A satellite stays in a circular orbit because gravity pulls it toward the centre with exactly the force needed for that circle. Gravity is the centripetal force, not an extra one.
This lesson is part of gravitation. It comes before using orbital relationships with consistent units.
What is the core idea?
Centripetal force is defined by its direction and its job, not by its origin. Any force that points to the centre of a circular path and keeps an object on it is a centripetal force.
For a satellite, gravity is that force. Setting GMm ÷ r² = mv² ÷ r gives the orbital speed, v = √(GM ÷ r).
Worked example: a satellite at 7.0 × 10⁶ m
A satellite of mass 500 kg orbits at a radius of 7.0 × 10⁶ m from Earth’s centre. Use G = 6.67 × 10⁻¹¹ N m² kg⁻² and M = 5.97 × 10²⁴ kg, so GM = 3.98 × 10¹⁴ N m² kg⁻¹.
- Orbital speed: v = √(3.98 × 10¹⁴ ÷ 7.0 × 10⁶) = √(5.69 × 10⁷) = 7.5 × 10³ m s⁻¹.
- Centripetal force: F = mv² ÷ r = 500 × 5.69 × 10⁷ ÷ 7.0 × 10⁶ = 4.06 × 10³ N.
- Check with gravity: GMm ÷ r² = 3.98 × 10¹⁴ × 500 ÷ (4.9 × 10¹³) = 4.06 × 10³ N.
Both routes give the same number, and that is the evidence that gravity and centripetal force are one force.
The mistake that costs marks
The common slip is to draw two inward arrows, gravity and centripetal force, or to draw an outward “centrifugal” force balancing gravity. Neither belongs on a correct diagram.
| Diagram or statement | Problem | Correct |
|---|---|---|
| Gravity and centripetal force both drawn | One force counted twice | One arrow, labelled gravity |
| Outward force balances gravity | No balance, the satellite accelerates inward | Net force is gravity, toward the centre |
| v depends on satellite mass | Mass cancels | v = √(GM ÷ r) |
A satellite in a circular orbit is not in equilibrium. Its speed stays constant, but its direction changes, so it accelerates toward the centre.
Check yourself
A satellite of mass 200 kg orbits at a radius of 8.0 × 10⁶ m. Use GM = 3.98 × 10¹⁴ N m² kg⁻¹. Find the gravitational force on it and its orbital speed.
Answer
Force = GMm ÷ r² = 3.98 × 10¹⁴ × 200 ÷ (6.4 × 10¹³) = 1.24 × 10³ N.
Speed = √(GM ÷ r) = √(3.98 × 10¹⁴ ÷ 8.0 × 10⁶) = √(4.975 × 10⁷) = 7.05 × 10³ m s⁻¹.
The mass of 200 kg is not needed for the speed, which is a useful check on your working.
What to study next
Move on to using orbital relationships with consistent units, where period and radius are linked. Test the chapter with the gravitation practice set.
To have a teacher go through your orbit working, see online one-to-one Physics tuition. The units and significant figure checker helps with the powers of ten.