Gravitational Force (F=mg) | Formula, Calculator + Notes

Choose the right gravity formula

F = mg is not restricted to Earth's surface. It gives the gravitational force wherever the appropriate local g is known. Treating g as a constant 9.8 N/kg is the near-Earth approximation. Substituting g = GM/r² gives F = GMm/r².

For the inverse-square formulas, model the bodies as point masses or use the external field of a spherically symmetric body. The same simple centre-distance expression is not generally valid inside a body or close to an irregular mass distribution.

Mass, weight and support force

Mass measures inertia and is expressed in kg. Weight is the gravitational force on that mass and is expressed in N. A bathroom scale senses a support force and may display the equivalent mass under a calibration assumption. In a stationary lift, support force usually equals weight; in an accelerating lift it need not. In free fall, support force can be nearly zero while gravity still acts.

How distance and mass change the force

Keep both masses fixed. Doubling their centre-to-centre distance makes r² four times larger, so the force becomes one-quarter. Tripling it gives one-ninth. Halving it makes the force four times larger.

F_new / F_old = (m₁,new / m₁,old) × (m₂,new / m₂,old) × (r_old / r_new)²

For example, doubling one mass and doubling the distance gives 2 ÷ 2² = one-half of the original force. Doubling both masses while doubling distance gives 2 × 2 ÷ 2² = 1: the force is unchanged.

Field lines are a drawing convention, not physical threads. Around a spherical source, the area of a sphere grows as 4πr²; spreading a fixed total field flux over that area illustrates the inverse-square dependence. With several sources, add the individual gravitational forces as vectors, including their directions.

Planet presets: keep the reference radius consistent

The values below match this calculator's presets. The model gravity column is calculated from the displayed mass and mean radius. It should not be confused with JPL's separate equatorial gravity column: for example, the Jupiter mean-radius model gives about 25.92 m/s², whereas its equatorial reference is 24.79 m/s².

Planet masses and mean radii follow NASA JPL's reference table, with Earth's radius rounded to a metre. The Moon and Sun presets are rounded teaching values. These nominal inputs support learning rather than precision ephemeris or spacecraft design. Reference links and data scope appear at the end.

A more massive planet does not automatically have a stronger field at its surface: a larger radius reduces GM/R². Saturn illustrates why both quantities matter. Do not mix a mean radius, an equatorial field and an independently rounded standard g as if they were one self-consistent model.

Gravity at altitude and weightlessness

At altitude h above a spherical body of radius R, use r = R + h:

g(h) = GM / (R + h)²

Take M = 5.972 × 10²⁴ kg and R = 6.371 × 10⁶ m. At a representative 400 km altitude, h = 400,000 m and r = 6,771,000 m. The model gives g ≈ 8.694 m/s², about 88.53% of its surface value, 9.820 m/s². That is still a strong gravitational field.

Astronauts float because they and their spacecraft are in continuous free fall together. Their apparent weight, the support force they feel, is nearly zero. Gravity has not switched off. Small residual accelerations make the real environment microgravity rather than perfect weightlessness.

Why these orbital formulas work

For a circular orbit, gravity supplies the centripetal force: GMm/r² = mv²/r. Cancelling m and rearranging gives v² = GM/r. One circuit covers distance 2πr, so T = 2πr/v. At a greater circular-orbit radius around the same central body, speed decreases and period increases.

Kepler's laws and the small-satellite approximation

In the ideal two-body model, bound orbits are ellipses with the common centre of mass at a focus, and the line joining the bodies sweeps equal areas in equal times. In the usual planet-and-small-satellite description, the central body is approximately at that focus. For the relative orbit, with semi-major axis a equal to the sum of the two barycentric semi-major axes, the more general period relation is T² = 4π²a³/[G(M + m)]. Replace M + m by M only when m is negligible; for a circle, a = r.

Geostationary is more than a 24-hour orbit

A geostationary satellite needs a circular, equatorial orbit in Earth's direction of rotation, with a period of approximately 86,164 s, one sidereal day. A 24-hour solar day is 86,400 s. Using 86,164 s and M = 5.972 × 10²⁴ kg gives a radius near 42,164 km from Earth's centre, roughly 35,786 km above the equator. Those rounded altitude figures use an equatorial Earth radius of about 6,378 km.

Some communications and weather satellites use geostationary orbit; many others use different orbits. A period match alone does not make an inclined or eccentric orbit appear fixed over one point.

Escape speed and energy

In this ideal isolated model, an object starting at radius r needs enough kinetic energy to reach infinity with zero remaining speed: ½mv² − GMm/r = 0. Hence v_esc = √2 v_c at the same radius. This is about 41.4% more speed, not twice the speed. It is an energy threshold for unpowered escape, not a rocket launch prescription.

Potential energy U = −GMm/r is negative because its reference is zero at infinity. A negative U alone does not prove an object is bound; the total energy K + U matters. For a circular orbit, K = GMm/(2r) and E = −GMm/(2r). The energy to unbind that orbit is |E|, not |U|. Near the surface over a small height change, ΔU ≈ mgΔh with nearly constant g.

For focused extensions, use the orbital velocity calculator, escape velocity calculator, Hohmann transfer orbit tool or delta-v calculator. The Mars rocket trajectory calculator provides another educational application; simplified calculators are not flight-planning systems.

G and g are different quantities

G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² is the central value of the measured gravitational constant in the 2022 CODATA recommendation. Its standard uncertainty is 0.00015 × 10⁻¹¹ in those units. G is not exact. Henry Cavendish's 1798 torsion-balance experiment determined Earth's mean density and enabled inference of G.

g is local gravitational field strength and varies with position. Conventional standard gravity g₀ = 9.80665 m/s² exactly is a defined reference, not the actual field at every place on Earth. Use 9.8 or 10 only when the question or required approximation calls for it.

The unit check is useful: kg × N/kg = N, and N/kg = m/s² because 1 N = 1 kg·m/s². For the powers of ten in astronomical calculations, review the scientific notation examples.

Six worked examples, from weight to orbit

1. Convert grams before using F = mg

A 750 g book has mass 0.750 kg. For g = 9.8 N/kg, F = 0.750 × 9.8 = 7.35 N downward. Entering 750 as kilograms would make the result 1,000 times too large.

2. Same mass, different weight

A 65 kg student has Earth weight 65 × 9.8 = 637 N. With lunar g = 1.62 N/kg, the weight is 65 × 1.62 = 105.3 N. The mass is still 65 kg.

3. Rearrange the formula

An object weighs 196 N where g = 9.8 N/kg. Divide both sides of F = mg by g: m = F/g = 196/9.8 = 20 kg. If mass and weight were known instead, g = F/m.

4. Two separated spheres

Model spherical 12 kg and 18 kg objects with centres 0.50 m apart. F = (6.67430 × 10⁻¹¹) × 12 × 18 / 0.50². The mass product is 216 kg² and the squared distance is 0.25 m², giving 5.7665952 × 10⁻⁸ N, or 5.77 × 10⁻⁸ N to three significant figures. Each sphere feels that force toward the other.

5. Field at a planet's surface

For M = 6.0 × 10²⁴ kg and R = 6.4 × 10⁶ m, g = GM/R² = (6.67430 × 10⁻¹¹)(6.0 × 10²⁴)/(6.4 × 10⁶)² = 9.77680664… m/s², approximately 9.8 m/s² to two significant figures. Squaring R means squaring its coefficient and its power of ten.

6. A circular orbit 400 km above Earth

Use M = 5.972 × 10²⁴ kg and R = 6.371 × 10⁶ m. First r = R + h = 6.771 × 10⁶ m. Then v = √(GM/r) ≈ 7,672.5 m/s, and T = 2πr/v ≈ 5,544.9 s = 92.416 min. Escape speed at the same radius is approximately 10,850.5 m/s. These ideal circular-orbit values are not an exact prediction of the ISS's changing trajectory.

Common mistakes to catch before calculating

• Weight in kg: give force in N and mass in kg
• Using altitude as r: add the body's radius first
• Forgetting to square distance: at twice the separation, force is one-quarter
• Mixing km and m: 400 km is 400,000 m; the error is squared in GM/r²
• Treating g₀ as universal: standard gravity is a convention; local g can differ
• Using circular speed for every orbit: elliptical speed changes around the orbit
• Calling orbit “no gravity”: orbiting objects are falling under gravity
• Rounding too early: keep extra digits while calculating, then round the final result

Practice problems with explained answers

Try each question before opening its answer. Use G = 6.67430 × 10⁻¹¹ N·m²/kg² where needed.

How this fits into a physics course

Start with units, F = mg and rearranging formulas. Then practise inverse-square ratios, field strength and centre-distance diagrams. Circular-orbit derivations and gravitational potential usually require additional mechanics. Follow your own specification and formula sheet: this page is not a claim that every course assesses every topic.

For related study, see the existing IB Physics revision notes and AP Physics 1 2025 past-paper resource. After marking relevant practice, the Physics Regents score calculator is a separate score-estimation tool.

Beyond the model

Newtonian gravity is useful for many introductory problems. General relativity describes gravity through spacetime geometry and is needed for effects such as gravitational time dilation, light deflection and strong-field motion near compact objects. Relativistic timing corrections matter for precision satellite navigation. The calculators here do not model those effects, extended-body tidal stresses or real mission trajectories.

Gravity calculations also support structural loads, satellite motion and geophysical measurements. Tides involve differences in gravitational pull across an extended body, so a single force number alone does not describe their full pattern.

Frequently asked questions