Gravitational Field

At 400 km altitude, 6.771 × 10⁶ m from Earth's centre, an 80.0 kg astronaut is pulled with 695 N and the 4.20 × 10⁵ kg station beside her with 3.65 × 10⁶ N. The forces differ by a factor of 5250, yet each divided by its own mass gives 8.69 N kg⁻¹. That common ratio is the field, g⃗ = F⃗/m. It is a statement about the point, set by Earth alone, and it is there when nothing is. Because the mass gravity acts on is also the mass that resists acceleration, the same ratio is the free-fall acceleration: a = F/m = g for anything falling freely there, which is why 1 N kg⁻¹ is exactly 1 m s⁻². One caution keeps the definition clean: the test mass must be too small to shift the sources, so strictly g⃗ is the limit of F⃗/m as m shrinks. With the field in hand a two-body problem splits in two. The sources set g⃗; any body then responds through F⃗ = m g⃗, and you never revisit the sources.

For a point mass, or any spherically symmetric body seen from outside, g⃗ = −(GM/r²) r̂. Read it in two parts. The unit vector r̂ points from the source to the field point, so the minus sign aims g⃗ back at the source's centre. The size GM/r² uses the distance to the centre, never the height above the surface. Earth's surface field is 9.82 N kg⁻¹ at R = 6.371 × 10⁶ m. A geostationary satellite orbits at r = 4.216 × 10⁷ m = 6.617R, so the field there is 9.82 ÷ 6.617² = 0.224 N kg⁻¹, found by ratio without G appearing again. When the source is not at the origin, skip the trigonometry. With r⃗ the vector from the source to the field point, g⃗ = −GM r⃗/r³, so gₓ = −GM(x − x₀)/r³ and likewise for y. The cube is not a new law: one power of r is spent turning r⃗ into the unit vector r̂, and the inverse square is what remains.

Forces on a test mass add as vectors, and dividing every one by the same m shows that fields do as well: g⃗ = g⃗₁ + g⃗₂ + …, each found as if the other sources were absent. Magnitudes add only when the fields are parallel. On the line between two bodies they are antiparallel, so at one point they cancel. Setting GM₁/x² = GM₂/(d − x)² and taking the square root gives x = d/(1 + √(M₂/M₁)) from M₁. For Earth and Moon, M₂/M₁ = 0.0123, its root is 0.1109 and d = 3.844 × 10⁸ m, so the null point lies 3.46 × 10⁸ m from Earth, nine tenths of the way to the Moon. It sits nearer the lighter body because the heavier body's field needs more distance to weaken to the same size, and the square root appears because the field is inverse-square. A null point is not a place without gravity nearby. Step off it along the line and the field points away from it; step off sideways and the field points back toward the line.

A field-line diagram follows three rules. The tangent to a line gives the direction of g⃗ there. The number of lines crossing a unit area held square to them is proportional to the field's size. And lines never cross, because g⃗ has one direction at each point. Gravitational lines begin nowhere in empty space; they come in from far away and end on mass, a fixed number per kilogram. The density rule works only because gravity is inverse-square. All N lines ending on an isolated sphere must pierce any larger sphere drawn round it, spread over 4πr², so their density N/(4πr²) falls exactly as GM/r² does. Double r and the same lines cover four times the area, so the field is a quarter. Two warnings. A sketch on flat paper thins as 1/r, not 1/r², so take strength from the argument and not from spacing on the page. And lines are not trajectories: they show acceleration, and a body with sideways velocity, the Moon for one, cuts across them all month.

At the surface, M = (4/3)πρR³ turns GM/R² into g₀ = (4/3)πGρR: at fixed density the surface field grows in proportion to R, not R² or R³. Earth, with ρ = 5514 kg m⁻³ and R = 6.371 × 10⁶ m, gives 9.82 N kg⁻¹. The Moon, with 3340 kg m⁻³ and 1.737 × 10⁶ m, gives 1.62 N kg⁻¹, and the ratio 0.165 is a density factor 0.606 times a size factor 0.273. Above the surface g(h) = g₀(1 + h/R)⁻², and for h ≪ R the binomial expansion gives g ≈ g₀(1 − 2h/R). That is what constant g means: the first term of a series. At the 8.8 km summit of Everest the correction is 2 × 8.8/6371 = 0.28%. At 400 km the linear formula says 12.6% where the exact drop is 11.5%, so the expansion itself is starting to fail. Direction behaves the same way. Across a laboratory 10 m wide, neighbouring field lines converge by only 10/(6.4 × 10⁶) rad, which is why they are drawn parallel and the field is called uniform.

Inside a spherically symmetric body the shells above you contribute nothing and the mass beneath acts from the centre. For uniform density that enclosed mass is Mr³/R³, so g = GMr/R³ = g₀ r/R: the field falls linearly to zero at the centre, where pulls from every direction cancel. Halfway down a uniform Earth it would be 4.9 N kg⁻¹. The real Earth is not uniform. Its dense core keeps g between 9.8 and 10.7 N kg⁻¹ all the way down to the core boundary at r = 0.55R, a reminder that the linear law belongs to the density model and not to gravity. Three further limits. A scale at the equator of a spinning planet reads, per kilogram, the field less the centripetal requirement ω²R, which is 0.034 m s⁻² on Earth. Beside an irregular body, a comet nucleus or a mountain range, the field need not point at the centre of mass, and the sum over pieces must be done in full. And the Newtonian field responds everywhere the instant a source moves, where in reality the change spreads at the speed of light.