5.4.2 - Newton's law of gravitation

The Law And Its Direction

In the previous gravitational-fields lesson, you met the idea that a spherical mass can be modelled as if its mass were concentrated at its centre. Newton's law of gravitation now gives the size and direction of the force between two point masses, or between two spherical masses treated as point masses at their centres.

The negative sign is a direction sign. If the positive radial direction is chosen outwards from the source mass, gravity acts inwards, so the force is negative in that coordinate system. When an OCR calculation asks for the magnitude of the gravitational force, use:

and then state the direction separately as towards the other mass.

The inverse-square part is the most important shape of the law. If $r$ doubles, $r^{2}$ becomes four times larger, so the force becomes one quarter as large. If one mass doubles, the force doubles. These changes combine multiplicatively: doubling one mass and tripling the separation changes the force by a factor of $2/9$.

[DIAGRAM: asset_slug: 5_04_2_newtons_law_of_gravitation__diagram_01; description: Equal and opposite gravitational forces act along the line joining the centres of the masses.; caption: Equal and opposite gravitational forces act along the line joining the centres of the masses.]

Using The Law

The value of $r$ is often the trap. It is not usually the height above a surface; it is the distance from the centre of one mass to the centre of the other. For a satellite near Earth, for example, $r$ is the Earth's radius plus the satellite's height above the surface.

Worked example: a $720\,\mathrm{kg}$ spacecraft is $4.00\times 10^{5}\,\mathrm{m}$ above the Earth's surface. Use $M_E = 5.97\times 10^{24}\,\mathrm{kg}$, $R_E = 6.37\times 10^{6}\,\mathrm{m}$, and $G = 6.67\times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}\,\mathrm{kg}^{-2}$ to calculate the magnitude of the gravitational force on the spacecraft.

First find the centre-to-centre separation:

Now substitute into Newton's law:

The force is towards the centre of the Earth. Notice that the force is still large because the Earth has a huge mass, even though $G$ is very small.

A useful check on any answer is proportionality. If the spacecraft moved further away, $r$ would increase and the force must decrease. An answer that increases just because the object is "higher" has the wrong physical direction.