Circular Motion and Gravitation — IB Physics SL Study Guide

For: IB Physics SL candidates sitting IB Physics SL.

Covers: IB Topic 6 — uniform circular motion (period, frequency, angular speed), centripetal acceleration and force, Newton's universal law of gravitation, gravitational field strength, orbital speed and period (Kepler's third law).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics SL style for educational use. They are not reproductions of past IBO papers.

1. Why Circular Motion Matters

Topic 6 ties Newton's laws (Topic 2) to orbital and rotational systems. About 8-12% of IB Physics SL touches Topic 6 directly, but the framework powers Topic 5 (electric forces in circular paths) and Topic 7 (atomic-scale orbits).

The core insight: an object moving in a circle at constant speed is accelerating — its velocity direction changes even if magnitude doesn't. The acceleration always points toward the centre.

2. Uniform circular motion

For an object moving in a circle of radius $r$ at constant speed $v$:

• Period $T$: time for one full revolution.
• Frequency $f=1/T$.
• Angular speed $\omega =2\pi /T=2\pi f$ (rad/s).
• Linear speed: $v=r\omega =2\pi r/T$.

Centripetal acceleration points toward the centre with magnitude: $$a_c=\frac{v^2}{r}=r{\omega }^2$$

This is not a new force — it's the result of whatever force is providing the centripetal effect (gravity, tension, friction, normal force).

3. Centripetal force

By Newton's 2nd law: $$F_c=m{a}_c=\frac{m{v}^2}{r}=mr{\omega }^2$$

What provides $F_c$ depends on the situation:

In a vertical circle (e.g. ball on string swung vertically), gravity contributes differently at different points — at the top, $T+mg=m{v}^2/r$; at the bottom, $T-mg=m{v}^2/r$.

4. Newton's law of gravitation

The gravitational force between two point masses $M$ and $m$ separated by distance $r$: $$F=\frac{GMm}{r^2},G=6.67\times 10^{-11}{\text{ Ncdotpm}}^2/{\text{kg}}^2$$

Always attractive. Inverse square — double the distance, quarter the force.

Gravitational field strength at distance $r$ from a point mass $M$: $$g=\frac{F}{m}=\frac{GM}{r^2}$$

Units N/kg or m/s². At Earth's surface, $g=9.81$ m/s² (using Earth's mass and radius).

5. Orbital motion

For a satellite in circular orbit at radius $r$ from a planet of mass $M$:

The gravitational force provides centripetal force: $$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$$

Solve for orbital speed: $v=\sqrt{GM/r}$.

For period: $T=2\pi r/v=2\pi \sqrt{r^3/(GM)}$, which gives Kepler's third law: $$T^2=\frac{4{\pi }^2}{GM}{r}^3$$

So $T^2\propto r^3$ — the cube of orbital radius is proportional to the square of period.

Geostationary orbit: $T$ = Earth's rotation period (24 hours = 86,400 s). Solving gives $r\approx 42,000$ km from Earth's centre (about 36,000 km altitude).

6. Worked Example

A 1500 kg car negotiates a circular track of radius 50 m at constant 20 m/s.

(a) Calculate the centripetal force required. (b) If the road is flat and friction is the only horizontal force, what is the minimum coefficient of static friction needed? (c) The track is now banked at angle $\theta$. Find $\theta$ such that no friction is needed.

Solution.

(a) $F_c=m{v}^2/r=1500\cdot 400/50=12000$ N.

(b) Friction provides $F_c$: $\mu mg=m{v}^2/r$, so $\mu =v^2/(rg)=400/(50\cdot 9.81)=0.815$. Need $\mu \ge 0.82$.

(c) On a banked road with no friction, the horizontal component of normal force provides $F_c$: $N\sin \theta =m{v}^2/r$ (horizontal) $N\cos \theta =mg$ (vertical)

Dividing: $\tan \theta =v^2/(rg)=0.815$, so $\theta =\arctan (0.815)=39.2°$.

7. Common Pitfalls

• "Centrifugal force": there is no centrifugal force in inertial reference frames — it's a pseudoforce that appears in rotating frames. AP/IB problems use inertial frames.
• Mixing radius and altitude: orbital radius $r$ is measured from the centre of the planet, not its surface. Geostationary altitude ≈ 36,000 km but $r\approx 42,000$ km.
• Forgetting the vertical circle gravity contribution: at the top, gravity helps (less tension needed); at the bottom, gravity opposes (more tension needed). Set up free-body separately.
• Period vs frequency confusion: $T$ is in seconds per cycle; $f$ is cycles per second. Reciprocals.

8. Practice Questions

1. A satellite orbits 6.4 × 10⁶ m above Earth's surface (Earth radius 6.4 × 10⁶ m, mass 6.0 × 10²⁴ kg). Find its orbital speed and period.
2. A 0.20 kg ball on a 0.50 m string is swung in a horizontal circle at 5.0 m/s. Find the tension in the string.
3. The Moon orbits Earth at 3.84 × 10⁸ m with period 27.3 days. Use Kepler's third law to estimate Earth's mass.

9. Quick Reference Cheatsheet

• $T=2\pi /\omega$, $\omega =2\pi f$, $v=r\omega$.
• Centripetal: $a_c=v^2/r=r{\omega }^2$, $F_c=m{v}^2/r$.
• Gravitation: $F=GMm/r^2$, $g=GM/r^2$.
• Orbital speed: $v=\sqrt{GM/r}$.
• Kepler 3: $T^2=(4{\pi }^2/GM)r^3$.
• Banking (no friction): $\tan \theta =v^2/(rg)$.

10. What's Next

Circular motion connects to Topic 5 (Electricity & Magnetism) for charged particles in magnetic fields — they trace circles with $r=mv/(qB)$. Topic 7 (Atomic & Nuclear) treats electron orbits classically with circular motion. Use Ollie for any specific orbital or banking problem: "Walk me through the velocity required for a satellite to escape" or "Why does a faster car need a higher banking angle?"