Orbital Period Calculator

Orbital radius (R)

Central body mass (M)

Orbital period (T)

Enter any two of the orbital radius, the central body's mass, and the orbital period, and leave the third blank. The calculator fills the missing one in place. Switch the unit on any field, including Earth and Solar masses and astronomical units.

Show orbital motion & dynamics

Add the orbital velocity's companions: escape velocity, the gravitational acceleration at this radius, the angular velocity, the number of orbits per day, and the standard gravitational parameter.

Use an elliptical orbit

Enter the closest (periapsis) and farthest (apoapsis) distances to get the semi-major axis, the eccentricity, the true period of the ellipse, and the speeds at closest and farthest approach.

Show altitude above the surface

Enter the central body's radius to convert the orbital radius into an altitude above the surface, and to read that body's surface gravity.

Please input values into the fields

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Show the period-vs-radius curve

Plot how the orbital period grows with radius for your central mass, and a table of the period and speed at a range of radii.

Satellites, moons and planets all follow the same law when orbiting a larger body in space. The length of time it takes to complete one orbit depends on only two factors: the distance of the orbit from the center, and the mass of the central body. If you input the orbital radius and the mass of the central body, this calculator will give you the period. It can be used to calculate things like how long it takes the International Space Station to go around Earth, or how long it takes Mars to go around the Sun, or to figure out what altitude a new satellite should be placed at. If you leave one of the three inputs blank, this calculator will solve for that value.

What is an orbital period?

The orbital period is the time required for a complete orbit. It can be predicted by Kepler's third law. For all objects orbiting the same central body, the ratio of the square of the period to the cube of the radius is constant. Johannes Kepler discovered this law in 1619 from original planetary data, and later Isaac Newton explained why it must be true based on his law of universal gravitation.

The most amazing thing about this law is what doesn't appear in the formula: the mass of the orbiting body does not matter at all. If two objects are at the same altitude, both a small satellite and much heavier Hubble Space Telescope will have exactly the same period because gravity gives them both the same acceleration. All that matters is the mass of the central body and the distance.

The formula for Kepler's third law is:

The period is equal to two pi times the square root of the cube of the radius divided by the product of the mass of the central body and the gravitational constant.

T = 2\pi \sqrt{\dfrac{R^3}{G M}}

where T is the orbital period, R is the radius of the orbit measured from the center of the central body, M is the mass of the central body and G is the gravitational constant. If both sides are squared, then one gets the more commonly quoted form of Kepler's law, which states that the square of the period is proportional to the cube of the radius.

T^2 = \dfrac{4\pi^2}{G M}\,R^3

The gravitational constant G

The constant G determines the strength of gravity throughout the universe, and relates mass and distance to a time measurement in seconds.

G = 6.6743 \times 10^{-11}\ \tfrac{\text{m}^3}{\text{kg}\cdot\text{s}^2}

Because G is so small, the period only matters if the mass of the central body is very large. For that reason, celestial mechanics deals with planets, stars and satellites. Astronomers often combine G and M into a single number, the standard gravitational parameter μ = G * M. The accuracy of this measurement is usually much higher than the accuracy one could achieve by measuring G or M separately.

How to use this calculator:

Select two known values and enter them into the corresponding fields using the units that are appropriate for your current problem. Leave the third input field blank.

If you enter the radius and mass of a central body, it will calculate the period. If you enter the period and mass, it will back-calculate the required radius for a given orbit, which is one method to place a satellite at a particular altitude. If you enter the period and radius, it will determine the mass of the central body. This is one way that astronomers measure the mass of a planet by using one of its moons. You do not have to select a mode as the calculator can detect what needs to be calculated from an empty input field.

Determine radius or mass:

Since this equation contains three variables in a single formula, any one of them can be converted and determined. The following is the form that the calculator internally uses:

R = \sqrt[3]{\dfrac{G M\,T^2}{4\pi^2}} \qquad M = \dfrac{4\pi^2 R^3}{G\,T^2}

To find	Formula	In words
Period T	T = 2 pi sqrt(R cubed / (G M))	Two pi times the square root of the radius cubed over G times the mass
Radius R	R = cube root of (G M T squared / 4 pi squared)	The cube root of G, the mass, and the period squared, over four pi squared
Mass M	M = 4 pi squared R cubed / (G T squared)	Four pi squared times the radius cubed, over G times the period squared

Example: International Space Station

For example, let's say we're dealing with the International Space Station (ISS). It orbits Earth at an altitude of about 400 km, but for our equation we need to know its distance from the center of the Earth. So we add the Earth's radius of 6371 km and get R = 6771 km, or 6,771,000 meters. The mass of Earth is about 5.972 × 10 to the power of 24 kg. Plugging those values into our equation, we get:

T = 2\pi \sqrt{\dfrac{(6.771\times10^{6})^3}{(6.6743\times10^{-11})(5.972\times10^{24})}}

The cube of the radius is about 3.10 times ten to the power of 20. The product of G and M is about 3.99 times ten to the power of 14. Dividing we get about 778,800, and taking the square root of that gives us about 882.5 seconds. Multiplying by twice the value of pi we get 5545 seconds, which is about 92.4 minutes. The actual orbital period of the ISS is about 92 minutes, so our calculation checks out. Our calculated result for the orbital speed is about 7.7 kilometers per second.

Orbital and escape velocity

This calculator also outputs the orbital velocity, which is the speed required to maintain a circular orbit at a particular radius.

v_{orbit} = \sqrt{\dfrac{G M}{r}}

When the Celestial Mechanics layer is enabled, this tool also calculates escape velocity, which is the speed required to completely break free from the gravity of the central celestial body. This is always the orbital velocity multiplied by the square root of two.

v_{esc} = \sqrt{\dfrac{2 G M}{r}}

At low Earth orbit the orbital speed is about 7.7 km/s and the escape velocity from the ground is about 11.2 km/s. The same panel also displays the gravitational acceleration and angular velocity at that radius as well as how many orbits an object would make per day.

Elliptical orbits

Most actual orbits are not perfect circles but ellipses. The point in which the object is closest to the central body is called periapsis and the farthest point is called apoapsis. Kepler's third law still applies, except that instead of using radius you use semimajor axis, which is the average of the periapsis distance and apoapsis distance.

a = \dfrac{r_{peri} + r_{apo}}{2}

The elliptical orbit plane takes these two distances and gives back the major axis, eccentricity (the stretch of the ellipse), actual period, and speeds at periapsis (closest point) and apoapsis (farthest point). Celestial bodies accelerate as they approach periapsis and decelerate as they move away from apoapsis. They also sweep out equal areas in equal times. This is Kepler's second law. The speed at any distance can be calculated using the vis-viva equation.

v = \sqrt{G M\left(\dfrac{2}{r} - \dfrac{1}{a}\right)}

Measurement from center:

A common mistake is to measure the orbit from the surface rather than from the center. The radius in Kepler's laws always refers to the distance between centers. For Earth satellites, the altitude must be added to the 6371 km Earth radius. An elevation plane does this step instead. If a central body radius is entered, it converts the orbital radius into an altitude above sea level and also gives the surface gravity of that body.

Application of orbital period:

Satellite engineers use this formula in reverse to determine the altitude that will allow a given period. Geostationary satellites are fixed at an altitude of about 35,786 km above the Earth's equator because they must match the Earth's rotational speed of 24 hours. The GPS satellites orbit much lower, at about 20,200 km, and have a period of 12 hours. This allows enough satellites to be visible from any point on Earth.

Astronomers use it in reverse. If the time taken for an exoplanet to transit a star is measured and the mass of the star is inputted then the orbit can be determined using this law. Alternatively they can observe the behavior of a satellite orbiting a planet and deduce the mass of the planet from that. This is the only way to determine the mass of worlds we will never visit. Mission planners also rely on this relationship when designing transfer orbits and calculating when two spacecrafts will actually meet in space.

Assumptions and limitations:

This formula assumes that the mass of the orbiting object is negligible compared to the central mass. This assumption holds for satellites, moons and planets. In binary star systems where both masses are roughly equal, then the sum of the two masses should be used for M. Also this formula treats the central object as a point mass or a homogeneous sphere so that the distance is measured from the center of the object.

For elliptical orbits, the major axis must be used instead of any individual distance. Atmospheric drag in low Earth orbit, gravitational effects from other bodies, and relativistic effects near very massive objects all cause small deviations. However, usually the difference between results and observed values is within about one percent.

Tips for accurate results:

Measure the radius of a celestial body from its center, not surface, and add that radius to any height. Be careful about significant digits. Since central masses can be 10 to the 24th kg or more, use units like "Earth mass" or "Sun mass" instead of typing out long strings of zeros. For interstellar distances, use Astronomical Units (AU) or light years. Don't mix units by hand; let a calculator tool convert them for you.

Frequently asked questions

Why is the mass of an orbiting body not important?

Gravity exerts the same acceleration on all objects regardless of their own mass, so a heavy and light satellite will move at the same speed in the same orbit at the same altitude. When deriving Kepler's third law, the mass of the orbiting body is canceled out, leaving only the mass and radius of the central body. This is why we can use one formula to describe everything from screws to space stations.

What is a geostationary orbit?

The period of a geostationary orbit is exactly 24 hours, so the satellite follows Earth's rotation and stays in the same position above the equator. If you input the mass of the Earth and the period of 24 hours, the tool calculates an orbital radius of about 42,164 km, which corresponds to a height of about 35,786 km above ground.

Does this also apply to elliptical orbits?

Yes it does. The third Kepler's law applies to all orbits, both circular and elliptical. Instead of the radius, simply use the major axis which is the average of the periapsis and apoapsis distance. If you enable the plane for elliptical orbits, then you can directly enter the periapsis and the apoapsis and get the major axis, eccentricity, period and velocity at both ends.

How accurate is the orbital period formula?

In most applications it is very accurate. Atmospheric drag in low orbits, the gravity of other bodies and relativistic effects that occur near objects with very large mass can cause small deviations. For Earth satellites, the difference between calculated values and actual observations is within about one percent. This is a sufficiently accurate value to plan and understand how an orbit works.

Can you determine a planet's mass by observing its moon?

Yes, that is exactly the method astronomers use to determine the mass of distant objects. If you input the orbital period and radius of a satellite's orbit and leave the central mass as an unknown variable, then a suitable tool will calculate the planet's mass using this formula: M = (4*pi squared*R cubed over G times period squared. And in the same way, you can figure out the mass of a star along with its planets.

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Disclaimer: This calculator is provided for general informational and educational purposes only. Our calculators are under active development, and results may be inaccurate, incomplete, or unsuitable for your situation. Always verify the figures independently and seek advice from a qualified professional before relying on them. We make no warranties and accept no liability for any loss or decision arising from use of this tool.

References

Wikipedia: Kepler's laws of planetary motion
The three laws, the derivation of the period-radius relationship, and the role of the semi-major axis for elliptical orbits.
Wikipedia: Orbital period
Definitions of the orbital period, the formula for a small body around a large one, and the binary-system form.
NASA: Kepler's Third Law
A plain-language explanation of Kepler's laws and how they govern the motion of planets, moons, and spacecraft.