Surface Gravity of a Planet Calculator

Background.

Surface gravity follows from just two numbers: how much mass a body has, and how far its surface sits from the centre. Newton's law gives g₀ = GM/R², and that single expression governs everything from the 0.0006 m/s² of a small asteroid to the 24.79 m/s² at Jupiter's cloud tops. This calculator takes a mass and a radius for any body — a planet, a moon, an asteroid, an exoplanet, or something hypothetical — and returns the field strength, its bulk density, and the amount its own rotation cancels. It also inverts, so you can ask what mass a body of a given size would need to reach a target gravity, or how large a body of a given mass would have to be.

The rotation term is the reason this page reports two gravities rather than one, and the distinction trips up more people than any other part of the subject. JPL's planetary tables carry a column headed 'Equatorial Gravity', and for Earth it reads 9.80 m/s². Compute GM/R² with Earth's equatorial radius and you get 9.7983 — so that published column excludes rotation entirely. What a scale at the equator actually responds to is smaller, because the ground beneath it is moving in a circle and part of the gravitational field is spent supplying the centripetal acceleration. Subtract ω²R = 0.0339 m/s² and you get 9.7644 m/s². Both numbers are correct answers to different questions, and this page shows the field, the correction and the result side by side so it is never ambiguous which is on screen.

For Earth the correction is a third of a percent and easy to ignore. For Jupiter it is not. Jupiter turns once every 9.925 hours, and at its equator ω²R comes to 2.21 m/s² — 8.9 % of the 24.79 m/s² gravitational field, leaving an effective 22.58 m/s². Push the spin further and the two terms meet: when ω²R equals GM/R² the effective gravity is zero and loose material at the equator is no longer bound. That is rotational break-up, and the calculator reports it rather than hiding a negative number. Earth would reach it at a rotation period of 1.408 hours.

The model has an honest limit that belongs here rather than in a footnote: it treats the body as a sphere. Real rotating planets bulge at the equator, and that bulge redistributes mass in a way a point-mass field cannot represent. Fed the WGS 84 standard's own defining parameters for Earth — GM = 3.986004418 × 10¹⁴ m³/s², a = 6 378 137.0 m, ω = 7.292115 × 10⁻⁵ rad/s — this calculator returns 9.76437 m/s² at the equator against the standard's normal equatorial gravity of 9.78033 m/s². The 0.163 % shortfall is precisely the oblateness term. So treat results for a fast rotator as good to a few parts in a thousand, not to the last displayed digit.

One more scope note. For the giant planets there is no solid surface to stand on. An agency 'radius' for Jupiter, Saturn, Uranus or Neptune is the level where the atmospheric pressure reaches one bar — a chosen reference, not ground — so 'surface gravity' for those worlds means 'gravity at the 1-bar level' and nothing more.

If what you actually want is your own weight on each planet rather than the physics of an arbitrary body, the Weight on Other Planets calculator does that from a single input and published gravity values, with no mass or radius to look up.

What is surface gravity of a planet calculator?

Surface gravity is the gravitational acceleration a body produces at its own surface, in metres per second squared. For a spherically symmetric mass it is g = GM/R², where G is the gravitational constant, M the mass and R the radius. The result does not depend on the mass of whatever is being attracted, which is why a feather and an anvil fall at the same rate in a vacuum.

Because g goes as M/R², density matters as much as size. Mars has about a tenth of Earth's mass but only half the radius, so its surface gravity is 3.71 m/s² — 38 % of Earth's rather than 10 %. Mercury, despite being much smaller than Mars, has almost identical surface gravity at 3.70 m/s², because it is far denser.

On a rotating body the quantity a scale measures is not g itself. Part of the gravitational field goes into holding the object on a circular path with the rotating ground, so the apparent weight is reduced by ω²R at the equator and by nothing at all at the poles. Geodesists call the field GM/R² and the measured quantity 'normal gravity' or 'surface acceleration', and the difference is why the same object weighs slightly more in Oslo than in Nairobi.

How to use this calculator.

1. 01Choose whether you want the surface gravity, or want to solve back for the mass or the radius.
2. 02Enter the mass in kilograms and the radius in metres. Earth's values are pre-filled from JPL.
3. 03Enter the sidereal rotation period in hours, or 0 to ignore rotation. A retrograde period may be entered negative — only its magnitude is used.
4. 04Read the Newtonian gravity if you want the value comparable with agency tables, and the effective gravity if you want what a scale at the equator would respond to.
5. 05Check the bulk density against a published value as a sanity check; to reproduce a tabulated density for a flattened planet, enter its volumetric mean radius rather than its equatorial one.
6. 06Read the regime note: it flags a rotation correction above 5 %, and flags rotational break-up when the effective gravity reaches zero.

The formula.

The gravitational field of a spherically symmetric body at its surface is g₀ = GM/R², a result Newton proved in the Principia by showing that a spherical shell attracts an external point exactly as if all its mass sat at the centre. The calculator evaluates that with G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² from CODATA 2022.

The rotation term comes from the rotating frame. A point at the equator of a body turning with angular velocity ω = 2π/P travels a circle of radius R, so it needs a centripetal acceleration ω²R, and that comes out of the gravitational field. The apparent gravity is therefore g₀ − ω²R at the equator, falling to g₀ at the poles where the circle has zero radius. Only ω² enters, so the sign of the rotation period is irrelevant — a retrograde rotator such as Venus, which agencies tabulate as −243.018 days, gives the same answer as +243.018 days.

All three solve modes operate on the non-rotating relation. Solving for the mass gives M = g₀R²/G; solving for the radius gives R = √(GM/g₀). Inverting the rotating relation instead would mean rooting the cubic ω²R³ + gR² − GM = 0, whose answer depends on which root you meant, so the calculator keeps the inverse unambiguous and reports the rotation correction at the geometry it lands on.

ROUNDING STAGE. Nothing is rounded part-way through. All arithmetic runs at 40 significant digits in a dedicated high-precision decimal context; rounding happens once, at the return boundary — 10 decimal places for accelerations, ratios and densities, and 12 significant digits for the mass and radius, which span many orders of magnitude.

SIGNIFICANT FIGURES. Two limits apply. First, a mass in kilograms inherits the 22-parts-per-million uncertainty of G, so absolute gravities are capped near five significant figures however precise your inputs look. Second, and far larger, the point-mass model itself is only good to a few parts in a thousand for a flattened body — see the WGS 84 comparison above. Do not quote the tenth decimal place of an answer as though it meant anything.

SIGN CONVENTION AND INVALID DOMAIN. Gravity and the centrifugal term are reported as positive magnitudes; the effective gravity is their difference and may legitimately be negative, which means rotational break-up. Mass, radius and target gravity must all be strictly positive: at R = 0 both GM/R² and the density are singular, and at g₀ = 0 the radius R = √(GM/g₀) is infinite. Those raise a labelled error against the offending field. A rotation period of 0 is read as 'ignore rotation' rather than as an infinite spin.

A worked example.

Frequently asked questions.

References& sources.

1. [1]NASA JPL Solar System Dynamics, 'Planetary Physical Parameters': column headers 'Equatorial Radius (km)', 'Mean Radius (km)', 'Mass (×10²⁴ kg)', 'Bulk Density (g cm⁻³)', 'Sidereal Rotation Period (d)' and 'Equatorial Gravity (m s⁻²)'. Earth: 6378.1366 km, 6371.0084 km, 5.97217, 5.5134, 0.99726968 d, 9.80 m/s². Jupiter: 71 492 km, 1898.125, 0.41354 d, 24.79 m/s². The gravity column is noted as 'derived from other referenced values and uncertainties in this table'. Independent, open access. Retrieved 2026-07-29.
2. [2]NIST, CODATA 2022 recommended value of the Newtonian constant of gravitation: G = 6.674 30(15) × 10⁻¹¹ m³ kg⁻¹ s⁻², relative standard uncertainty 2.2 × 10⁻⁵. Independent, open access. Retrieved 2026-07-29.
3. [3]BIPM, 3rd CGPM (1901) Resolution 2, 'Declaration on the unit of mass and on the definition of weight; conventional value of gₙ': adopts 980.665 cm/s² as the conventional standard acceleration due to gravity, and defines weight as a force equal to mass times that acceleration. Independent, open access. Retrieved 2026-07-29.
4. [4]NGA Office of Geomatics, World Geodetic System 1984, standard NGA.STND.0036_1.0.0_WGS84 (revision of 8 July 2014). DEFINING parameters verified on the NGA WGS 84 page and used directly in the worked example: semi-major axis a = 6 378 137.0 m, flattening 1/f = 298.257223563, angular velocity ω = 7 292 115 × 10⁻¹¹ rad/s, geocentric gravitational constant GM = 3.986 004 418 × 10¹⁴ m³/s². The normal equatorial gravity of 9.780 325 3359 m/s² quoted alongside it is a DERIVED constant from the same standard; that derived table was not machine-readable at retrieval, so treat it as a standards-document reference rather than a fetched figure. Retrieved 2026-07-29.
5. [5]Newton, I. (1687). Philosophiae Naturalis Principia Mathematica, Book I, Propositions LXXI and LXXIII — the shell theorem, which is what licenses treating a spherically symmetric planet as a point mass at its centre. Print/bibliographic reference.