0.0000e+0
kg
0
M⊕
0
M_J
—
km/s
1.0000e+0
AU³/yr²
0.0000e+0
kg
0
M⊕
0
M_J
—
km/s
1.0000e+0
AU³/yr²
The Planet Mass Calculator determines the mass of a planet orbiting a star using Kepler's Third Law of planetary motion. By inputting the orbital period, semi-major axis, and host star mass, the calculator derives the planet's mass from the fundamental gravitational relationship governing orbital mechanics.
This method is central to exoplanet characterization. When combined with radial velocity measurements or transit timing variations, Kepler's Third Law allows astronomers to determine planetary masses even for worlds orbiting distant stars. The technique has been instrumental in discovering thousands of exoplanets since the first confirmed detection in 1995.
The calculator applies Kepler's Third Law in its Newtonian form:
$$T^2 = \frac{4\pi^2 a^3}{G(M_* + m_p)}$$
Solving for the total mass of the system:
$$M_* + m_p = \frac{4\pi^2 a^3}{G T^2}$$
The planet mass is then:
$$m_p = \frac{4\pi^2 a^3}{G T^2} - M_*$$
where $$T$$ is the orbital period in seconds, $$a$$ is the semi-major axis in meters, $$G = 6.674 \times 10^{-11}$$ N·m²/kg² is the gravitational constant, and $$M_*$$ is the star mass.
The orbital velocity is calculated as:
$$v = \frac{2\pi a}{T}$$
The Kepler consistency check computes $$a^3/T^2$$ in AU³/yr², which should equal the total system mass in solar masses for a valid orbit.
Note that for real exoplanet detection, the planet mass is typically much smaller than the star mass, so the measurement precision of $$a$$ and $$T$$ must be extremely high to extract $$m_p$$ from the difference.
The planet mass result represents the gravitationally inferred mass assuming a circular orbit. For Earth orbiting the Sun (T = 365.25 days, a = 1 AU, M* = 1 M☉), the calculator recovers Earth's mass. The Kepler consistency value should approximately equal the star mass in solar masses for planetary-mass companions.
If the computed planet mass is negative, it indicates the input parameters are inconsistent — the given period is too short for the specified semi-major axis and star mass, which is physically impossible. The orbital velocity gives the mean speed of the planet in its orbit, useful for understanding energy budgets and escape conditions.
Inputs
Results
Earth's orbital parameters recover its known mass of 1 M⊕ and orbital velocity of ~29.78 km/s.
Inputs
Results
Jupiter's 11.86-year period and 5.2 AU distance yield its mass of approximately 1 M_J and orbital speed of ~13 km/s.
In practice, the planet mass is a tiny fraction of the star mass (Earth is only 3×10⁻⁶ M☉). Extracting $$m_p$$ from $$M_* + m_p$$ requires knowing both $$a$$ and $$T$$ with extraordinary precision. Real exoplanet masses are typically measured through radial velocity (Doppler) shifts or transit timing variations, which are more sensitive to the planet's gravitational influence on the star.
Kepler's Third Law states that the square of a planet's orbital period is proportional to the cube of its semi-major axis. In Newton's generalization, the constant of proportionality depends on the total mass of the system: $$T^2 = 4\pi^2 a^3 / [G(M_* + m_p)]$$. For our Solar System with masses in solar units and distances in AU, this simplifies to $$a^3/T^2 \approx 1$$.
Yes. Kepler's Third Law applies to elliptical orbits — the semi-major axis $$a$$ is half the longest diameter of the ellipse. The orbital period depends only on the semi-major axis and total mass, not on the eccentricity. However, the orbital velocity shown is the mean circular velocity; actual velocity varies along an elliptical orbit.
Enter the orbital period in Earth days, the semi-major axis in astronomical units (1 AU = Earth-Sun distance), and the star mass in solar masses. The calculator handles all unit conversions internally using SI constants.
Yes, but the interpretation changes. For two stars of comparable mass, the 'planet mass' output represents the companion star mass. This is actually how Kepler's Third Law is most accurately applied — binary star systems provide the most precise stellar mass measurements in astrophysics.
A negative result means the input combination is physically impossible: the orbital period is too short for the given distance and star mass. This would require more total mass than the star alone provides. Check your inputs for unit errors or reconsider the star mass estimate.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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