Library / Foundations: Physics / FPH-41
Foundations: Physics
Newton’s law of gravitation
Attraction between two masses falls with the square of the distance between their centres.
- Ref
- FPH-41
- Rev
- v1
- Field
- Foundations: Physics
- Result
- F [N]
M and m are the two masses and r is the distance between their centres, not between their surfaces. The force acts equally and oppositely on both bodies. The default is a 1000 kg satellite at about 400 km altitude, which is 6 771 km from the centre of the Earth.
Attraction between two masses falls with the square of the distance between their centres. Doubling the distance cuts the force to a quarter.
Calculator
- Gravitational field strength GM/r²8.694 m/s²
- Force if the distance is doubled2,174 N
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At 400 km up the pull is still about 89 % of the weight at the surface (8.7 m/s² against 9.8 m/s²). Astronauts float because they are falling freely, not because gravity has gone.
- The force on the satellite is exactly the force on the Earth. The Earth hardly moves because its mass is so large.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| M Mass of the large body | 1e20 to 1e31 kg | 0.1456 | 1.456e10 | 1.456e10 | +10% | +10 % |
| r Distance between centres | 100,000 to 1e9 m | 0.3986 | 3.986e7 | 3.986e7 | +10% | −17.4 % |
| m Mass of the small body | 1 to 1,000,000 kg | 8.694 | 8,694,000 | 8,694,000 | +10% | +10 % |
Sweep of Distance between centres
| Distance between centres (m) | Gravitational force (N) |
|---|---|
| 100,000 | 3.986e7 |
| 316,200 | 3,986,000 |
| 1,000,000 | 398,600 |
| 3,162,000 | 39,860 |
| 1e7 | 3,986 |
| 3.162e7 | 398.6 |
| 1e8 | 39.86 |
| 3.162e8 | 3.986 |
| 1e9 | 0.3986 |
Sweep one variable
Variables and how each one contributes
Gravitational force
What the equation gives you. Everything else feeds it.
Mass of the large body
Mass of the large body, such as a planet. Default is Earth. Directly proportional: a body with twice the mass pulls twice as hard.
Mass of the small body
Mass of the small body, such as a satellite. Directly proportional: the same pull on a heavier satellite is simply a bigger force.
Distance between centres
Distance from the centre of one body to the centre of the other. Inverse square: doubling r cuts the force to a quarter. Because the default r is large, a 10 % rise in r lowers F by about 17 %.
Worked example with the values above
- Numerator: G M m = 6.674×10−11 × 5.972×1024 × 1,000 = 3.986×1017.
- Denominator: r² = (6,771,000)² = 4.585×1013.
- F = 3.986×1017 / 4.585×1013 = 8,694 N.
Assumptions and limits
- Spherical bodies (or point masses), so that r is measured between centres.
- Newtonian gravity: speeds are far below the speed of light and the fields are weak.
Common mistakes
- Using the altitude above the surface as r. Add the radius of the planet (6 371 km for Earth).
- Entering r in km. The formula needs metres.
Statistics and uncertainty
F depends on M and m to the first power and on r to the inverse square, so the distance is the term that matters. A 1 % error in r gives a 2 % error in F, while 1 % errors in M and m each give 1 %. Treating them as independent, the combined error is about √(εM² + εm² + 4εr²). G itself is known only to about 5 significant figures, so quoting more digits is false precision.
Where it is used
- Load on a satellite structure or its attitude control.
- Comparing the pull of different bodies at a given distance.
Maths you need
- Inverse-square law: a quantity that falls as 1/r²
- Standard form: writing 5.972 × 10²⁴