Physics 9702 · Paper 4

Circular Orbits Under Gravity

An orbit is circular motion (chapter 12) whose centripetal force is gravity (this chapter). Equating the two solves every orbit problem.

The bridge equation

The two-mark explain: the gravitational force acts perpendicular to the motion, and it provides the centripetal acceleration (9702/41/O/N/24 Q1(b)(i)).
The mark scheme accepts any of the three equivalent lines: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , or $= mr\omega^2$ with $\omega = 2\pi/T$ , or $= \dfrac{4\pi^2 m r}{T^2}$ (9702/42/F/M/24 Q1(b)(ii)).
Cancel and rearrange: $v^2 = \dfrac{GM}{r}$ and $T^2 = \dfrac{4\pi^2}{GM}r^3$ . The satellite mass cancels: every object in a given orbit moves at the same speed, astronaut and spacecraft together.

Worked example

Smallest case: the Moon's orbital speed

The Moon orbits 3.84 × 10⁸ m from the Earth's centre (Earth mass 6.0 × 10²⁴ kg). Find its orbital speed (9702-style, from the coursebook).

$v^2 = \dfrac{GM}{r} = \dfrac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{3.84 \times 10^{8}} = 1.04 \times 10^{6}$ .
$v = 1.0 \times 10^{3} \text{ m s}^{-1}$ : about 1 km per second.

Answer

v \approx 1.0 \times 10^{3} \text{ m s}^{-1}

$T^2 \propto r^3$ means closer orbits are faster and shorter. The 2024 paper turned this into a straight-line exercise, plotting $T^{2/3}$ against orbit height and extracting the planet's mass from the gradient and its radius from the intercept (9702/41/O/N/24 Q1(b)).

Common mistake

No minus sign in the bridge equation: the gravitational force and the centripetal resultant point the same way (at the centre), so the magnitudes are equated directly.

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