Circular Orbits | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Circular Orbits

Velocity is tangent to the orbit: it points along the circle. Gravitational force points to the centre, at 90° to the velocity. The radius r is measured from the planet's centre to the satellite.

Gravity provides the centripetal force

The satellite moves in a circle, so it needs a pointing at the centre.
The only force on it, gravity, points exactly at the centre. So gravity is that centripetal force.
The force is at 90° to the velocity, so it changes the direction but not the speed. The orbit is a circle at steady speed.
Exam wording (why circular): the gravitational force is perpendicular to the direction of motion and provides the centripetal acceleration (9702/41/O/N/24 Q1(b)(i)).

Gravity equals the centripetal force

Set gravity equal to the centripetal force from the Circular Motion chapter. The satellite mass m cancels.
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$

v^2 = \dfrac{GM}{r}

Symbols

$v$ = orbital speed (m s⁻¹)
$G$ = gravitational constant (N m² kg⁻²)
$M$ = mass of the planet (kg)
$r$ = orbit radius from the centre (m)

The satellite mass m cancels. Therefore, all satellites at the same radius move at the same speed, even when their own masses are different.
Closer orbits (small r) are faster.

Worked example

Calculate the Moon's orbital speed

The Moon orbits the Earth at 3.84 × 10⁸ m from the centre. Find its orbital speed. Mass of the Earth 6.0 × 10²⁴ kg.

Substitute the Earth's mass and the orbit radius.
$GM=4.0\times10^{14}\ \text{m}^3\text{ s}^{-2}$
$v = \sqrt{\dfrac{4.0\times10^{14}}{3.84\times10^8}}$
Evaluate the expression.
$v = 1.0 \times 10^{3} \text{ m s}^{-1}$

Answer

1.0\times10^3\ \text{m s}^{-1}

An astronaut and a spacecraft at the same orbit radius move at the same orbital speed.

Your turn9702-style

Find the orbital speed of a satellite 200 km above the Earth's surface. Radius of Earth 6.4 × 10⁶ m, mass 6.0 × 10²⁴ kg.

Show worked answer

First find the orbit radius from the centre.
$r = 6.4 \times 10^{6} + 2.0 \times 10^{5}$
$r = 6.6 \times 10^{6} \text{ m}$
Use the orbit radius in the speed equation.
$GM=4.0\times10^{14}\ \text{m}^3\text{ s}^{-2}$
$v = \sqrt{\dfrac{4.0\times10^{14}}{6.6\times10^6}}$
$v=7.8\times10^3\ \text{m s}^{-1}$

Answer: 7.8 × 10³ m s⁻¹ (about 7.8 km s⁻¹).

Common mistake

r is the orbit radius from the centre, not the height above the surface. Add the planet's radius to the height first.