Circular Motion and Gravitation — Study Notes

1. Circular Motion — Key Definitions

When an object moves in a circle at constant speed, it is in uniform circular motion. Although the speed is constant, the velocity is not — it continuously changes direction. This means the object is constantly accelerating, requiring a resultant force directed toward the centre of the circle.

• Angular displacement ($\theta$): The angle subtended at the centre, measured in radians. One full circle = $2\pi$ rad.
• Angular velocity ($\omega$): The rate of change of angular displacement: $\omega = \frac{\Delta\theta}{\Delta t}$, measured in rad s⁻¹.
• Period ($T$): The time for one complete revolution: $T = \frac{2\pi}{\omega}$
• Frequency ($f$): Number of revolutions per second: $f = \frac{1}{T}$, measured in Hz.

The relationship between linear speed $v$ and angular velocity $\omega$: $$v = r\omega$$

where $r$ is the radius of the circle. This shows that for the same angular velocity, objects at greater radius move with greater linear speed.

2. Centripetal Acceleration

Any object undergoing circular motion experiences centripetal acceleration directed toward the centre: $$a = \frac{v^2}{r} = r\omega^2 = \frac{4\pi^2 r}{T^2}$$

"Centripetal" means "centre-seeking." This acceleration is always perpendicular to the velocity (which is tangential), so it changes direction but not speed.

Derivation concept: Over a small time $\delta t$, velocity changes from $\vec{v_1}$ to $\vec{v_2}$ (same magnitude, slightly different direction). The change $\delta\vec{v}$ points toward the centre, giving centripetal acceleration.

3. Centripetal Force

By Newton's Second Law ($F = ma$), a resultant force must cause the centripetal acceleration: $$F = \frac{mv^2}{r} = mr\omega^2$$

This centripetal force is not a separate type of force — it is whatever force(s) in the situation act toward the centre. Examples:

Centrifugal force is a fictitious force experienced in a rotating reference frame — it does not exist in an inertial frame. In exam answers, always explain circular motion using centripetal force, not centrifugal force.

4. Vertical Circular Motion

For an object on a string moving in a vertical circle, the required centripetal force changes with position because the weight component changes.

At the top of the circle: both tension $T$ and weight $mg$ point downward (toward centre): $$T + mg = \frac{mv^2}{r} \quad \Rightarrow \quad T = \frac{mv^2}{r} - mg$$

At the bottom: tension points upward (toward centre) and weight points downward (away from centre): $$T - mg = \frac{mv^2}{r} \quad \Rightarrow \quad T = \frac{mv^2}{r} + mg$$

The minimum speed at the top for the object to maintain contact (T ≥ 0): $$\frac{mv^2}{r} \geq mg \quad \Rightarrow \quad v_{min} = \sqrt{gr}$$

graph TD
    A[Object in Circular Motion] --> B[Centripetal Acceleration: v²/r toward centre]
    B --> C[Newton's 2nd Law: F = mv²/r]
    C --> D{What provides the force?}
    D --> E[Gravity: orbiting bodies]
    D --> F[Tension: string/rod]
    D --> G[Friction: car on road]
    D --> H[Normal Force: banked track]

5. Newton's Law of Universal Gravitation

Every two masses in the universe attract each other with a gravitational force: $$F = \frac{Gm_1 m_2}{r^2}$$

Where: - $G = 6.67 \times 10^{-11}$ N m² kg⁻² (Universal Gravitational Constant) - $m_1, m_2$ = masses of the two objects (kg) - $r$ = distance between their centres (m)

The force is attractive, acts along the line joining the centres, and obeys an inverse-square law — doubling the distance reduces the force by a factor of four.

6. Gravitational Field Strength

The gravitational field strength $g$ at a point is the gravitational force per unit mass experienced by a small test mass placed at that point: $$g = \frac{F}{m} = \frac{GM}{r^2}$$

Near the Earth's surface ($r \approx R_E = 6.37 \times 10^6$ m), $g \approx 9.81$ N kg⁻¹. At greater heights, $g$ decreases with $1/r^2$.

7. Orbital Motion and Kepler's Third Law

For a satellite in circular orbit of radius $r$ around a planet of mass $M$, the gravitational force provides the centripetal force: $$\frac{GMm}{r^2} = \frac{mv^2}{r} \quad \Rightarrow \quad v = \sqrt{\frac{GM}{r}}$$

The period of orbit: $$T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{r}{GM}} = \frac{2\pi r^{3/2}}{\sqrt{GM}}$$

This gives Kepler's Third Law: $T^2 \propto r^3$, or $\frac{T^2}{r^3} = \frac{4\pi^2}{GM}$ (constant for all satellites orbiting the same body).

Geostationary orbits have period T = 24 hours and orbit above the equator at altitude ~35,800 km. The satellite appears stationary from Earth's surface — ideal for communications satellites.

8. Escape Velocity

The minimum launch speed needed to escape a planet's gravitational field (reaching infinity with zero KE): $$\frac{1}{2}mv_{esc}^2 = \frac{GMm}{R} \quad \Rightarrow \quad v_{esc} = \sqrt{\frac{2GM}{R}}$$

For Earth: $v_{esc} \approx 11.2$ km s⁻¹. For the Moon ($g \approx 1.6$ m s⁻²): $v_{esc} \approx 2.4$ km s⁻¹. Black holes have escape velocities exceeding the speed of light ($3 \times 10^8$ m s⁻¹).