The law of gravitation

In a nutshell

Newton's law of gravitation gives the gravitational force acting between two separate masses separated by a distance. Newton's law of gravitation can also be used to find the gravitational field strength of a point mass at a distance away. Both equations are inverse square laws which means that the values of force and gravitational field strength decreases exponentially with distance.

Equations

Description	Equation
Newton's law of gravitation	$F=(-)G \frac {Mm}{r^2}$
Gravitational field strength of a point source	$g=(-)G \frac M{r^2}$

Constants

name	symbol	value
$gravitational \ constant$	$G$	$6.67 \times 10^{-11} \, N\,m^2\,kg^{-2}$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$gravitational \ force$	$F$	$N$	$kg\,m\,s^{-2}$
$larger \ point \ mass$	$M$	$kg$	$kg$
$smaller \ point \ mass$	$m$	$kg$	$kg$
$distance \ between \ centres \ of \ masses$	$r$	$m$	$m$
$gravitational \ field \ strength$	$g$	$N\,kg^{-1}$	$m\,s^{-2}$

Newtons law of gravitation

Newtons law of gravitation is an equation which is used to calculate the force acting on a point mass within a gravitational field.

The equation is given as:

$F=-G \frac {Mm}{r^2}$

$G$ is the gravitational constant and is given in the exam data booklet. The equation has a minus sign in it because the gravitational force is always attractive and hence is negative.

Note: $M$ denotes the mass of the object which is creating the gravitational field and $m$ is the point mass within the field. This varies depending on context.

Example

When considering the force of the Earth on the Moon, $M$ would be the mass of the Earth as it is creating the gravitational field and $m$ is the mass of the Moon.

However if considering the gravitational force of the Sun on the Earth, $M$ would be the mass of the Sun and $m$ would be the mass of the Earth.

Example

Calculate the gravitational force on a Starlink satellite which has a mass of $1250 \, kg$ and orbits at a distance of $530\, km$ above Earth's surface.

$R_{Earth}=6.37 \times 10^6 \, m \newline \\[0.1in]M_{Earth}=5.97 \times 10^{24} \, kg$ 

Firstly, write down the known values:

$M=5.97 \times 10^{24} \, kg \newline \\[0.1in]m=1250 \, kg \newline \\[0.1in]r_{orbit}=530 \, km = 530 \, 000 \, m \newline \\[0.1in]R_{Earth}=6.37 \times 10^6 \, m$

Next, write down the equations needed and rearrange if necessary:

$F=-G \frac {Mm}{(R_{Earth}+r_{orbit})^2}$

Note: The orbit of the satellite is $530 \, km$ above the surface of the Earth, so the distance of the satellite to the centre of the Earth is $R_{Earth}+r_{orbit}$ 

Then, substitute the values into the equation:

$F=-6.67\times 10^{-11} \times \frac {5.97 \times 10^{24} \times 1250}{(6.37 \times 10^6+530 \, 000)^2} \newline \\[0.1in]F=-10454...$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$gravitational \ force, \ F=(-)10 \, 500\,N$

The gravitational force of the Earth on the Starlink satellite is $\underline{(-)10\,500 \, N}.$

Note: Technically the value of the force is negative, but you won't get penalised in an exam if you state the force a positive magnitude of force.

The inverse square law

The gravitational force is inversely proportional to the distance squared of the point mass within a gravitational field.

This means that the gravitational force is strongest at the surface of a planet or star and decreases exponentially until it gets to $0 \, N$ at an infinite distance away.

Physics; Gravitational Fields; KS5 Year 12; The law of gravitation

Example

The Starlink satellite has a gravitational force of roughly $10 \, 000 \, N$ at a distance of $6.9 \times 10^6 \,m$ . If another satellite with the same mass were at twice the distance away, at $1.38 \times 10^7 \, m$ , the gravitational force would be 4 times less at roughly $2500 \, N$ .

Mathematically this can be shown:

$F \propto \dfrac {1}{r^2} \newline \\[0.1in]$

If distance $r$ is doubled, then:

$F\propto \frac 1 {2^2} \newline \\[0.1in]F \propto \frac 14$ 

The force is a quarter of the original value, or 4 times less.

Gravitational field strength

Gravitational field strength is defined as:

$g=\frac Fm$

Substituting Newton's law of gravitation for $F$ :

$g= - \frac {GMm}{mr^2} \newline \\[0.1in]g=- \frac {GM}{r^2}$

This equation can be used to find the gravitational field strength of a point mass at a distance of $r \ metres$ away from the point mass.

Note: Similar to before, the mass in this equation is the object which is causing the gravitational field. (The most massive object in the scenario).

Graphically, how the gravitational field strength of a point mass varies with the distance away from the point mass looks like this:

Physics; Gravitational Fields; KS5 Year 12; The law of gravitation

It is another inverse square relationship.

Example

The following graph is for Jupiter.

Physics; Gravitational Fields; KS5 Year 12; The law of gravitation

The radius of Jupiter is $69 \, 900 \, km$ . Calculate the mass of Jupiter.

Firstly, write down the known values:

$r_{Jupiter}=69 \,900 \, km = 69\,900\,000 \, m \newline \\[0.1in]$

The graph can be used to obtain the gravitational field strength at the surface of Jupiter:

$g_{Jupiter}=24.8 \, N\,kg^{-1}$ 

Next, write down the equations needed and rearrange if necessary:

$g=\frac {GM}{r^2} \rightarrow M=\frac {gr^2}{G}$

Then, substitute the values into the equation:

$M=\frac {24.8\times 69\,900\,000^2}{6.67\times 10^{-11}} \newline \\[0.1in]M=1.81669...\times 10^{27} \ kg$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$mass \ of \ Jupiter, \ M=1.82 \times 10^{27} \, kg$

The mass of Jupiter is $\underline{1.82 \times 10^{27} \, kg}.$ 