Circular motion and gravitation

IB Physics SL· Unit 1: Space, time and motion, Topic 4· 45 min read

1. Uniform Circular Motion and Centripetal Acceleration★★☆☆☆⏱ 10 min

📘 Definition

Uniform Circular Motion

Motion of an object traveling at constant speed (constant magnitude of velocity) along a circular path

Example:

A car turning a level corner at a constant 30 m/s

Even though the speed of the object is constant, the direction of its velocity changes continuously as it moves around the circle. This means the object has non-zero acceleration, which always points toward the center of the circle. This acceleration is called centripetal (center-seeking) acceleration.

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

Where $v$ is tangential speed, $r$ is the radius of the circle, and $ω$ is angular speed in radians per second.

📐 Worked Example

A child rides a merry-go-round at a distance of 2.5 m from the center, moving with a constant angular speed of 0.4 rad s⁻¹. Calculate the magnitude of their centripetal acceleration.

1
Identify known values:
2
$r = 2.5 \ \text{m}, \omega = 0.4 \ \text{rad s}^{-1}$
3
Substitute into the centripetal acceleration formula:
4
$a_c = \omega^2 r = (0.4)^2 \times 2.5 = 0.16 \times 2.5 = 0.4 \ \text{m s}^{-2}$

Exam tip:

Always remember that for uniform circular motion, the magnitude of centripetal acceleration is constant, but its direction changes continuously to always point toward the center.

2. Centripetal Force★★☆☆☆⏱ 12 min

📘 Definition

Centripetal Force

F_c

The net force acting on an object to keep it moving in uniform circular motion, always directed toward the center of the circle

From Newton's second law ( $F_{n e t} = ma$ ), we get the magnitude of centripetal force:

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

📐 Worked Example

A 900 kg car turns a flat circular corner of radius 45 m at a constant speed of 12 m/s. What is the minimum coefficient of static friction between the tires and road required to avoid slipping?

1
The centripetal force is provided entirely by static friction, so $F_{f} = F_{c}$ :
2
$\mu_s N = \frac{mv^2}{r}$
3
Normal reaction equals the car's weight, so $N = m g$ , mass cancels out on both sides:
4
$\mu_s g = \frac{v^2}{r}$
5
Rearrange for $μ_{s}$ and substitute values ( $g = 9.81 m s^{- 2}$ ):
6
$\mu_s = \frac{v^2}{rg} = \frac{12^2}{45 \times 9.81} \approx 0.33$

3. Newton's Law of Universal Gravitation★★★☆☆⏱ 15 min

📘 Definition

Newton's Law of Universal Gravitation

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

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, $M$ and $m$ are the two masses, and $r$ is the distance between the centers of the masses. For uniform spherical masses, this law applies directly.

📐 Worked Example

Calculate the gravitational force between Earth (mass $5.97 \times 1 0^{24}$ kg) and a 70 kg person standing at Earth's surface, where Earth's radius is $6.37 \times 1 0^{6}$ m.

1
Substitute all values into the gravitational force formula:
2
$F = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(70)}{(6.37 \times 10^6)^2}$
3
Calculate numerator and denominator:
4
$F \approx \frac{2.79 \times 10^{16}}{4.06 \times 10^{13}} \approx 687 \ \text{N}$
5
This matches the expected weight $m g = 70 \times 9.81 = 687$ N, confirming the result.

4. Gravitation and Circular Orbital Motion★★★☆☆⏱ 15 min

For a stable circular orbit of a smaller mass around a larger central mass, gravitational attraction provides exactly the centripetal force required to maintain the circular motion. This relationship lets us derive key properties of orbits.

🔬 Derivation

Goal:

Derive Kepler's third law for circular orbits

Starting from:

Equating gravitational force to centripetal force

1
1. Equate force: $G \frac{M m}{r ^{2}} = m \frac{v ^{2}}{r}$ . The orbiting mass $m$ cancels out:
2
$v^2 = \frac{G M}{r}$
3
1. Orbital speed is $v = \frac{2 π r}{T}$ , substitute into the equation:
4
$\frac{4 \pi^2 r^2}{T^2} = \frac{G M}{r}$
5
1. Rearrange to get the relationship between $T$ and $r$ :

Result:

$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ → the square of the orbital period is proportional to the cube of the orbital radius, for any orbit around the same central mass M.

📐 Worked Example

The ISS orbits Earth at 400 km altitude. Earth's radius = 6370 km, mass = $5.97 \times 1 0^{24}$ kg. Calculate the ISS orbital period in minutes.

1
1. Calculate orbital radius (add altitude to Earth's radius):
2
$r = 6370 + 400 = 6770 \ \text{km} = 6.77 \times 10^6 \ \text{m}$
3
1. Substitute into Kepler's third law:
4
$T^2 = \frac{4 \pi^2 r^3}{G M} = \frac{4 \pi^2 (6.77 \times 10^6)^3}{(6.67 \times 10^{-11})(5.97 \times 10^{24})} \approx 3.07 \times 10^7$
5
1. Solve for T and convert to minutes:
6
$T = \sqrt{3.07 \times 10^7} \approx 5540 \ \text{s} = \frac{5540}{60} \approx 92 \ \text{minutes}$

5. Common Pitfalls

Wrong move:

Treating centripetal force as an extra separate force on free body diagrams

Why:

Centripetal force is the net force, not a new interaction force. Adding it leads to incorrect force balances.

Correct move:

Draw only actual forces (tension, friction, gravity) then sum forces toward the center and set equal to $\frac{m v ^{2}}{r}$ .

Wrong move:

Using altitude instead of orbital radius for gravitational calculations

Why:

Orbital radius is measured from the center of the central body, not the surface. This leads to large errors in results.

Correct move:

Always add the radius of the central body to the altitude to get $r$ .

Wrong move:

Saying centripetal acceleration points outward from the center

Why:

Confusion with fictitious centrifugal force in rotating reference frames, which are not used in IB Physics.

Correct move:

For inertial reference frames (the standard frame for IB exams), centripetal acceleration always points toward the center of the circle.

Wrong move:

Canceling both masses $M$ and $m$ when deriving orbital speed

Why:

Students often accidentally cancel the central mass $M$ , leading to wrong formulas.

Correct move:

Only the orbiting mass $m$ cancels. The central mass $M$ always remains in the final formula.

Wrong move:

Using inverse proportionality $F \propto \frac{1}{r}$ instead of inverse square for gravity

Why:

Simple memorization error that changes all results.

Correct move:

Remember gravitational force follows the inverse square law: $F \propto \frac{1}{r ^{2}}$ .

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Centripetal acceleration	$a_{c} = \frac{v ^{2}}{r} = ω^{2} r$	Points toward center of circle
Centripetal force	$F_{c} = \frac{m v ^{2}}{r} = m ω^{2} r$	Net force, not an extra force
Newton's gravitation	$F = G \frac{M m}{r ^{2}}$	$r$ = distance between centers
Orbital period (circular)	$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$	$M$ = mass of central body
Orbital speed	$v = \frac{GM}{r}$	Independent of orbiting mass $m$

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.

2023 · 2
Orbital speed calculation question
2022 · 1
Centripetal force direction MCQ
2021 · 2
Derivation of Kepler's third law

Going deeper

official documentIB Physics Data BookletContains value of G and formula list

What's Next

Circular motion and gravitation form a core foundation of classical mechanics, underpinning topics from rotational motion to astrophysics. Mastery of this subtopic is critical for exam success, as it appears regularly in both multiple-choice and extended-response questions in IB Physics SL. Understanding how gravitational force provides centripetal force for stable orbits is the basis for all astrophysical calculations, which you will explore further in the IB Astrophysics option. The principles of circular motion also extend to HL topics like rotational dynamics, where you will extend these ideas to rotating rigid bodies. Next, you will build on these mechanics concepts to study energy changes and work in moving systems.