Circular Motion & Gravitation

Uniform Circular Motion

An object moving in a circle at constant speed has a continuously changing velocity direction, meaning it accelerates.

• Centripetal acceleration (directed toward the center): $a_c = \frac{v^2}{r} = \omega^2 r$

• Centripetal force (net force toward center): $F_c = \frac{mv^2}{r}$

• Period: $T = \frac{2\pi r}{v}$

Centripetal force is not a new type of force — it is the net inward force provided by tension, gravity, friction, normal force, etc.

Newton's Law of Universal Gravitation

$F_g = \frac{Gm_1 m_2}{r^2}$ where $G = 6.674 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2$.

Gravitational Field Strength

$g = \frac{GM}{r^2}$ At Earth's surface, $g \approx 9.8\ \text{m/s}^2$.

Orbital Motion

For a satellite in circular orbit, gravity provides the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ $v_{\text{orbit}} = \sqrt{\frac{GM}{r}}$

Kepler's Third Law

$T^2 \propto r^3 \quad \Rightarrow \quad T^2 = \frac{4\pi^2}{GM}r^3$

Problem: A $1000\ \text{kg}$ car travels around a flat circular track of radius $50\ \text{m}$ at $15\ \text{m/s}$. What minimum coefficient of static friction is needed?

Solution:

The friction force provides centripetal force: $f_s = \frac{mv^2}{r} = \frac{1000(15)^2}{50} = 4500\ \text{N}$

Normal force on flat ground: $F_N = mg = 1000(9.8) = 9800\ \text{N}$.

$\mu_s = \frac{f_s}{F_N} = \frac{4500}{9800} \approx 0.46$

• Centripetal acceleration is $v^2/r$, always directed toward the center.
• Centripetal force is the net inward force — identify which real force(s) provide it.
• Gravity follows an inverse-square law: $F \propto 1/r^2$.
• Orbital speed depends on the central mass and orbital radius, not the satellite's mass.