Gravitational Field Strength

There is a universal force of attraction between all matter with mass
- This force is known as the ‘force due to gravity’ or the weight
The Earth’s gravitational field is responsible for the weight of all objects on Earth
A gravitational field is defined as:

A region of space where a test mass experiences a force due to the gravitational attraction of another mass

The direction of the gravitational field is always towards the centre of the mass causing the field
- Gravitational forces are always attractive

Gravity has an infinite range, meaning it affects all objects in the universe
- There is a greater gravitational force around objects with a large mass (such as planets)
- There is a smaller gravitational force around objects with a small mass (almost negligible for atoms)

Gravitational Attractive Force, downloadable AS & A Level Physics revision notes

The Earth's gravitational field produces an attractive force. The force of gravity is always attractive

The gravitational field strength at a point is defined as:

The force per unit mass experienced by a test mass at that point

This can be written in equation form as:

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

Where:
- g = gravitational field strength (N kg⁻¹)
- F = force due to gravity, or weight (N)
- m = mass of test mass in the field (kg)

This equation shows that:
- On planets with a large value of g, the gravitational force per unit mass is greater than on planets with a smaller value of g

An object's mass remains the same at all points in space
- However, on planets such as Jupiter, the weight of an object will be greater than on a less massive planet, such as Earth
- This means the gravitational force would be so high that humans, for example, would not be unable to fully stand up

gravitational field strength, downloadable AS & A Level Physics revision notes

A person’s weight on Jupiter would be so large that a human would be unable to fully stand up

Factors that affect the gravitational field strength at the surface of a planet are:
- The radius r (or diameter) of the planet
- The mass M (or density) of the planet
This can be shown by equating the equation F = mg with Newton's law of gravitation:

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

Substituting the force F with the gravitational force mg leads to:

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

Cancelling the mass of the test mass m leads to the equation:

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

Where:
- G = Newton's Gravitational Constant
- M = mass of the body causing the field (kg)
- r = distance from the mass where you are calculating the field strength (m)

This equation shows that:
- The gravitational field strength g depends only on the mass of the body M causing the field
- Hence, objects with any mass m in that field will experience the same gravitational field strength
- The gravitational field strength g is inversely proportional to the square of the radial distance, r²

Worked example

Calculate the mass of an object with weight 10 N on Earth.

Answer:

Worked example

The mean density of the Moon is $\frac{3}{5}$ times the mean density of the Earth. The gravitational field strength on the Moon is $\frac{1}{6}$ the gravitational field strength on Earth.

Determine the ratio of the Moon's radius $r_{M}$ to the Earth's radius $r_{E}$ .

Answer:

Step 1: Write down the known quantities

$g_{M}$ = 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

$ρ_{M} = \frac{3}{5} ρ_{E}$

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

Step 2: Write down the equations for the gravitational field strength, volume and density

Gravitational field strength: $g = \frac{G M}{r^{2}}$

Volume of a sphere: $V = \frac{4}{3} π r^{3} \Rightarrow V \propto r^{3}$

Density: $ρ = \frac{M}{V} \Rightarrow M = ρ V = \frac{4}{3} π ρ r^{3} \Rightarrow M \propto ρ r^{3}$

Step 3: Substitute the relationship between M and r into the equation for g

$g \propto ρ \frac{(r^{3})}{r^{2}} \Rightarrow g \propto ρ r$

Step 4: Find the ratio of the gravitational field strength

$g_{M} \propto ρ_{M} r_{M}$

$g_{E} \propto ρ_{E} r_{E}$

$g_{M} = \frac{1}{6} g_{E} \Rightarrow ρ_{M} r_{M} = \frac{1}{6} ρ_{E} r_{E}$

Step 5: Substitute the ratio of the densities into the equation

$(\frac{3}{5} ρ_{E}) r_{M} = \frac{1}{6} ρ_{E} r_{E}$

$\frac{3}{5} r_{M} = \frac{1}{6} r_{E}$

Step 6: Calculate the ratio of the radii

$\frac{r_{M}}{r_{E}} = \frac{1}{6} \div \frac{3}{5} = \frac{5}{18} = 0.28$

Exam Tip

There is a big difference between g and G (sometimes referred to as ‘little g’ and ‘big G’ respectively), g is the gravitational field strength and G is Newton’s gravitational constant. Make sure not to use these interchangeably!

Remember the equation $d e n s i t y = \frac{m a s s}{v o l u m e}$ , which may come in handy with some calculations. The equation for the volume of common shapes is in your data booklet.

Gravitational Field Strength (HL IB Physics)

Revision Note