Gravitational Force Converter
Convert gravitational force between units instantly. Includes conversion tables, common equivalents, and calculation formulas.
Gravitational Force Converter
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Formula: F = G * m1 * m2 / r^2
Worked example โ Gravitational force = 686.95 N (about 154.4 lbf)
Formula
F = G * m1 * m2 / r^2
The gravitational force F in Newtons equals the gravitational constant G (6.674e-11) multiplied by both masses (m1 and m2 in kg) divided by the square of the distance r between their centers (in meters). The result can be converted to dynes (multiply by 1e5), pounds-force (divide by 4.44822), or kilogram-force (divide by 9.80665).
Worked Examples
Example 1: Weight of a Person on Earth
Problem:Calculate the gravitational force on a 70 kg person on Earth's surface (mass 5.972e24 kg, radius 6,371 km).
Solution:F = G * m1 * m2 / r^2 F = 6.674e-11 * 5.972e24 * 70 / (6.371e6)^2 F = 2.789e16 / 4.059e13 = 686.95 N
Result:Gravitational force = 686.95 N (about 154.4 lbf)
Example 2: Earth-Moon Gravitational Force
Problem:Find the gravitational force between Earth (5.972e24 kg) and Moon (7.342e22 kg) at 384,400 km apart.
Solution:F = G * m1 * m2 / r^2 F = 6.674e-11 * 5.972e24 * 7.342e22 / (3.844e8)^2 F = 2.926e37 / 1.478e17 = 1.98e20 N
Result:Earth-Moon gravitational force = 1.98 x 10^20 N
Frequently Asked Questions
What is Newton's Law of Universal Gravitation?
Newton's Law of Universal Gravitation states that every object with mass attracts every other object with mass through a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is F = G * m1 * m2 / r^2, where G is the gravitational constant (6.674e-11 N*m^2/kg^2). This law accurately describes gravitational interactions for most everyday and astronomical scenarios, though Einstein's General Relativity provides corrections needed for extreme conditions.
Why does gravitational force decrease with the square of distance?
The inverse-square relationship arises from geometry. Gravitational influence radiates outward from a mass in all directions, spreading over the surface of an expanding sphere. Since the surface area of a sphere is 4*pi*r^2, the gravitational field strength (force per unit area) decreases proportionally to 1/r^2. Doubling the distance reduces the force to one-quarter, tripling it reduces force to one-ninth, and so on. This same principle applies to light intensity and other radially spreading phenomena.
How strong is Earth's gravitational pull on a person?
For a 70 kg person standing on Earth's surface (radius 6,371 km from center, Earth mass 5.972e24 kg), Newton's law gives a gravitational force of about 686 Newtons or roughly 154 pounds-force. This matches the familiar calculation of weight = mass times gravitational acceleration (70 kg * 9.81 m/s^2 = 686.7 N). The force decreases with altitude: at the International Space Station's orbit (about 408 km altitude), it is still about 89% of surface gravity.
What is the gravitational constant G and how was it measured?
The gravitational constant G equals 6.67430e-11 N*m^2/kg^2 and was first measured by Henry Cavendish in 1798 using a torsion balance experiment. He suspended a horizontal bar with small lead balls from a thin wire and measured the tiny twist caused by placing large lead spheres nearby. Modern measurements use similar principles with improved precision. G remains one of the least precisely known fundamental constants because gravitational forces between laboratory-scale objects are extremely small.
What is the difference between mass and weight in this calculation?
Mass in kilograms is an intrinsic property that never changes, while weight is the gravitational force acting on that mass and is reported here in newtons. The kilogram-force output divides the force by the standard gravity of exactly 9.80665 m/s squared, giving the reading a mass-calibrated scale would display. The pound-force output divides by 4.44822 N instead. Move the same mass to Mars and its kilograms are unchanged while all the force outputs shrink.
What is a dyne and why is the newton multiplied by 100,000?
The dyne is the CGS unit of force, defined as the force that accelerates one gram at one centimetre per second squared. Working through the prefixes, one dyne is 10^-5 newtons, so converting newtons to dynes multiplies by 100,000. The unit survives in older physics literature and in surface-tension figures quoted as dynes per centimetre. Its small size makes it convenient for the tiny forces that laboratory gravitation experiments actually measure.
Why do the two accelerations shown differ so enormously?
Newton's third law makes the force on each body identical in magnitude and opposite in direction, but acceleration is a = F / m, so it is inversely proportional to each body's own mass. With the default inputs the 70 kg person accelerates at about 9.8 m/s squared while the Earth accelerates toward the person at roughly 10^-22 m/s squared. The mutual attraction is perfectly symmetric; only the resulting motion is lopsided.
Why is the distance measured between centres rather than between surfaces?
Newton's shell theorem shows that a spherically symmetric body attracts an outside object exactly as if all of its mass were concentrated at its centre. That is why the default distance for a person on Earth is the 6,371,000 m planetary radius rather than zero. Entering a surface-to-surface gap instead makes r far too small and, because the force goes as one over r squared, inflates the answer dramatically.
When should you enter distances in astronomical units or light-years?
Use the astronomical unit, about 1.496 x 10^11 m and roughly the mean Earth-Sun separation, for planetary system problems, and the light-year, about 9.461 x 10^15 m, for interstellar ones. Both simply scale your input before the inverse-square step. Because the force depends on r squared, choosing the wrong length unit does not shift the answer by that factor but by its square, which is how gravitational estimates end up wrong by many orders of magnitude.
When does Newton's law stop being accurate enough?
It breaks down where gravity is strong or speeds approach that of light. The classic case is Mercury, whose perihelion advances about 43 arcseconds per century more than Newtonian gravity predicts, a discrepancy general relativity resolved. Satellite navigation also needs relativistic clock corrections to stay accurate. For laboratory masses, planets, spacecraft trajectories, and ordinary engineering, the inverse-square law used here is more than sufficient.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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