Gravitational fields and Newton's law of gravitationEdexcel A-Level Physics: Revision notes

A gravitational field is a region of space in which a mass experiences a force because of another mass. Gravitational forces are always attractive, so field lines point towards the mass creating the field.

Gravitational field strength at a point is the force per unit mass on a small mass placed there:

$g = \frac{F}{m}$

It is measured in N kg⁻¹ and is a vector, pointing towards the source. Around a point mass or a sphere the field is radial: the lines point to the centre and get further apart as you move away, so the field gets weaker. Over a small region near a planet's surface the lines are parallel and equally spaced, so the field is uniform.

Every point mass attracts every other point mass with a force that is directly proportional to the product of the masses and inversely proportional to the square of their separation:

$F = \frac{G m_1 m_2}{r^2}$

$G = 6.67\times10^{-11}$ N m² kg⁻² is the gravitational constant. The law is an inverse square law: doubling the separation reduces the force to a quarter; trebling it reduces the force to a ninth.

The two forces form a Newton's third law pair: they are equal in size and opposite in direction, and act on different bodies. A uniform sphere acts as if all its mass were at its centre, so $r$ is always measured centre to centre.

Combine $g = F/m$ with Newton's law for a small mass $m$ at distance $r$ from a mass $M$:

$g = \frac{F}{m} = \frac{GM}{r^2}$

The mass $m$ cancels, so the field strength depends only on the mass creating the field and the distance from its centre. This is why all objects fall with the same acceleration in a vacuum at a given place.

The field also follows an inverse square law, so if $r$ doubles, $g$ falls to a quarter. At the Earth's surface $g = 9.81$ N kg⁻¹.

Worked example: Earth mass 5.97×10²⁴ kg, radius 6.37×10⁶ m. $g = \frac{6.67\times10^{-11}\times5.97\times10^{24}}{(6.37\times10^{6})^2} = 9.81$ N kg⁻¹.

The force on any mass $m$ at a point where the field strength is $g$ is $F = mg$. This force is the weight of the object. Weight changes with position because $g$ changes, but mass does not.

To find the force between two bodies, either use $F = Gm_1m_2/r^2$ directly, or find $g$ for one body and multiply by the mass of the other. Both give the same answer.

Astronauts in orbit still have weight: at 400 km up $g$ is about 8.7 N kg⁻¹. They appear weightless because they and the spacecraft are in free fall together, so there is no contact force between them.

Fields are vectors, so where two bodies both create a field you add the field strengths as vectors. Between the Earth and the Moon, the Earth's field points towards the Earth and the Moon's field points towards the Moon, so they act in opposite directions along the line between them.

At the neutral point the two fields are equal and opposite, so the resultant is zero:

$\frac{GM_1}{x^2} = \frac{GM_2}{(d-x)^2}$

Take square roots to give $x/(d-x) = \sqrt{M_1/M_2}$. The neutral point lies closer to the less massive body.