9 minute lesson · Physics · Original foundation material
Before you start: Newton's second law; Square roots; Circumference of a circle.
An orbit is continuous free fall. A satellite keeps moving sideways while gravity bends its path towards the planet. Begin with a circular orbit so that the radius and speed stay constant.
Measure radius from the centre
For a spherical planet of mass M, the gravitational force on an external satellite of mass m is F = GMm/r². Here r is the distance between their centres. If the planet has radius R and the satellite is at height h above its surface, use r = R + h.
Divide the force by m to find the gravitational acceleration: g = GM/r². Doubling r therefore divides g by 4. A satellite is not beyond gravity simply because it is in space.
Let gravity provide the turn
Circular motion needs an inward acceleration v²/r. Equating this to gravity gives GM/r² = v²/r, so v = √(GM/r). This approximation treats the planet as much more massive than the satellite and ignores drag and other bodies.
One orbit has length 2πr. Its period is T = 2πr/v, giving T² = 4π²r³/(GM). Around the same planet, v scales as 1/√r and T scales as r^(3/2). A larger circular orbit is slower and takes longer.
A worked example
Satellite A has a circular orbital radius of 8,000 km and a period of 2 hours. Satellite B moves in a circular orbit around the same planet at a radius of 32,000 km. Find B's period, treating both satellite masses as negligible.
- Compare radii from the planet's centre: rB/rA = 32,000/8,000 = 4.
- For the same planet, (TB/TA)² = (rB/rA)³ = 4³ = 64.
- Periods are positive, so TB/TA = √64 = 8.
- Multiply A's period by 8: TB = 8 × 2 = 16 hours.
16 hours
Check your understanding
At the same orbital radius around the same planet, would doubling a small satellite's mass change its circular orbital speed?
Show the explanation
No. In GMm/r² = mv²/r, m cancels. Under the negligible-satellite-mass approximation, v depends on M and r.