The big idea: A satellite or planet moves in a circle because gravity pulls it toward the central body.
That inward pull is the centripetal force — the single force that keeps any object turning in a circle instead of flying off straight.
Nothing pushes the satellite forward — it just keeps 'falling' around the central body.
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Define: centripetal: Centripetal means 'toward the centre'. The centripetal force is whatever points inward and bends the path into a circle.
For an orbit, that force is gravity — there is no separate 'orbit force'.
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Because gravity is the centripetal force, we set the two equal. The gravitational force on the orbiting mass m is its weight in the field (g = GM/r², so the force is mg = GMm/r²), and the centripetal force needed is mv²/r:
- orbital speed (m s⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻² (given)
- mass of the central body, e.g. Earth or the Sun (kg)
- orbit radius — centre of the central body to the orbiting body (m)
- gravitational field strength (N kg⁻¹)
- gravitational force (N)
- mass feeling the force (kg)
- gravitational constant (given)
- mass of the central body (kg)
- distance from the centre of the central body (m)
Cancel the m and one factor of r, then make v the subject. This gives the orbital speed — notice the mass of the satellite has vanished, so a heavy and a light satellite at the same radius orbit at the same speed:
- orbital speed (m s⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻² (given)
- mass of the central body, e.g. Earth or the Sun (kg)
- orbit radius — centre of the central body to the orbiting body (m)
From speed to period: In one orbit the satellite travels a full circumference 2πr in one period T, so its speed is also v = 2πr ÷ T (given in the data booklet for circular motion).
Putting the two expressions for v together gives Kepler's third law: T² = (4π²/GM) r³ — period squared is proportional to radius cubed.
- orbital speed (m s⁻¹)
- orbit radius (m)
- orbital period — time for one full orbit (s)
IB-style question — orbital speed and period
A satellite orbits Earth (M = 6.0 × 10²⁴ kg) at a radius of 7.0 × 10⁶ m from Earth's centre. Find (this part) its orbital speed. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
Solution
- Gravity provides the centripetal force, which gives the orbital-speed result:
- Put in the numbers (G, M, r):
- Work out the inside, then square-root — keep the unit:
Final answer
orbital speed v ≈ 7.6 × 10³ m s⁻¹ (about 7.6 km s⁻¹).
IB-style question — the period
For the same satellite (r = 7.0 × 10⁶ m, v = 7.6 × 10³ m s⁻¹), find its orbital period T.
Solution
- Use the given circular-motion speed and make T the subject:
- Put in the numbers:
- Work it out — keep the unit:
Final answer
period T ≈ 5.8 × 10³ s (about 96 minutes — a typical low Earth orbit).
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How this is tested: Orbits appear both as a quick multiple-choice item and as an extended Paper 2 calculation.
- Paper 1A / 2: state that the satellite's acceleration points toward the central body (gravity is centripetal), or find an orbital speed with v = √(GM/r). - Paper 2: 'show that' the gradient of a T² against r³ graph is 4π²/GM, then use it to find the mass of the Sun from a planet's motion, or the height of a satellite from its period.
Classic trap: r is measured from the centre of the planet, so a satellite's HEIGHT above the surface is r − (planet radius), not r itself.
Weighing the central body: Kepler's third law T² = (4π²/GM) r³ contains the central mass M but not the orbiting mass.
So measuring any orbit's T and r lets you rearrange for M = 4π²r³ ÷ (GT²) — that is how astronomers find the mass of the Sun from the planets' motion.
IB-style question — (a) direction of acceleration
A satellite moves at constant speed in a circular orbit around Earth. State the direction of its acceleration, and explain why it has an acceleration even though its speed is constant.
Solution
- The only force on the satellite is gravity, which points toward Earth's centre.
- By Newton's second law the acceleration is in the same direction as that force — so it points toward Earth's centre (centripetal).
- Velocity is a vector: the direction is changing all the time even though the speed is fixed, so the satellite is accelerating.
Final answer
The acceleration points toward the centre of the Earth (centripetal). The speed is constant but the direction of motion keeps changing, so the velocity changes — that is an acceleration.
IB-style question — (b) mass of the central body
A planet orbits a star in a circle of radius 2.0 × 10¹¹ m with a period of 2.0 × 10⁷ s. Determine the mass of the star. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
Solution
- Gravity as the centripetal force gives Kepler's third law:
- Rearrange to make the star's mass M the subject:
- Put in the numbers:
- Work it out — keep the unit:
Final answer
mass of the star M ≈ 1.2 × 10³¹ kg (about six times the Sun's mass).