Mass & weight
Gravitational field strength
Gravitational field strength
The gravitational field strength (\( g \)) at a point is the gravitational force acting per unit mass placed at that point.
\[ g = \frac{W}{m} \]
SI unit: newtons per kilogram (N/kg)
It is a vector quantity — it points in the direction the gravitational force acts (downward on Earth).
It is a vector quantity — it points in the direction the gravitational force acts (downward on Earth).
At the Earth's surface: \( g \approx 10\,\text{N/kg} \)
This means: every kilogram of mass experiences a downward gravitational force of 10 N.
Note: \( g \) is also the free-fall acceleration (10 m/s²) — the value is the same because of Newton's second law (\( F = ma \) gives \( W = mg \), so \( g \) in N/kg numerically equals \( g \) in m/s²).
This means: every kilogram of mass experiences a downward gravitational force of 10 N.
Note: \( g \) is also the free-fall acceleration (10 m/s²) — the value is the same because of Newton's second law (\( F = ma \) gives \( W = mg \), so \( g \) in N/kg numerically equals \( g \) in m/s²).
Gravitational field strength on other bodies
The value of \( g \) depends on the mass and radius of the planet (or moon). Larger, denser bodies have stronger gravitational fields.
| Body | \( g \) (N/kg) | Compared with Earth |
|---|---|---|
| Earth | 10 | 1× (reference) |
| Moon | 1.6 | ≈ 1/6 of Earth |
| Mars | 3.7 | ≈ 0.37× |
| Jupiter | 25 | ≈ 2.5× |
| Deep space (far from all planets) | ≈ 0 | Effectively zero |
Worked example — weight on the Moon
An astronaut has a mass of 80 kg. Calculate their weight (a) on Earth and (b) on the Moon (\( g_{\text{Moon}} = 1.6\,\text{N/kg} \)).
(a) \( W_{\text{Earth}} = mg = 80 \times 10 = 800\,\text{N} \)
(b) \( W_{\text{Moon}} = mg = 80 \times 1.6 = 128\,\text{N} \)
The mass remains 80 kg in both cases. The weight changes because \( g \) is different.
Worked example — find g from weight and mass
On Mars, a 5.0 kg rock has a weight of 18.5 N. Calculate the gravitational field strength on Mars.
\( g = \dfrac{W}{m} = \dfrac{18.5}{5.0} = 3.7\,\text{N/kg} \)
Gravitational field direction
Around a planet or moon, the gravitational field points radially inward — towards the centre of the planet from every direction. This is why objects on the surface of a sphere all fall "down" (towards the centre of the sphere, not all in the same absolute direction).
Near Earth's surface the field is approximately uniform — the field lines are parallel and vertical, and \( g = 10\,\text{N/kg} \) is the same everywhere at the surface. This is the model used at IGCSE.
Near Earth's surface the field is approximately uniform — the field lines are parallel and vertical, and \( g = 10\,\text{N/kg} \) is the same everywhere at the surface. This is the model used at IGCSE.
An object in free fall (e.g. a satellite in orbit) is not weightless — it still has weight (\( W = mg \)) and \( g \) is still significant at orbital altitude. It feels weightless because the satellite and everything in it fall together with the same acceleration. True zero weight occurs only infinitely far from all masses.
Do not confuse "gravitational field strength" (\( g \), measured in N/kg) with "gravitational force" (\( W \), measured in N). The field strength is a property of the location; the force depends on the mass of the object placed there.
Quick-reference summary
| Quantity | Symbol | Unit | Changes with location? |
|---|---|---|---|
| Mass | \( m \) | kg | No |
| Weight | \( W \) | N | Yes (\( W = mg \)) |
| Gravitational field strength | \( g \) | N/kg | Yes (property of location) |