g and its link with G
The acceleration due to gravity (g) is the acceleration of a body falling freely under Earth's pull. Take a body of mass m on the surface. Earth's pull on it is GMm/R². By Newton's second law this force = mg. So
mg = GMm/R² → g = GM/R²
The mass m cancels, so a stone and a feather fall with the same g (when there is no air). With M = 6 × 10²⁴ kg and R = 6.4 × 10⁶ m, g ≈ 9.8 m/s².
Weight = mg (a force, in N). Mass (kg) is the amount of matter and stays the same everywhere. Weight changes when g changes.
Change of g with height
At height h above the surface, the distance from the centre is (R + h). So
g_h = GM/(R + h)² = g R²/(R + h)²
For small heights (h ≪ R)
- g_h = g (1 + h/R)⁻²
- Binomial rule: (1 + x)⁻² ≈ 1 − 2x when x is small.
- g_h ≈ g (1 − 2h/R)
So g falls by a fraction 2h/R. At h = 32 km, 2h/R = 1%, so g falls by 1%.
Use the exact formula when h is not small (like h = R or h = R/2).
Change of g with depth
Go down a mine to depth d. Now the distance from the centre is (R − d).
- The outer shell of thickness d pulls you equally from all sides, so it gives no net force.
- Only the inner sphere of radius (R − d) pulls you. Its mass (uniform density ρ) is M' = M (R − d)³/R³.
- g_d = GM'/(R − d)² = GM (R − d)/R³
- g_d = g (1 − d/R)
This is exact for a uniform Earth. It is a straight line: g falls evenly as you go down.
At the centre
d = R gives g = 0. At the centre, the Earth pulls you equally in every direction. You would feel weightless (but squeezed by huge pressure!).
The full picture: g against distance r
- Inside (r < R): g = g_s r/R, a straight line from 0 at the centre.
- At the surface (r = R): g is the largest, g_s ≈ 9.8 m/s².
- Outside (r > R): g = g_s R²/r², a curve falling toward zero.
So g decreases both when you go up and when you go down. For the same small distance x, going up (2x/R) lowers g twice as much as going down (x/R).
Other small changes (for interest)
Earth is slightly flat at the poles and it spins. So g is about 9.83 m/s² at the poles and about 9.78 m/s² at the equator.
Try it: measure g with a pendulum
Tie a small heavy nut to a 1 m thread. Let it swing gently (small angle). Time 20 swings with a phone stopwatch and divide by 20 to get T. Then g = 4π²L/T². With L = 1 m you should get T ≈ 2.0 s and g close to 9.8 m/s². In the 3D, slide the ball to 1.0 R and check that the readout shows 9.8.
Key formulas and definitions
- g = GM/R²
- GM = gR²
- g_h = g R²/(R + h)²
- g_h ≈ g(1 − 2h/R) for h ≪ R
- g_d = g(1 − d/R)
- g = 0 at the centre
- Weight W = mg
Worked examples
1. Find g on the surface. M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹.
g = GM/R² = 4.0 × 10¹⁴ / 4.096 × 10¹³ ≈ 9.77 m/s² ≈ 9.8 m/s².
2. Find g at a height h = R/2.
g_h = g R²/(1.5R)² = g/2.25 = 9.8/2.25 ≈ 4.36 m/s².
3. Find g at a height of 64 km (R = 6400 km).
h ≪ R, so g_h ≈ g(1 − 2h/R) = 9.8 (1 − 128/6400) = 9.8 × 0.98 ≈ 9.60 m/s².
4. Find g at a depth of 1600 km.
g_d = g(1 − d/R) = 9.8 (1 − 1600/6400) = 9.8 × 0.75 = 7.35 m/s².
5. A man weighs 600 N on the surface. What is his weight at a height equal to R?
g_h = g R²/(2R)² = g/4. Weight = 600/4 = 150 N.
6. At what height does g become 1% less than at the surface? (R = 6400 km)
2h/R = 0.01 → h = 0.005 R = 32 km.
7. Find the depth at which g equals g at a height of 10 km (small h).
g(1 − d/R) = g(1 − 2h/R) → d = 2h = 20 km.
8. Mass of a planet is 2 times Earth's and its radius is 2 times Earth's. Find g there.
g' = G(2M)/(2R)² = (2/4) g = g/2 = 4.9 m/s².
Common mistakes
- Using g(1 − 2h/R) for big heights like h = R. It only works when h is much smaller than R; otherwise use g R²/(R + h)².
- Thinking g becomes larger when you go down a mine. It becomes smaller: g_d = g(1 − d/R).
- Saying mass changes on the Moon. Mass stays the same; only weight (mg) changes.
- Thinking g depends on the mass of the falling body. The m cancels: g = GM/R² depends only on the planet.