Physics 9702 · Paper 4

Geostationary Orbits

One special orbit keeps a satellite over the same ground point forever. It is the most recycled scenario in recent Paper 4 gravity questions, usually re-set on another planet.

The three conditions

A satellite must have: (1) a period equal to the planet's rotation period (24 h for Earth); (2) an orbit in the equatorial plane; (3) motion in the same direction as the planet's spin (west to east) (9702/41/M/J/25 Q1(c)).
Period alone is not enough: a 24-hour polar orbit, or one running east to west, drifts over the ground. Naming the missing condition is a standard one-marker (9702/42/M/J/23 Q1(d)(ii)).

The geostationary orbit at true scale: r = 4.2 × 10⁷ m ≈ 6.6 Earth radii from the centre, above the equator, moving west to east with a 24-hour period.

Worked example

Exam version: the radius of the orbit

Find the radius of the Earth's geostationary orbit, and the height of the satellite above the surface (Earth mass 6.0 × 10²⁴ kg, radius 6.4 × 10⁶ m). (9702/42/M/J/23 Q1(c))

T = 24 h = 86 400 s. From the bridge equation: $r^3 = \dfrac{GMT^2}{4\pi^2} = \dfrac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times (86\,400)^2}{4\pi^2} = 7.57 \times 10^{22} \text{ m}^3$ .
$r = 4.2 \times 10^{7} \text{ m}$ from the centre: about 6.6 Earth radii.
Height above the surface: $h = 4.2 \times 10^{7} - 6.4 \times 10^{6} = 3.6 \times 10^{7} \text{ m}$ . Subtracting the radius is the final mark.

Answer

r = 4.2 \times 10^{7} \text{ m}, \; h = 3.6 \times 10^{7} \text{ m}

The re-settings: the same question on Mars (period 25 h — the deduction mark is that Mars rotates once in 25 h) (9702/41/M/J/25 Q1(c)(i)); on an invented planet with T = 0.72 Earth days (9702/42/F/M/24 Q1(b)(ii)); on “Artemis” with the radius read off a potential graph (9702/42/F/M/23 Q1(b)(iv)). The method never changes.
And the potential link from lesson 13.06: constant r means constant φ, a horizontal line for the whole day (9702/42/F/M/25 Q2(b)).

Geostationary questions are the bridge equation plus one extra fact: T equals the planet's day. Convert the day to seconds, cube-root carefully, and check whether the answer wants r or the height.

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