A.6 Circular motion and gravitation (AHL)

IB Physics HL· IB Physics HL Topic A: A.6 AHL· 20 min read

1. Uniform Circular Motion Dynamics★★★☆☆⏱ 5 min

📘 Definition

Uniform circular motion (UCM)

UCM

Motion of an object traveling at constant speed along a circular path of fixed radius. Speed is constant, but velocity changes direction, so acceleration is non-zero.

Example:

A car rounding a circular bend at constant 30 km/h

Acceleration in UCM is always directed towards the center of the circle, called centripetal acceleration. Its magnitude is given by:

a_c = \frac{v^2}{r} = \omega^2 r

By Newton's second law, the net force causing this acceleration (centripetal force) is also directed towards the center, with magnitude:

F_c = m a_c = \frac{m v^2}{r} = m \omega^2 r

📐 Worked Example

A 1000 kg car drives around a circular bend of radius 50 m at constant speed 15 m s⁻¹. Calculate the required centripetal force and state what provides it.

1
Write the formula for centripetal force:
2
$F_c = \frac{m v^2}{r}$
3
Substitute given values:
4
$F_c = \frac{1000 \times 15^2}{50} = \frac{225000}{50} = 4500 \text{ N}$
5
State the origin of the force: Friction between the car's tires and the road surface provides the centripetal force.

Exam tip:

Centripetal force is the net resultant force towards the center, not a separate force. Never add it as an extra force in free-body diagrams.

2. Newton's Law of Universal Gravitation★★★☆☆⏱ 5 min

📘 Definition

Newton's Law of Universal Gravitation

Every mass attracts every other mass with a force proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.

Example:

Force between the Sun and the Earth, force between you and the Earth

F = G \frac{M m}{r^2}

Where $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ is the universal gravitational constant, and $r$ is the distance between the centers of the two masses. For spherical masses, the formula applies directly.

📐 Worked Example

Calculate the gravitational force between Earth ( $M = 5.97 \times 1 0^{24} kg$ ) and a 70 kg person standing on Earth's surface (Earth radius $r = 6.37 \times 1 0^{6} m$ ).

1
Substitute values into the gravitation formula:
2
$F = (6.67 \times 10^{-11}) \frac{(5.97 \times 10^{24})(70)}{(6.37 \times 10^6)^2}$
3
Calculate numerator and denominator:
4
$F ≈ \frac{2.79 \times 10^{16}}{4.06 \times 10^{13}} ≈ 687 \text{ N}$
5
This matches the expected weight of ~ $70 g$ , confirming the formula works for objects on a planet's surface.

3. Kepler's Laws of Planetary Motion★★★★☆⏱ 5 min

Kepler derived three empirical laws describing planetary motion before Newton developed his theory of gravitation. Newton later proved these laws follow directly from universal gravitation:

Law of Orbits: All planets move in elliptical orbits with the Sun at one focus.
Law of Areas: A line joining a planet to the Sun sweeps out equal areas in equal time intervals.
Law of Periods: The square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

For circular orbits, where the semi-major axis $a = r$ (orbital radius), Kepler's third law becomes:

T^2 = \left( \frac{4 \pi^2}{G M} \right) r^3

📐 Worked Example

Earth's orbital period is 1 year, with semi-major axis 1 AU. Mars has a semi-major axis of 1.52 AU. Calculate Mars' orbital period.

1
For objects orbiting the same central body, $T^{2} / a^{3}$ is constant:
2
$\frac{T_E^2}{a_E^3} = \frac{T_M^2}{a_M^3}$
3
Rearrange for $T_{M}$ :
4
$T_M = T_E \sqrt{\left( \frac{a_M}{a_E} \right)^3}$
5
Substitute values:
6
$T_M = 1 \times \sqrt{(1.52)^3} ≈ 1.87 \text{ years}$

Exam tip:

Only use the $T^{2} / a^{3} = constant$ ratio when both objects orbit the same central body. The proportionality constant depends on the central mass, so it changes for different central bodies.

4. Orbital Motion and Escape Speed★★★★☆HL only⏱ 5 min

For a satellite in circular orbit, gravitational force provides the centripetal force required to maintain the orbit. We can derive the orbital speed directly from this equality:

🔬 Derivation

Goal:

Derive orbital speed for a circular orbit

Starting from:

Gravitational force = centripetal force

1
Equate the two force expressions:
2
$G \frac{M m}{r^2} = \frac{m v^2}{r}$
3
Cancel $m$ and one $r$ from both sides:
4
$v^2 = \frac{G M}{r}$
5
Rearrange for $v$ :

Result:

$v = \frac{GM}{r}$ : orbital speed depends only on the central mass and orbital radius, not the satellite mass.

Escape speed is the minimum speed required for an object to escape a planet's gravitational pull, reaching infinite distance with zero remaining kinetic energy. Its derivation from energy conservation gives:

v_{esc} = \sqrt{\frac{2 G M}{R}}

📐 Worked Example

Calculate escape speed from Earth's surface, given $M_{E} = 5.97 \times 1 0^{24} kg$ , $R_{E} = 6.37 \times 1 0^{6} m$ .

1
Substitute into the escape speed formula:
2
$v_{esc} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}}$
3
Calculate the value inside the square root:
4
$v_{esc} ≈ \sqrt{1.25 \times 10^8} ≈ 11200 \text{ m s}^{-1} = 11.2 \text{ km s}^{-1}$

5. Common Pitfalls

Wrong move:

Treat centripetal force as an extra separate force in free-body diagrams.

Why:

Centripetal force is the net resultant force towards the center, provided by real forces like gravity, friction, or tension.

Correct move:

Identify all real forces, then take the component towards the center as the centripetal force.

Wrong move:

Use the $T^{2} / a^{3} = constant$ ratio for planets orbiting different stars.

Why:

The proportionality constant depends on the mass of the central body, so it changes when the central mass changes.

Correct move:

Only use the ratio for objects orbiting the same central body, otherwise use the full formula $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ .

Wrong move:

Use height above a planet's surface instead of distance from the planet's center in gravitational calculations.

Why:

Newton's law of gravitation uses the distance between the centers of the two masses, not surface height.

Correct move:

Always add the planet's radius to the height above the surface to get the total $r$ .

Wrong move:

Claim orbiting objects are weightless because gravity does not act on them.

Why:

Gravity is still acting on the orbiting object, and provides the centripetal force for orbit.

Correct move:

Weightlessness is apparent weightlessness: the object and its reference frame are both in free fall.

Wrong move:

Confuse orbital speed and escape speed, use the wrong formula.

Why:

Orbital speed is for stable circular orbit, escape speed is to leave the gravitational field entirely.

Correct move:

Remember $v_{esc} = 2 v_{or bi t}$ at the same radius.

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Centripetal acceleration	$a_{c} = \frac{v ^{2}}{r} = ω^{2} r$	Always directed towards center
Centripetal force	$F_{c} = \frac{m v ^{2}}{r}$	Net force, not an extra force
Gravitational force	$F = G \frac{M m}{r ^{2}}$	r = distance between centers
Kepler's 3rd Law (circular)	$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$	M = mass of central body
Orbital speed	$v = \frac{GM}{r}$	Independent of satellite mass
Escape speed	$v_{esc} = \frac{2 GM}{R}$	R = radius of planet

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Gravitational force between two masses
2024 · Paper 2
Orbital speed of a geostationary satellite
2023 · Paper 1
Kepler's third law application

What's Next

This subtopic forms the foundation for advanced gravitational concepts and astrophysics in IB Physics HL. It connects directly to gravitational field theory, where you will extend these ideas to calculate gravitational potential and field strength around extended masses. Understanding circular motion dynamics is also critical for rotational motion and simple harmonic motion later in the course. Derivations of orbital speed and Kepler's third law are common exam questions, so mastering the steps here will give you easy marks in assessments.