Derived by combining g = F/m with F = GMm/\(r^{2}\)
Gravitational Fields - OCR A-Level Physics
$$g = \frac{GM}{r^2}$$
- Derived by combining $g = F/m$ with F = GMm/\(r^{2}\): $g = (GMm/r^2)/m = GM/r^2$.
- This gives the field strength at any point outside a spherical mass.
- At the surface of a planet: $g_{surface}$ = GM/\(R^{2}\) where R is the planet's radius.
- Above the surface: g decreases with the square of the distance from the centre.
- Inside a uniform sphere: g decreases linearly with distance from the centre (not on the OCR spec but useful context).
Worked Example
Calculate 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. at the surface of Mars. Mass of Mars = $6.42 \times 10^{23}$ kg, radius = $3.39 \times $\(10^{6}\)$ m.
Show Solution
1
$g = GM/r^2.$
2
$g = (6.67 \times 10^{-11} \times 6.42 \times 10^{23}) / (3.39 \times 10^6)^2.$
3
$g = (4.28 \times 10^{13}) / (1.15 \times 10^{13}).$
4
$g = 3.72 N kg^{-1} \approx 3.7 m s^{-2}.$
Answer
g_{Mars} ≈$ 3.7 $N kg^{-1}