Syllabus Edition

First teaching 2023

First exams 2025

Gravitational Field Strength (CIE A Level Physics)

Revision Note

Test yourself

Author

Leander

Last updated

30 October 2024

Deriving gravitational field strength (g)

There are two situations where gravitational field strength is considered:
- at a point
- due to a mass

Gravitational field strength due to a point

The gravitational field strength at a point describes how strong or weak a gravitational field is at that point

The gravitational field strength at a point is defined as

The force per unit mass of a gravitational force on an object

Gravitational field strength at a point is given by the equation:

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

Where:
- g = gravitational field strength measured in newtons per kilogram (N/kg)
- F = gravitational force measured in newtons (N)
- m = mass of object in gravitational field measured in kilograms (kg)

Gravitational field strength due to a point mass

The gravitational field strength due to a point mass within a gravitational field can be derived from
- combining the equations for Newton’s law of gravitation
- the definition of a gravitational field

Newton’s law of gravitation states that the attractive force F_G between two masses M and m with separation r is equal to:

$F_{G} = \frac{G M m}{r^{2}}$

Rearrange the definition of gravitational field strength at a point to make force F the subject:

$F = m g$

Equate the gravitational force and the force due to gravitational field strength:

$F = F_{G}$

$m g = \frac{G M m}{r^{2}}$

Cancel out the mass, m, on each side:

$m g = \frac{G M m}{r^{2}}$

The equation for gravitational field strength due to a point mass is:

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

Where:
- g = gravitational field strength (N kg^-1)
- G = Newton’s Gravitational Constant
- M = mass of the body producing the gravitational field (kg)
- r = distance between point source (mass, m) and position in field (m)

Examiner Tip

It is important to recognise the difference between the two gravitational field strength situations:

gravitational field strength at a point due to the object creating the gravitational field $g = \frac{F}{m}$
gravitational field strength due to a point mass placed in a the gravitational field of a bigger object is $g = \frac{G M}{r^{2}}$

Calculating g

Gravitational field strength, g, is a vector quantity
The direction of g is always towards the centre of the body creating the gravitational field
- This is the same direction as the gravitational field lines

Gravitational field strength, g, and orbital radius, r, have an inverse square law relationship:

$g \propto \frac{1}{r^{2}}$

Where:
- g decreases as r increases by a factor of 1/r²

Worked example

The mean density of the moon is ⅗ times the mean density of the Earth. The gravitational field strength is ⅙ on the Moon than that on Earth.

Determine the ratio of the Moon’s radius r_M and the Earth’s radius r_E.

Answer:

Step 1: Write down the known quantities

$ρ_{M} = \frac{3}{5} ρ_{E}$

$g_{M} = \frac{1}{6} g_{E}$

g_M = gravitational field strength on the Moon, ρ_M = mean density of the Moon
g_E = gravitational field strength on the Earth, ρ_E = mean density of the Earth

Step 2: The volumes of the Earth and Moon are equal to the volume of a sphere

$V = \frac{4}{3} π r^{3}$

Step 3: Write the density equation and rearrange for mass M

$ρ = \frac{M}{V}$

$M = ρ V$

Step 4: Write the gravitational field strength equation

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

Step 5: Substitute M in terms of ρ and V

$g = \frac{G ρ V}{r^{2}}$

Step 6: Substitute the volume of a sphere equation for V, and simplify

$g = \frac{G ρ 4 π r^{3}}{3 r^{2}} = \frac{G ρ 4 π r}{3}$

Step 7: Find the ratio of the gravitational field strengths

$\frac{g_{M}}{g_{E}} = \frac{G ρ_{M} 4 π r_{M}}{3} \div \frac{G ρ_{E} 4 π r_{E}}{3} = \frac{ρ_{M} r_{M}}{ρ_{E} r_{E}}$

Step 8: Rearrange and calculate the ratio of the Moon’s radius r_M and the Earth’s radius r_E

_{$\frac{r_{M}}{r_{E}} = \frac{ρ_{E} g_{M}}{ρ_{M} g_{E}} = \frac{ρ_{E} (\frac{1}{6} g_{E})}{(\frac{3}{9} ρ_{E}) g_{E}}$}

_{$\frac{r_{M}}{r_{E}} = \frac{5}{3} \times \frac{1}{6} = \frac{5}{18} = 0.28 (2 s . f .)$}

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Test yourself Next topic

Did this page help you?