Circular Motion and Gravitation

By the end of this lesson, you will be able to analyze circular motion using centripetal acceleration and force, apply Newton's law of universal gravitation, and calculate orbital speeds and periods.

Circular Motion

An object moving in a circle at constant speed is not in uniform motion — its direction (and therefore velocity) is constantly changing. This requires a net force toward the center.

Centripetal acceleration: directed toward the center of the circle.

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

Centripetal force: the net inward force providing centripetal acceleration.

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

$\omega$ = angular velocity (rad/s)

Important: centripetal force is NOT a new force — it's the net result of existing forces (gravity, tension, normal force, friction) directed toward the center.

Example 1 — Car on a Curve

A 1200 kg car rounds a flat curve of radius 80 m at 20 m/s. What friction force is required?

Friction provides centripetal force (flat, circular curve):

$f = \frac{mv^2}{r} = \frac{1200 \times 400}{80} = 6000$ N

Required $\mu = f/N = 6000/(1200 \times 9.8) = 0.51$

Example 2 — Vertical Circular Motion

At the top of a loop-the-loop of radius 5 m, what minimum speed keeps the car on the track?

At the top, gravity + normal force both point down (toward center):

$mg + N = mv^2/r$

Minimum speed: $N = 0$:

$mg = mv^2_{min}/r \implies v_{min} = \sqrt{rg} = \sqrt{5 \times 9.8} = 7.0$ m/s

Newton's Law of Universal Gravitation

Every mass attracts every other mass with a gravitational force:

$$F_g = \frac{Gm_1m_2}{r^2}$$

• $G = 6.674 \times 10^{-11}$ N·m²/kg²
• $r$ = distance between centers of mass

Inverse-square law: doubling distance → force is $1/4$ as strong.

Gravitational Field

$$\vec{g} = \frac{GM}{r^2}$$ (directed toward mass $M$)

Near Earth's surface: $g = 9.8$ m/s² (using $M_\oplus = 5.97 \times 10^{24}$ kg, $R_\oplus = 6.37 \times 10^6$ m).

Orbital Mechanics

For a circular orbit at radius $r$ around mass $M$:

Gravity provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$

Orbital speed: $v = \sqrt{\frac{GM}{r}}$

Orbital period: $T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}$ (Kepler's Third Law)