What is a gravitational field?
A mass does not need to touch another mass to pull it. We say the mass makes a field in the space around it. Put any other mass in that space and it feels a force.
The gravitational field strength g at a point is the force on a test mass divided by that mass: g = F / m. Its unit is N/kg (the same as m/s²). It is a vector: it has a size and a direction. The direction is always towards the mass that makes the field.
Field of a point mass or a sphere: g = GM/r²
Put Newton's law F = GMm/r² into g = F/m. The test mass m cancels:
g = GM / r²
Double r and g becomes one quarter. This holds for a point mass, and for a uniform sphere at any point outside it (measure r from the centre). For Earth, M = 5.97 × 10²⁴ kg and R = 6.37 × 10⁶ m, so g at the surface is about 9.8 N/kg.
Inside a uniform solid planet, only the mass closer to the centre than you pulls net. There g = GM r / R³: it falls in a straight line to zero at the centre.
Fields of several masses: add the vectors
Fields add. To find the field at a point P from two or more masses:
- Find the size of each field with g = GM/r².
- Draw each as an arrow from P towards its mass.
- Split each arrow into x and y parts (components).
- Add the x parts, add the y parts.
- Size = √(gx² + gy²). Direction from the angle tan θ = gy/gx.
If two arrows are in a straight line, just add or subtract. Between two masses the arrows are opposite, so they can cancel. The point where g = 0 is called a neutral point. For masses M₁ and M₂ a distance d apart it is at distance x from M₁ where x/(d − x) = √(M₁/M₂). The bigger mass sits farther from the neutral point's middle: the neutral point is closer to the smaller mass.
Effect on the motion of bodies in the field
A mass m in a field g feels force F = mg. By Newton's second law its acceleration is a = F/m = g. So every body, light or heavy, accelerates the same way at the same spot. This is why a feather and a hammer fall together on the Moon.
The acceleration is in the direction of the field. A body released from rest moves along a field line towards the mass. Near Earth's surface g is nearly constant and we get the usual equations of motion. Far from Earth g changes with r, so the acceleration is not constant and we need energy ideas (see escape speed and orbits).
Try it: add the arrows yourself
In the 3D, go to the last step. First guess: if M₂ is made 0, where will the red arrow point? Then move the M₂ slider to 0 and check. Now set M₂ equal to M₁ and swing the probe round: find the angle where the red arrow is longest and where it is shortest. Watch the yellow dot (the g = 0 spot) move towards the smaller mass when you change M₂.
At home: drop a coin and a heavy key from the same height at the same time. They land together. The field gives both the same acceleration.
Key formulas and definitions
- g = F / m (unit N/kg = m/s²)
- g = GM / r² (outside a sphere, r from the centre)
- g = GM r / R³ (inside a uniform sphere, r < R)
- Net field: g = g₁ + g₂ + … (vector sum)
- Acceleration of a body in a field: a = g
- Neutral point between M₁ and M₂: x / (d − x) = √(M₁ / M₂), x measured from M₁
Worked examples
1. Find g at Earth's surface. (M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg²)
g = GM/R² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (6.37 × 10⁶)² = 3.98 × 10¹⁴ / 4.06 × 10¹³ ≈ 9.8 N/kg, directed towards Earth's centre.
2. Find g at a height equal to Earth's radius above the surface. Take g₀ = 9.8 N/kg at the surface.
The distance from the centre is r = 2R. g = g₀ (R/r)² = 9.8 × (1/2)² = 2.45 N/kg.
3. The Moon has M = 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Find g on its surface.
g = GM/R² = 6.67 × 10⁻¹¹ × 7.35 × 10²² / (1.74 × 10⁶)² = 4.90 × 10¹² / 3.03 × 10¹² ≈ 1.62 N/kg, about one sixth of Earth's.
4. At a point P, mass A makes a field of 2.0 N/kg towards the east and mass B makes 1.5 N/kg towards the north. Find the net field.
The two are at right angles. g = √(2.0² + 1.5²) = √6.25 = 2.5 N/kg. The angle north of east: tan θ = 1.5/2.0 = 0.75, so θ ≈ 37°.
5. Masses M and 4M are 6 m apart. Where on the line between them is g = 0?
Let x be the distance from M. Then GM/x² = G(4M)/(6 − x)². So (6 − x)² = 4x², and 6 − x = 2x, giving x = 2 m from M (4 m from 4M). It is nearer the smaller mass.
6. Surface g of a uniform planet is 9.8 N/kg. Find g at a depth where r = R/2 from the centre.
Inside a uniform sphere g ∝ r. So g = 9.8 × (1/2) = 4.9 N/kg.
Common mistakes
- Writing g = GMm/r². The test mass m cancels: the field depends only on the source mass M and distance r.
- Measuring r from the surface instead of the centre of the planet. For height h use r = R + h.
- Adding field magnitudes without thinking about direction. Fields are vectors: arrows at an angle need components.
- Using g = GM/r² inside a planet. That formula is only for points outside (or on) the sphere; inside, g = GMr/R³ and it falls to zero at the centre.