Example 1
Reference case: Earth at 5.972×10²⁴ kg and a 70 kg person, centers 6.3710e+6 m apart. The person’s acceleration is 9.82 m/s², not the conventional 9.80665 m/s².
F = 687.4 N · a₂ = 9.82 m/s²
Gravitational Force Calculator
Enter two masses in kilograms and the distance between their centers in meters. You may leave G blank: the calculator then uses 6.67430×10⁻¹¹. The result is the force F = G·m₁·m₂/r². Earth and a 70 kg person at r = 6371 km give 687.4 N, an acceleration of 9.82 m/s² for that person. The model does not return g = GM/r² without a second mass or orbital speed as the primary result.
Result
Enter both masses and the center-to-center distance. The force, the accelerations and the F(r) chart appear here.
For the field of a single mass, use the g = GM/r² calculator; for potential energy near the surface at constant g, Ep = m·g·h. On a circular orbit, this same F supplies the centripetal force. For motion at constant g, use the free fall calculator.
How it works
m₁ · m₂5.972×10²⁴ kg · 70 kg
÷ r²6371 km
F687.4 N
Newton’s law of universal gravitation says that two masses attract each other along the line of centers with a force F = G·m₁·m₂/r². The force is mutual: the same magnitude acts on both masses, while the accelerations a = F/m differ. The default constant G is the CODATA 2018 value, 6.67430×10⁻¹¹ N·m²/kg². A blank G field means that number.
For an Earth mass of 5.972×10²⁴ kg, a 70 kg person and r = 6.371×10⁶ m, the product G·m₁·m₂ divided by r² is 687.4 N. The person’s acceleration is then a₂ = F/m₂ = 9.82 m/s². That is not the conventional standard gravity of 9.80665 m/s²: the calculator uses Newton’s law with these three numbers, not the product 70 × 9.80665. One kilogram at the same r gives 9.82 N.
The force falls with the square of the distance. Doubling r lowers 687.4 N to 171.8 N. At an altitude of 400 km, with r = 6.771×10⁶ m, the same 70 kg person is pulled with 608.6 N, not zero. Weightlessness on an orbital station is shared free fall along the orbit, not the disappearance of gravity.
Two 1,500 kg cars whose centers are 3 m apart attract each other with 1.6686×10⁻⁵ N. Two 1 kg masses 0.10 m apart give 6.6743×10⁻⁹ N. Those forces stay invisible without a sensitive balance.
What the result panel shows. The calculator reports F, the acceleration of the second mass a₂ = F/m₂, the acceleration of the first mass a₁ = F/m₁, the force at twice the distance, the r, and the G it used. For the reference values that is 687.4 N, 9.82 m/s², 1.1510×10⁻²² m/s² and 171.8 N. The chart draws F against r at fixed masses: the curve falls as 1/r². The field of one mass, g = GM/r², is covered by the gravitational acceleration calculator, and circular-orbit speed is covered by the first cosmic velocity calculator.
The calculator uses three required numbers: mass m₁, mass m₂ and distance r, plus an optional G. The masses and the distance must be greater than zero. A distance entered in kilometers instead of meters makes the force a million times too large, because r is squared in the denominator. The header unit switch changes masses to pounds, distance to feet and force to pounds-force. G stays in N·m²/kg². A comma and a period in the fields are equivalent. The button “Fill this example” enters the numbers from the chosen example and refreshes the panel and the chart.
Formula
F = G · m1 · m2 / r2
a2 = F / m2 = G · m1 / r2; a1 = F / m1
Model: Newton’s law of universal gravitation (Isaac Newton, Philosophiae Naturalis Principia Mathematica, 1687). By default G = 6.67430×10−11 N·m2/kg2 (CODATA 2018). The calculator finds the force between two masses and both accelerations. It does not find orbital speed or potential.
Theory, derivation and domain
Definition of gravitational force and the inverse-square law
Gravitational force on this calculator is the mutual attraction of two masses. Newton’s law says the magnitude is proportional to the product of the masses and inversely proportional to the square of the distance between centers: F = G·m₁·m₂/r². The direction lies on the segment that joins the centers. The senses are opposite, so each mass is pulled toward the other. The SI unit of force is the newton.
The formula this calculator uses includes the constant G. When the field is blank, G is 6.67430×10⁻¹¹ N·m²/kg². For an Earth mass of 5.972×10²⁴ kg, a 70 kg person and r = 6.371×10⁶ m, the force is 687.4 N. That pair, Earth and 70 kg at a radius of 6371 km, comes back in the later examples. Doubling r makes the force four times smaller, because (2r)² = 4r². Rounded to one decimal place, the panel shows 171.8 N.
History: Newton, Cavendish and the constant G
Isaac Newton set out the law of universal gravitation in Philosophiae Naturalis Principia Mathematica (1687). The same force was meant to explain the fall of bodies near the surface and the motion of the planets. The inverse-square dependence was a hypothesis that Newton tied to Kepler’s laws for orbits. That book does not yet contain the modern number G in SI units. Newton compared proportions, not newtons per kilogram squared.
In 1798 Henry Cavendish measured the density of the Earth with a torsion balance. He did not publish G in the form used today. G was later inferred from that density and from the size of the Earth. The recommended value in the blank field of this calculator is CODATA 2018: 6.67430×10⁻¹¹ N·m²/kg², with a standard uncertainty of 0.00015×10⁻¹¹. The name newton comes from the SI. The BIPM SI Brochure defines the newton as kg·m/s². You can replace G with your own number, for example 6.67×10⁻¹¹, and the result then belongs to that constant, not to CODATA.
Deriving the acceleration a = F/m
Newton’s second law says the acceleration of a body is the net force divided by its mass. The gravitational force therefore gives two accelerations: a₂ = F/m₂ = G·m₁/r² and a₁ = F/m₁ = G·m₂/r². For the reference values, a₂ = 687.4 / 70 = 9.82 m/s², and a₁ = 687.4 / 5.972×10²⁴ ≈ 1.1510×10⁻²² m/s². Earth hardly accelerates, because its mass is huge. The person accelerates as in the school notion of weight, but the number 9.82 comes from this G, this mass and this r.
Conventional standard gravity is 9.80665 m/s². The product 70 × 9.80665 is about 686.5 N, and that is not the path of this calculator. The path is one formula: F = G·m₁·m₂/r², then a = F/m. The field of one mass, without a second mass in the numerator of the force, is covered by the g = GM/r² calculator. Here both masses stay, and the result is a force in newtons.
Assumptions: point masses, spheres and center-to-center distance
The formula applies to point masses or to spherically symmetric bodies. Newton showed that outside a uniform sphere or a thin spherical shell the force is the same as if all the mass sat at the center. That is why r is measured to the center, not to the surface. For Earth and a person, a height of about 2 m disappears next to 6,371,000 m. A gap between the surfaces would be the wrong r and would overstate the force.
The calculator does not add Earth’s flattening, mountains, local geology or tides. It does not subtract air resistance and it does not apply a relativistic correction. It also does not separate “scale weight” from gravitational force: a scale in an elevator or in orbit reads a contact force, while this formula reports the attraction. Mass and r must be positive. r = 0 is meaningless, because the denominator vanishes.
Mutual force and a circular orbit
Newton’s third law requires the forces to have the same magnitude. Swapping m₁ and m₂ does not change F. It only changes which mass receives which acceleration. On a circular orbit this force supplies the centripetal force: G·M·m/r² = m·v²/r, so v = √(GM/r). This calculator does not find v. Circular-orbit speed is covered by the first cosmic velocity calculator, and the force m·v²/r is covered by the centripetal-force calculator. Escape speed, √(2GM/r), is covered by the second cosmic velocity calculator.
At an altitude of 400 km the radius from Earth’s center is about 6,771,000 m. For 70 kg the calculator finds 608.6 N, and the acceleration falls to 8.69 m/s². That is still a large force. Weightlessness on a station follows from the station and the astronaut falling together along the orbit. It does not follow from F dropping to zero. Free fall at constant g, the formula h = ½gt², is a separate page and assumes g does not change over a short distance.
Units, orders of magnitude and attraction around us
The force comes out in newtons. With the header set to US units, the panel shows pounds-force and, in parentheses, newtons. Masses switch to pounds, distance to feet, and G stays in N·m²/kg², because the constant is defined in SI. The notations 5.972e24 and 5,972e24 mean the same mass. Kilometers in the r field, instead of meters, multiply the force by one million.
The orders of magnitude below are the result of this formula for the stated masses and r, not a measurement from a mission. Two 1,500 kg cars 3 m apart: 1.6686×10⁻⁵ N. Two kilograms 0.10 m apart: 6.6743×10⁻⁹ N. Earth and a Moon of mass 7.35×10²² kg at r = 3.844×10⁸ m: 1.9827×10²⁰ N. A Sun of mass 1.989×10³⁰ kg and Earth at r = 1.496×10¹¹ m: 3.5424×10²² N. Your own G = 6.67×10⁻¹¹, Earth and 1 kg at r = 6371 km give 9.81 N, a little less than 9.82 N with the CODATA value. That is why the G field is optional and visible: the result belongs to the constant the calculator used.
Sources
The definition, the derivation and the history come from textbooks and primary sources: I. Newton, Philosophiae Naturalis Principia Mathematica, London 1687; H. Cavendish, Experiments to determine the density of the Earth, Philosophical Transactions of the Royal Society 88 (1798); D. Halliday, R. Resnick and J. Walker, Fundamentals of Physics, gravitation chapter; H. D. Young and R. A. Freedman, University Physics, gravitation chapter; R. P. Feynman, R. B. Leighton and M. Sands, The Feynman Lectures on Physics, Vol. I, Chapter 7; CODATA 2018 recommended value of G (NIST); BIPM, SI Brochure, 9th edition (the newton); NASA Planetary Fact Sheet for Earth, as the order of magnitude of the mass and radius used on this page, rounded to 5.972×10²⁴ kg and 6371 km. This page is for educational purposes.
How to use
- Leave G blank unless the problem gives another constant. A blank field means 6.67430×10⁻¹¹ N·m²/kg².
- Enter m₁ and m₂ in kilograms. For Earth and a person: 5.972e24 and 70.
- Enter r between the centers in meters, for example 6371000. That is Earth’s radius, not the gap between the surfaces.
- Click Calculate, or wait for the live result. For these numbers the calculator shows 687.4 N.
- Check the panel: both accelerations, the force at twice the distance, the r and the G that was used. Clear removes the inputs and hides the panel.
Symbol glossary
F is the mutual force, G the gravitational constant, m₁ and m₂ the masses, and r the distance between centers. For Earth, 70 kg and r = 6371 km the force is 687.4 N.
- F
- Mutual force in newtons. For 5.972×10²⁴ kg, 70 kg and 6.371×10⁶ m it is 687.4 N.
- G
- Constant 6.67430×10⁻¹¹ N·m²/kg² (CODATA 2018). A blank field uses that value. Do not enter g = 9.82 here.
- m₁
- First mass in kilograms. In the example Earth is 5.972×10²⁴ kg.
- m₂
- Second mass. Swapping it with m₁ does not change F, only which acceleration is larger.
- r
- Center-to-center distance in meters. A value in kilometers makes the force a million times too large.
- a
- Acceleration a = F/m. For the reference values a₂ = 9.82 m/s² and a₁ ≈ 1.1510×10⁻²² m/s².
Worked examples
Example 2
A 0.20 kg mass on Earth’s surface, same radius. The force is 350 times smaller than at 70 kg, because F is proportional to m₂.
F = 1.96 N
Example 3
Earth and a Moon of mass 7.3500e+22 kg at r = 3.8440e+8 m. This is the force that holds the Moon in orbit, not a person’s weight.
F = 1.9827e+20 N
Example 4
Two cars of 1,500 kg each, centers 3 m apart. The force is far too small to notice by hand.
F = 1.6686e-5 N
Example 5
A 500 kg satellite at an altitude of 400 km, r = 6.7710e+6 m. The acceleration of m₂ falls to 8.69 m/s².
F = 4,347.0 N · a₂ = 8.69 m/s²
Example 6
An 80 kg person on that same orbit, r = 6.7710e+6 m. This is not zero: weightlessness is shared free fall, not a missing force.
F = 695.5 N
Example 7
Two 1 kg masses 0.10 m apart. That force shows up only on a torsion balance.
F = 6.6743e-9 N
Example 8
The Sun at 1.9890e+30 kg and Earth at r = 1.4960e+11 m. This force holds Earth on its yearly orbit.
F = 3.5424e+22 N
Example 9
Two 10 kg dumbbells, centers 0.5 m apart. The attraction is smaller than the weight of a speck of dust.
F = 2.6697e-8 N
Example 10
A rounded textbook G = 6.67×10⁻¹¹ for Earth and a 1 kg mass at r = 6371 km. The rounded constant gives a slightly different force from the CODATA default.
F = 9.81 N
Common questions
What does this calculator compute, and which formula does it use?
The calculator finds the gravitational force from Newton’s law, F = G·m₁·m₂/r². For an Earth mass of 5.972×10²⁴ kg, a 70 kg person and r = 6.371×10⁶ m, the force is 687.4 N.
Which units should the masses, the distance and G use?
Enter masses in kilograms and the center-to-center distance in meters. The force is in newtons, and G stays in N·m²/kg². With the header set to US units, masses are in pounds, distance is in feet and force is in pounds-force; the panel also shows the SI value.
Do I have to fill in G?
No. A blank field means the CODATA 2018 value, 6.67430×10⁻¹¹ N·m²/kg². Your own number, for example 6.67e-11, replaces that default in the formula.
Is the force the same on both masses?
Yes. Newton’s third law: the forces have the same magnitude and opposite direction along the line of centers. The accelerations differ because a = F/m. At 687.4 N a 70 kg person accelerates at 9.82 m/s², while Earth accelerates at about 1.1510×10⁻²² m/s².
Why is the distance squared?
This is the inverse-square law. Twice the distance gives four times less force, and three times the distance gives nine times less force. For the reference values, doubling r lowers 687.4 N to 171.8 N.
Is r the gap between the surfaces?
No. In the formula r is the distance between centers. Outside a spherically symmetric body the result is the same as if all the mass sat at the center. A person’s height disappears next to a radius of 6371 km. A distance entered in kilometers instead of meters makes the force a million times too large.
How do I get acceleration from the force?
The acceleration of the second mass is a₂ = F/m₂, and the acceleration of the first mass is a₁ = F/m₁. The field of one mass alone, g = GM/r², is covered by the gravitational acceleration calculator. At 70 kg and 687.4 N, a₂ is 9.82 m/s².
Why is the result not 70 times 9.80665?
Standard gravity, 9.80665 m/s², is a different convention. This calculator uses Newton’s law with the G, masses and r you enter. With Earth’s mass 5.972×10²⁴ kg, r = 6371 km and the default G, the acceleration is 9.82 m/s², not 9.80665 m/s².
Does weightlessness in orbit mean the force is zero?
No. At an altitude of 400 km, with r = 6.771×10⁶ m, a 70 kg person is pulled with 608.6 N. Weightlessness on a station is shared free fall along the orbit, not the disappearance of gravity.
What does this formula leave out?
The calculator does not compute tides, Earth’s flattening, local geology, air resistance or a relativistic correction. It also does not compute orbital speed or escape speed. On a circular orbit this same force supplies the centripetal force, and the speed v = √(GM/r) is covered by the first cosmic velocity calculator.
Are a comma and a period equivalent?
Yes. The calculator reads 5,972e24 and 5.972e24 as the same Earth mass. The fields are empty when you open the page. A hint in a field is not an entered value.
Why is there no result for zero mass or r = 0?
A zero mass is not a body in this formula, and r = 0 would divide by zero. Both masses and the distance must be greater than zero. Your own G, if you enter one, must be greater than zero as well.
How do I calculate gravitational force with F = G·m₁·m₂/r²?
Multiply G by both masses and divide by r². With Earth 5.972×10²⁴ kg, 70 kg, and r = 6,371,000 m you get about 687.4 N.
What is F between Earth and the Moon?
About 1.98×10²⁰ N for masses 5.972×10²⁴ kg and 7.35×10²² kg and a mean center distance near 3.84×10⁸ m.
How is this different from the gravitational acceleration calculator?
There the result is g = GM/r² with no second mass. Here the result is the force between two masses. The second mass’s acceleration is F/m₂.
How do I convert kilometers to meters for r?
Multiply by 1,000. Earth radius 6,371 km is 6,371,000 m. Entering kilometers without converting overstates the force about one million times.
What is F between two 70 kg people one meter apart?
About 3.3×10⁻⁷ N. Everyday masses pull weakly; planetary masses make the same formula give hundreds of newtons.
Sources
The definition, the unit, the history and the derivation below come from recognized physics and metrology sources.
- Wikipedia — Newton’s law of universal gravitation
- Wikipedia PL — prawo powszechnego ciążenia
- NIST — CODATA recommended value of the gravitational constant G
- BIPM — The International System of Units (SI), 9th edition
- NASA — Planetary Fact Sheet, Earth
- Feynman Lectures on Physics, Vol. I, Ch. 7 — The Theory of Gravitation
- NIST SP 330 — The International System of Units
- Newton, I., Philosophiae Naturalis Principia Mathematica, London 1687.
- Cavendish, H., Experiments to determine the density of the Earth, Philosophical Transactions of the Royal Society 88, 1798.
Page updated November 3, 2026.