Orbital Period

Did you know that a day on Earth has not always been 24 hours long? When the Moon and the Earth were just 30,000 years old, a day lasted only six hours! When the Earth-Moon system was 60 million years old, a day lasted ten hours. The gravitational force of the Moon on the Earth has (through complex tidal interactions) been slowing the Earth's rotation. Due to the conservation of energy, Earth's rotational energy is converted into orbital energy for the Moon. This interaction has consequently increased the Moon's distance from the Earth and therefore made its orbital period longer. Over time, this phenomenon has moved the Moon gradually away from the Earth, at a minuscule rate of \(3.78\, \mathrm{cm}\) per year.

Have you ever thought about why a year on Earth has 365 days? Is it 365 days for every planet or for just the Earth? We know that the Earth rotates about its axis 365.25 times for every full orbit around the Sun. In this article we will study the concept of the orbital period and speed, so we can understand why every planet has a different amount of days in a year.

Orbital speed definition

We can think of the orbital speed as the speed of an astronomical object as it orbits another celestial body.

There are several useful formulas and derivations associated with calculating the orbital speed of an object and other associated quantities.

Tangential velocity and centripetal acceleration

A satellite's tangential velocity is what stops it from simply returning to the Earth. When an object is in orbit, it is always in free fall toward the central body. However, if the tangential velocity of the object is large enough then the object will fall towards the central body at the same rate as it curves. If we know the constant speed \(v\) of a satellite in a circular orbit of the Earth and its distance \(r\) from its center, we can determine the centripetal acceleration \(a\) of the satellite, where the acceleration due to gravity acts toward the center of mass of the Earth,

\[a=\frac{v^2}r.\]

We can prove the expression for centripetal acceleration by analyzing the geometry of the system and using the principles of calculus. If we compare the triangles formed by the position and velocity vectors, we find that they are similar triangles.

Fig 1 - Triangle formed by position vectors and \(\triangle{\vec{r}}\) in a circular orbit. It has two equal sides and two equal angles, so it is an isosceles triangle.

Fig 2 - Triangle formed by velocity vectors and \(\triangle{\vec{v}}\) in a circular orbit. It has two equal sides and two equal angles, so it is an isosceles triangle.

The position vectors are perpendicular to the velocity vectors, and the velocity vectors are perpendicular to the acceleration vectors, so the triangle has two equal angles. The magnitude of the orbital distance and velocity vectors are constant for an object in a circular orbit, so each of these triangles also has two equal sides.

For any circular orbit, the triangles have the same shape, but their sizes will differ, so we can state the proportion as,

$$\begin{align}\frac{\triangle v}v=&\frac{\triangle r}r,\\\triangle v=&\frac vr\triangle r.\end{align}\\$$

We can differentiate the expression to determine the instantaneous acceleration,

$$\frac{\triangle v}{\triangle t}=\frac vr\lim_{\triangle t\rightarrow0} \frac{\triangle r}{\triangle t}.$$

Then we can prove the equation for centripetal acceleration using the principles of calculus,

$$\begin{align}a=&\frac vr\lim_{\triangle t\rightarrow0} \frac{\triangle r}{\triangle t},\\a=&\frac{v^2}r.\end{align}$$

Orbital speed derivation

The gravitational force \(F_g\) is the net force on the satellite which can be expressed as,

\[F_g=\frac{GMm}{r^2},\]

where \(G\) 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 satellite's mass in kilograms \(\mathrm{kg}\), and \(r\) is the distance between the satellite and the center of the Earth in meters \(\mathrm m\).

Fig. 3 - A satellite orbits the Earth. The gravitational force acts on the satellite, in the direction of the Earth's center. The satellite orbits at a constant speed.

We can apply Newton's Second Law to find the formula for the orbital speed.

$$\begin{align*}F_g&=ma,\\\frac{GMm}{r^2}&=\frac{mv^2}r,\\\frac{GMm}r&=mv^2.\end{align*}$$

If we multiply both sides of the equation by \(1/2\), we find an expression for the kinetic energy \(K\) of the satellite:

$$\begin{align*}\frac12mv^2&=\frac12\frac{GMm}r,\\K&=\frac12\frac{GMm}r.\end{align*}$$

To find the formula for the orbital speed we just solve the above equation for \(v\):

$$\begin{align*}\cancel{\frac12}\cancel mv^2&=\cancel{\frac12}\frac{GM\cancel m}r,\\v^2&=\frac{GM}r,\\v&=\sqrt{\frac{GM}r}.\end{align*}$$

Changing orbits and speed

Recall our scenario from earlier, if a satellite was in a circular orbit at a distance \(r_1\) from the center of the Earth and mission control wanted to maneuver the satellite to orbit at a closer distance \(r_2\) to the Earth, how would they determine the amount of energy required to do so? Mission control would have to evaluate the total energy (kinetic and potential) of the Earth-Satellite system before and after the orbital maneuver and calculate the difference.

We know that the only force acting on the system is the force of gravity. This force is conservative, such that it only depends on the object's initial and final position with respect to the radial distance from the center of the celestial body. As a consequence, we can determine the gravitational potential energy \(U\) of the object using calculus,

\[\begin{align}U&=-\int\overset\rightharpoonup F_{g}\cdot\overset\rightharpoonup{\,\mathrm dr},\\ &=-\left(\frac{-GMm}{r^2}\;\widehat r\right)\cdot\left(\mathrm{d} r\;\widehat r\right),\\ &=\int_r^\infty\frac{GMm}{r^2}\mathrm{d}r,\\ &=\left.GMm\;\frac{r^{-2+1}}{-1}\right|_r^\infty,\\ &=-\lim\limits_{r\to\infty}\frac{GMm}{r}- \left(-\frac{GMm}r\right),\\ &=\frac{GMm}r.\end{align}\]

The sum of the kinetic energy \(K\) and the gravitational potential energy \(U\) of an orbiting object is equal to the mechanical energy \(E\) and will always be constant. Therefore, by increasing the kinetic energy of an orbiting object its gravitational potential energy will decrease proportionally,

$$\begin{align*}E&=K\;+\;U,\\E&=\text{constant},\\W&=\triangle E.\end{align*}$$

Recall the expression for the satellite's kinetic energy from the previous section. Alongside our new expression for gravitational potential energy we can determine the total energy of the system:

$$\begin{align*}E&=\frac12\frac{GmM}r-\frac{GmM}r,\\E&=-\frac12\frac{GmM}r.\end{align*}$$

Now we can study the mechanical energy \(E_1\) and \(E_2\) of the satellite as its orbital distance changes from \(r_1\) to \(r_2\). The change in total energy \(\triangle{E}\) is given by,

$$\begin{align*}\triangle E&=E_2-E_1,\\\triangle E&=-\frac12\frac{GmM}{r_2}+\frac12\frac{GmM}{r_1}.\end{align*}$$

Because \(r_2\) is a smaller distance than \(r_1\), \(E_2\) will be larger than \(E_1\) and the change in energy \(\triangle{E}\) will be negative,

$$\begin{align*}\triangle E&<0.\end{align*}$$

Because the work done on the system is equal to the change in energy, we can infer that the work done on the system is negative.

$$\begin{align*}W&=\triangle E,\\W&<0,\\\overset\rightharpoonup F\cdot\overset\rightharpoonup{\triangle r}&<0.\end{align*}$$

Orbital period definition

The planets of the solar system have different orbital periods. For example, Mercury has an orbital period of 88 Earth days, while Venus has an orbital period of 224 Earth days. It is important to note that we often specify orbital periods in Earth days (which have 24 hours) for consistency because the length of a day is different for each respective planet. Even though Venus takes 224 Earth days to complete an orbit around the Sun, it takes 243 Earth days for Venus to complete one full rotation on its axis. In other words, a day on Venus is longer than its year.

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. This is due to Kepler's Third Law, which can also be derived thanks to the equation for the orbital period, as we will see in the next section.

The other reason why different planets have different orbital periods is that there exists an inversely proportional relationship between the orbital period and the orbital speed. Planets with larger orbital periods require lower orbital speeds.

Fig. 4 - From left to right in order from their distance to the Sun: Mercury, Venus, Earth, and Mars. NASA

Orbital Period formulas

Since we now know how to calculate orbital speed, we can easily determine the orbital period. For circular motion, the relationship between orbital period \(T\) and orbital speed \(v\) is given by,

$$v=\frac{2\pi r}T.$$

In the above equation, \(2\pi r\) is the total distance in one complete revolution of an orbit, as it is the circumference of a circle. We can solve for the orbital period \(T\) by substituting the equation for the orbital speed,

$$\begin{align*}v&=\frac{2\pi r}T,\\T&=\frac{2\pi r}v,\\T&=\frac{2\pi r}{\sqrt{\displaystyle\frac{GM}r}},\\T&=2\pi r\sqrt{\frac r{GM}},\\T&=\frac{2\pi r^{3/2}}{\sqrt{GM}}.\end{align*}$$

We can rearrange the expression above to derive Kepler's Third Law, which states the square of the orbital period is proportional to the cube of the semi-major axis (or radius for a circular orbit).

$$\begin{align*}T^2&=\left(\frac{2\pi r^{3/2}}{\sqrt{GM}}\right)^2,\\T^2&=\frac{4\pi^2}{GM}r^3,\\T^2&\propto r^3.\end{align*}$$

The mass of the orbiting body \(m\) is not relevant in many scenarios. For example, if we want to calculate the orbital period of Mars around the Sun, we should only consider the mass of the Sun. The mass of Mars is not relevant in the calculation as its mass is insignificant compared to the Sun. In the next section, we will determine the orbital period and speed of various planets in the Solar System.

The orbital period of Mars

Let's calculate the orbital period of Mars by using the equation derived in the previous section. Let us approximate that the radius of Mars' orbit around the Sun is approximately \(1.5\;\mathrm{AU}\), and is a perfectly circular orbit, and the mass of the Sun is \(M=1.99\times10^{30}\;\mathrm{kg}\).

First, let's convert \(\mathrm{AU}\) to \(\mathrm{m}\),

\[1\;\mathrm{AU}=1.5\times10^{11}\;\mathrm m.\]

Then use the equation for the time period and substitute in the relevant quantities,

$$\begin{align*}T&=\frac{2\pi r^{3/2}}{\sqrt{GM}},\\T&=\frac{2\pi\;\left(\left(1.5\;\mathrm{AU}\right)\left(1.5\times10^{11}\;\mathrm m/\mathrm{AU}\right)\right)^{3/2}}{\sqrt{\left(6.67\times10^{-11}\;\frac{\mathrm m^3}{\mathrm s^2\mathrm{kg}}\right)\left(1.99\times10^{30}\;\mathrm{kg}\right)}},\\T&=5.8\times10^7\;\mathrm s.\end{align*}$$

Since \(1\;\text{second}=3.17\times10^{-8}\;\text{years}\), we can express the orbital period in years.

$$\begin{align*}T&=\left(5.8\times10^7\;\mathrm s\right)\left(\frac{3.17\times10^{-8}\;\mathrm{yr}}{1\;\mathrm s}\right),\\T&=1.8\;\mathrm{yr}.\end{align*}$$

The orbital speed of Jupiter

Now we will calculate the orbital speed of Jupiter, considering its radius of orbit around the Sun can be approximated to a circular orbit of \(5.2\;\mathrm{AU}\).

$$\begin{align*}v&=\sqrt{\frac{GM}r},\\v&=\sqrt{\frac{\left(6.67\times10^{-11}\;\frac{\mathrm m^3}{\mathrm s^2\mathrm{kg}}\right)\left(1.99\times10^{27}\;\mathrm{kg}\right)}{\left(5.2\;\mathrm{AU}\right)\left(1.49\times10^{11}\;{\displaystyle\frac{\mathrm m}{\mathrm{AU}}}\right)},}\\v&=13\;\frac{\mathrm{km}}{\mathrm s}.\end{align*}$$