Gravitational field strengthAQA A-Level Physics: Revision notes

Gravitational field strength, $g$, at a point is the gravitational force per unit mass acting on a small test mass placed at that point: $g = F/m$. It is a vector that points in the direction of the force, and its unit is N kg⁻¹, which is equivalent to m s⁻².

So the weight of a mass is $W = mg$. The test mass must be small so that it does not disturb the field it is measuring.

A gravitational field is drawn using field lines. The arrow on a line gives the direction of the force on a mass, and the spacing shows the strength: closer lines mean a stronger field.

Outside a spherical mass the field is radial, with lines pointing towards the centre of the mass. Close to the surface over a small region the lines are parallel and equally spaced, which represents a uniform field of constant $g$.

For a point mass, or a uniform sphere outside its surface, combining $F = GMm/r^2$ with $g = F/m$ gives

$g = \dfrac{GM}{r^2}$

where $M$ is the mass producing the field and $r$ is the distance from its centre. The test mass cancels. $g$ is an inverse square relationship, so doubling $r$ divides $g$ by 4.

On the Earth's surface, $g = GM/R^2 = 9.81$ N kg⁻¹ with $M = 5.97\times10^{24}$ kg and $R = 6.37\times10^{6}$ m.

Mars has mass $6.42\times10^{23}$ kg and radius $3.39\times10^{6}$ m. $g = \dfrac{6.67\times10^{-11}\times6.42\times10^{23}}{(3.39\times10^{6})^2} = 3.7$ N kg⁻¹.

For a planet with mass $4M_E$ and radius $1.6R_E$: $g/g_E = 4/1.6^2 = 1.56$, so $g = 15.3$ N kg⁻¹. Ratios avoid recalculating $G$.

Between two bodies the fields point in opposite directions and subtract, so the resultant is zero where $GM_1/x^2 = GM_2/(d-x)^2$.

A satellite or astronaut in orbit is still in a gravitational field with $g$ only slightly smaller than at the surface. They feel weightless because they are in free fall with the same acceleration as their surroundings, so no contact force acts. The gravitational force has not disappeared.