Advanced · 13 min
Gravitation beyond the surface
Apply inverse-square relationships using distance from a body's centre.
An inverse-square attraction
Newton's law gives gravitational force magnitude F = G × m₁ × m₂ ÷ r². The distance r is between the centres of point masses or spherically symmetric bodies that do not overlap. Force grows with either mass and decreases with the square of separation. Doubling separation gives one quarter of the original force. Both bodies experience equal force magnitudes in opposite directions.
Worked example
Two bodies attract with a force of 18 N at separation r. At separation 3r, with masses unchanged, force = 18 ÷ 3² = 2 N.
Field strength changes with distance
Outside a spherical body, gravitational field strength is g = G × M ÷ r². Near a planet's surface, small changes in height barely change r, so constant g is useful. Farther away, use the centre-to-centre distance. At height h above a planet of radius R, the distance is R + h, not h. Circular orbits also need inward acceleration: gravity can supply the required centripetal force.
Worked example
If surface field strength is 12 N/kg, then at distance 2R from the centre: g = 12 × (R ÷ 2R)² = 3 N/kg
Use the correct distance
- For a planet with surface field strength 16 N/kg, find the field strength at distances 2R and 4R from its centre.
- Express each distance as a height above the surface. Explain why substituting that height directly into the inverse-square formula would be wrong.