Orbital Period Calculator

Created by Davide Borchia
Last updated: Jun 22, 2022

Planets, stars, and satellites move according to the rules dictated by gravitational force: the orbital period calculator will teach you one of the features of movements high above us.

With our orbital period calculator, you will learn how to calculate the period of revolution of a body. What are you waiting for? Jump on your spaceship; we are setting the course! In this spacey journey, you will learn:

• What is an orbit, and how do they define an orbital period;
• How to calculate the orbital period for a satellite (small orbital height approximation); and
• How to calculate the period of revolution of two bodies with comparable mass (we will calculate the Moon's orbit).

Walking in ellipses

An orbit is a path (trajectory) described by an object during its motion around another object. There are several types of orbits, but we are concerned with the closed orbits: circular and elliptic.

Orbits involve any type of body out there, from two massive black holes to a kilogram and a half CubeSat in orbit around Earth. The rules are fundamentally the same, as are the parameters. Here we will focus on the orbital period.

What is the orbital period?

The orbital period is the time required by an object to complete an orbit. The definition of an orbit varies (what if the central body moved during that time?): the general analysis in our calculator will assume we are in the reference frame of one of the bodies; thus, the orbit is properly closed.

🙋 You can find more about astrodynamics at our escape velocity calculator and orbital velocity calculator!

How to calculate the orbital period

Orbital mechanics is a fine art of approximation: the quantities involved vary so wildly that knowing when to neglect something is necessary.

We see one of the first and most common approximations when we consider the motion of a satellite around the Earth. The masses of the two bodies involved are more than 20 orders of magnitude away, making the satellite effectively weightless.

The formula for the orbital period of a body of negligible mass orbiting close to the surface of a massive body is:

$T = \sqrt{\frac{3\cdot\pi}{G\cdot\rho}}$

Where:

• $G$ is the gravitational constant; and
• $\rho$ is the density of the central body.

Yes! You read it correctly: in certain situations, the orbital period is defined only by the composing "material" of the central body!

If you are not in the mood for approximations, you can use the formula for the orbital period in a two-body system:

$T = 2\cdot \pi\sqrt{\frac{a^3}{G\cdot(M_1+M_2)}}$

Where:

• $a$ is the semi-major axis of the orbit; and
• $M_1$ and $M_2$ are the masses of the bodies involved in the orbit.

Remember that in a two-body system, if the masses are not so far off from each other, the two bodies orbit themselves around a common center of mass defined by the law of conservation of . Switch to a reference frame integral with one of the two bodies; it will be helpful!

How to use our orbital period calculator

Simply insert the parameters you know in the correct section of our calculator. We will give you the results of the desired equation for the orbital period.

Example of orbital period calculations

Let's take a look at a couple of examples.

Satellite in low Earth orbit

Low Earth orbit (LEO) is the most commonly used orbit for commercial satellites. Given the extreme difference in mass, and the closeness to the surface of the planet, we can use the first equation for the orbital period (small height approximation):

\begin{align*} T &= \sqrt{\frac{3\cdot\pi}{G\cdot\rho}}\\ &= \sqrt{\frac{3\cdot\pi}{G\cdot 5520}} = 84.3\ \text{min} \end{align*}

This value is close to the lower bound for the LEO orbits period, $90$ minutes. You can learn more about satellites orbiting Earth at our Earth's orbit calculator!

Moon's orbit

We can also calculate the Moon's orbit period around the Earth. Input in the second section of the calculator the following values:

• Semi-major axis: $384,748\ \text{km}$;
• First body mass: $1\ \text{Earth mass}$; and
• Second body mass: $1/82\ \text{Earth mass}$.
\begin{align*} T &= 2\cdot \pi\sqrt{\frac{a^3}{G\cdot(M_1+M_2)}} \\ &= 2\cdot \pi\sqrt{\frac{384,748^3}{G\cdot(1+\frac{1}{82})}}\\ &= 27\ \text{d}\ 7\ \text{h} \end{align*}

This is the real value of the Moon's period. Who would say that science works! Notice how we couldn't use the first approximation for the orbital period: the Moon is massive enough to exert a noticeable on the Earth (think of tides).

Davide Borchia
Satellite around central sphere
Central body density
lb/cu ft
Orbital period
hrs
Binary System
Semi-major axis
au
First body mass
Suns
Second body mass
Earths
Period
days
hrs
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