3.7.2.2
Gravitational field strength in a radial field: $g = GM/r^{2}$
Gravitational Fields & Orbits — AQA A-Level Physics
$$g = \frac{GM}{r^2}$$
- $g$: gravitational field strengthThe gravitational force per unit mass at a point in a gravitational fieldA region of space in which a mass experiences a gravitational force.. Measured in N kg⁻¹. (N kg⁻¹)
- $G$: Newton's gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²)
- $M$: mass of the body producing the field (kg)
- $r$: distance from the centre of the mass (m)
- This equation comes directly from combining $g = F/m$ with $F = GMm/r^{2}$.
- g is a vector. Its direction is always towards the centre of the mass creating the field.
- On Earth's surface, g = 9.81 N kg⁻¹. Outside the surface, g decreases with 1/r².
- Inside a uniform sphere (r < R), g is directly proportional to r. Outside (r > R), g follows the inverse square law.
The g-r graph
- For r < R (inside the planet): g increases linearly with r.
- At r = R (the surface): g has its maximum value.
- For r > R (outside the planet): g decreases as 1/r². This gives an 'L-shaped' curve showing rapid decrease.
g against r graph
Graph showing g increasing linearly from 0 to R on the x-axis, then decreasing as 1/r² for r > R. A vertical dashed line marks r = R (planet surface). The maximum g occurs at R.
Worked Example
The mean densityMass per unit volume of a material. Measured in kg m⁻³. of the Moon is 3/5 that of the Earth. The gravitational field strengthThe gravitational force per unit mass at a point in a gravitational fieldA region of space in which a mass experiences a gravitational force.. Measured in N kg⁻¹. on the Moon is 1/6 that on Earth. Determine the ratio r_M / r_E.
Show Solution
1
Write known ratios
$\rho_M = \frac{3}{5}\rho_E$ and $g_M = \frac{1}{6}g_E$
2
Express g in terms of densityMass per unit volume of a material. Measured in kg m⁻³.
Substitute $M = \rho V = \rho \times \frac{4}{3}\pi r^3$ into $g = \frac{GM}{r^2}$:
$$g = \frac{G \rho \frac{4}{3}\pi r^3}{r^2} = \frac{4G\pi \rho r}{3}$$3
Form the ratio
$$\frac{g_M}{g_E} = \frac{\rho_M r_M}{\rho_E r_E}$$
4
Substitute and solve
$$\frac{r_M}{r_E} = \frac{\rho_E g_M}{\rho_M g_E} = \frac{\rho_E \times \frac{1}{6}g_E}{\frac{3}{5}\rho_E \times g_E} = \frac{5}{3} \times \frac{1}{6} = \frac{5}{18}$$
Answer
$r_M / r_E = 5/18 = 0.28$ (2 s.f.)
Examiner Tips and Tricks
- Be comfortable drawing the g-r graph.
- The 1/r² section should look steeper and decay faster than the V-r graph (which is only 1/r).
- This distinction matters in sketch questions.