocity. In contrast, as the planet is moving further away, the gravity of the Sun gradually decelerates the body and it is slowed down.

Kepler's Second Law - A planet in orbit about the Sun sweeps out equal areas {\displaystyle A} in the same time interval {\displaystyle t} .

A line that divides an ellipse in half and passes through the widest part of the ellipse is called the major axis. A line perpendicular to this axis and dividing the ellipse in half is called the minor axis. Half the length of the major axis is called the semi-major axis, and is represented by {\displaystyle a} . The period required for a planet to complete one full orbit is represented by {\displaystyle P}. The relationship between the period P and the length of the semimajor axis {\displaystyle a} is known as Kepler's Third Law, and can be represented as follows:

{\displaystyle P^{2}\propto a^{3}}

where the symbol ∝ means "proportional to", and implies that there is a direct mathematical relationship between the period squared and the length of the semi-major axis cubed.

The Second and Third Laws provide a basis for calculating the period of any planet orbiting the Sun, as well as determining where the planet will be located along the orbital path.

Eccentricity and Orbital Paths

The ratio of the distance of a focus from the center of an ellipse to the semi-major axis is called the eccentricity of the orbit. When the two foci of the ellipse are on top of each other, the eccentricity is exactly 0.0 and the shape is a circle. As the eccentricity increases, the orbiting planet moves much further away than at the closest approach. The orbital eccentricities for planets in our Solar system vary from as much as 0.21 for Mercury down to 0.0068 for Venus.

The scientific name for the point of closest approach is the periapsis, while the most distance separation is the apoapsis. In the case of planets orbiting the Sun, these are called the perihelion and aphelion, respectively. (The -helion suffix comes from the Greek name for the Sun deity, Helios. This word is also the source of the name for the element Helium.)

Two elliptical orbits with the same major axis {\displaystyle a} but different eccentricity.

Perhaps the most counter-intuitive aspect of the Third Law is that for any two identical bodies orbiting the Sun with the same semi-major axis, the orbital period is the same. This is true even if one is orbiting in a perfect circle and the other has an orbit that is highly elliptical (has a relatively high eccentricity). The elliptical shape will fit entirely within the circle except at two points (the ends of the major axis, at which the two curves will be tangent), so it is actually a shorter orbital path. However the aphelion of the ellipse will be located further from the Sun, so the planet will spend more time traversing the distant section of the orbit. The shorter orbit and the slower traverse of the aphelion compensate for each other, resulting in an identical period with the circular orbit.

Some Examples Using Kepler's Third Law

Originally, Kepler's Third Law was used to describe the motions of the planets around the Sun. As it turns out, it also works very well with regards to other two-body orbital systems like the orbits of moons around Jupiter, or the orbits of binary stars about the center of mass of their system. In all of these cases the period of the orbit squared is proportional to the length of the semi-major axis cubed, with the differences between the orbital systems reflected in the constant of proportionality.

In this section we will consider the special case of the planets going around the Sun. If we choose to measure the length of the semi-major axis of an orbit in astronomical units (abbreviated AU, where 1 AU is the distance from the Earth to the Sun) and we measure the orbital period in years (abbreviated as {\displaystyle y}), then we can express Kepler's Third Law as

{\displaystyle P^{2}=\left({\frac {y^{2}}{AU^{3}}}\right)a^{3}}

where {\displaystyle P} is measured in years and {\displaystyle a} is measured in astronomical units. What follows are a few example of how to use this equation.

Mars' Orbital Period

Repeated measurements of Mars' orbit have determined that the semi-major axis of its orbit has a length of 1.52 AU. How long does it take for Mars to orbit the Sun once?

Solution: In this question, we are being asked for the orbital period {\displaystyle P} of Mars. We know from reading the question that the length of the semi-major axis {\displaystyle a} is 1.52 AU. Solving Kepler's Third Law for the period gives us

{\displaystyle P=\left({\frac {y}{\sqrt {AU^{3}}}}\right){\sqrt {a^{3}}}=\left({\frac {y}{\sqrt {AU^{3}}}}\right){\sqrt {(1.52AU)^{3}}}=1.87y}

which tells us that it takes 1.87 years for Mars to go around the Sun once.

An Unknown Asteroid's Orbital Semi-Major Axis

An amateur astronomer spends several months tracking an asteroid and is able to determine that it takes approximately 3/4 of a year for it to orbit once around the Sun. What is the semi-major axis of this asteroid's orbit?

Solution: Reading the question tells us that the orbital period of the asteroid is 3/4 of a year, or 0.75 {\displaystyle y} . We are expected to find the length of the semi-major axis {\displaystyle a}, so we need to solve Kepler's Third Law for that. Doing so gives us

{\displaystyle a=\left({\frac {AU}{\sqrt[{3}]{y^{2}}}}\right){\sqrt[{3}]{P^{2}}}=\left({\frac {AU}{\sqrt[{3}]{y^{2}}}}\right){\sqrt[{3}]{(0.75y)^{2}}}=0.83AU}

which means that is has a semi-major axis somewhere between that of Venus' orbit and Earth's orbit.