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Gravity & Motion

Syllabus reference

Unit 3, Topic 1 — 22 hours (including practicals)

Vector analysis (2D)

In Unit 3, vector analysis extends to two dimensions. Any vector can be resolved into two perpendicular components:

\[ F_x = F\cos\theta \qquad F_y = F\sin\theta \]

To find the resultant of multiple vectors: resolve each into components, add the components, then recombine:

\[ F_R = \sqrt{F_x^2 + F_y^2} \qquad \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) \]

Projectile motion

A projectile is any object moving freely under the influence of gravity alone (ignoring air resistance). The key principle is the independence of horizontal and vertical motion:

  • Horizontal: constant velocity (\(a_x = 0\))
  • Vertical: constant acceleration (\(a_y = g = 9.8\) m s⁻²)

Key formulas

Horizontal: \(s_x = u_x t\), where \(u_x = u\cos\theta\)

Vertical: \(s_y = u_y t + \tfrac{1}{2}g t^2\), where \(u_y = u\sin\theta\)

At any time, the resultant velocity: \(v = \sqrt{v_x^2 + v_y^2}\)

Worked example: Horizontal projection

Question: A ball is launched horizontally at 15 m s⁻¹ from a cliff 45 m high. How far does it land from the base?

Solution: Find time of flight from vertical motion (\(u_y = 0\)):

\(45 = \tfrac{1}{2}(9.8)t^2 \implies t = 3.03\) s

Horizontal distance: \(s_x = 15 \times 3.03 = 45.5\) m

Inclined planes

On an inclined plane at angle \(\theta\) to the horizontal, the weight force is resolved into:

  • Parallel to slope (causes acceleration down the slope): \(F_\parallel = mg\sin\theta\)
  • Perpendicular to slope (balanced by the normal force): \(F_\perp = mg\cos\theta\)

With friction (\(f\)), the net force along the slope: \(F_{net} = mg\sin\theta - f\)

Circular motion

An object moving in a circle at constant speed is accelerating because the direction of its velocity is continuously changing. This acceleration is directed toward the centre of the circle (centripetal).

Key formulas

[ a_c = \frac{v^2}{r} \qquad F_c = \frac{mv^2}{r} ] [ v = \frac{2\pi r}{T} ]

The centripetal force is not a new force — it is provided by an existing force (gravity, tension, friction, etc.) acting toward the centre.

Gravitation

Newton's law of universal gravitation

Key formula

\[ F = \frac{Gm_1 m_2}{r^2} \]

where \(G = 6.67 \times 10^{-11}\) N m² kg⁻²

Gravitational field strength at distance \(r\) from a mass \(M\):

\[ g = \frac{GM}{r^2} \]

Orbital motion

For a satellite in circular orbit, gravity provides the centripetal force:

\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}} \]

Orbital period:

\[ T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}} \]

Kepler's third law: \(T^2 \propto r^3\) — the square of the orbital period is proportional to the cube of the orbital radius.

Gravitational potential energy (in a field)

\[ E_p = -\frac{Gm_1 m_2}{r} \]

The negative sign indicates that energy must be added (work done) to move the object further from the mass. At infinite separation, \(E_p = 0\).

Simulations and videos

Simulations:

Crash Course Physics:

External resources: