Kepler's Third Law Calculator

This Kepler's third law calculator uses Kepler's third law equation to estimate the basic parameters of a planet's motion around the Sun, such as the orbital period and radius. It is based on the fact that the appropriate ratio of these parameters is constant for all planets in the same planetary system.

You can directly use our Kepler's third law calculator on the left-hand side or read on to find out what is Kepler's third law if you've just stumbled here. In the following article, you can learn about Kepler's third law equation, and we will present you with a Kepler's third law example involving all of the planets in our Solar system.

The orbital period calculator offers the estimation of the orbital period using two other equations. Be sure to check it!

The 17^th century German astronomer, Johannes Kepler, made a number of astronomical observations. Upon the analysis of these observations, he found that the motion of every planet in the Solar system followed three rules. You can read more about them in our orbital velocity calculator. Here, we focus on the third one:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

And that's what Kepler's third law is. Simple, isn't it? This sentence reflects the relationship between the distance from the Sun of each planet in the Solar system and its corresponding orbital period (also known as the sidereal period that we described in the synodic period calculator). Note that, since the laws of physics are universal, the above statement should be valid for every planetary system!

We can easily prove Kepler's third law of planetary motion using Newton's Law of gravitation. All we need to do is make two forces equal to each other: centripetal force and gravitational force (you can find more information about the latter in the gravitational force calculator). We obtain:

m × r × ω² = G × m × M / r²,

where:

• m – Mass of the orbiting planet;
• r – is the orbital radius;
• ω – is the angular velocity, ω = v/r for circular motion (v – linear velocity);
• G – is the Gravitational constant, G = 6.67408 × 10⁻¹¹ m³ / (kg·s); and
• M – is the mass of the central star.

If we substitute ω with 2 × π / T (T - orbital period), and rearrange, we find that:

R³ / T² = 4 × π²/(G × M) = constant.

That's the basic Kepler's third law equation. There is also a more general derivation that includes the semi-major axis, a, instead of the orbital radius, or, in other words, it assumes that the orbit is elliptical. Since the derivation is more complicated, we will only show the final form of this generalized Kepler's third law equation here:

a³ / T² = 4 × π²/[G × (M + m)] = constant.

As you can see, the more accurate version of Kepler's third law of planetary motion also requires the mass, m, of the orbiting planet. To picture how small this correction is, compare, for example, the mass of the Sun M = 1.989×10³⁰ kg with the mass of the Earth m = 5.972×10²⁴ kg. That's a difference of six orders of magnitude!

Learn more about ellipses in the ellipse calculator that helps to analyze the properties of such mathematical figures.

The Kepler's third law calculator is straightforward to use, and it works in multiple directions. Just fill in two different fields, and we will calculate the third one automatically.

If you're interested in using the more exact form of Kepler's third law of planetary motion, then press the advanced mode button, and enter the planet's mass, m. Note that the difference would be too tiny to notice, and you might need to change the units to a smaller measure (e.g., seconds, kilograms, or feet).

In our Kepler's third law calculator, we, by default, use astronomical units and Solar masses to express the distance and weight, respectively (you can always change it if you wish). It's very convenient since we can still operate with relatively low numbers. To test the calculator, try entering M = 1 Suns and T = 1 yrs, and check the resulting a. Is it another number one? It should be! That's proof that our calculator works correctly – this is the Earth's situation.

If you'd like to see some different Kepler's third law examples, take a look at the table below. Here, you can find all the planets that belong to our Solar system. Do they fulfill Kepler's third law equation?

Kepler's third law describes the relationship between the distance of the planets from the Sun, or their semi-major axis a, and their orbital periods, T. The formula for Kepler's third law is:

a³/T² = G(M + m)/4π² = constant

where G is the gravitational constant, M is the star mass, and m is the planet mass.

Kepler's third law implies that the greater the distance of a planet from the Sun, the longer the period of that planet's orbit around the Sun. Thus, Mercury - the planet closest to the Sun - makes an orbit every 88 days. By contrast, Saturn, the sixth planet in the solar system from the Sun, will take as many as 10,759 days to do so. You can also use the third law equation to find the masses of bodies in a given system.

No, it applies to all objects, planets, asteroids, comets, etc., that are in the Solar System. In 1643, the Flemish astronomer Godefroy Wendelin noted that Kepler's third law is fulfilled not only by the planets, but also by Jupiter's moons as well. In general, Kepler's third law describes the motion of any two bodies in gravitational orbits around each other.

You can use Kepler's third law to calculate the mass of the Sun using data for the Earth:

1. Write the Kepler's third law as:
   GM/4π² = r³/T².
2. Multiply by 4π²/G to get:
   M = 4r³π²/T²G.
3. Insert T for Earth as 1 year and r as 1 au.
4. The mass of the Sun is about 2×10³⁰ kg.

1.881 years or 687 days, knowing that the semi-major axis of Mars is 1.52 au and the mass of the Sun is 2×10³⁰ kg. To find it:

1. Use Kepler's third law:
   T² = 4π²a³/GM.
2. Square the equation:
   T = √(4π²a³/GM).
3. Convert 1.52 au to meters:
   1.52×149,597,870,700 m = 227,388,763,464 m.
4. Calculate the T:
   T = √(4×(3.14)²×(227,388,763,464)³/6,67×10⁻¹¹×2×10³⁰) = 5.93×10⁷ s ≈ 687 days.