When an object moves in a circle, it always needs a force that pulls it toward the centre. That force causes a special kind of acceleration called centripetal acceleration. It’s not about speeding up or slowing down; it’s about changing direction.
The magnitude of centripetal acceleration is given by:
\$a_c = \frac{v^2}{r}\$
Where \$v\$ = speed of the object and \$r\$ = radius of the circular path.
Imagine you’re on a merry‑go‑round. Even if you’re moving at a constant speed, you feel a pull toward the centre. That feeling is due to centripetal acceleration. The faster you spin (higher \$v\$) or the tighter the circle (smaller \$r\$), the stronger that pull feels.
\$a_c = \frac{(20)^2}{50} = 8 \text{ m/s}^2\$
That 8 m/s² is the acceleration pulling the car toward the centre of the curve.
Gravity is a special case of centripetal force. The Earth’s gravity keeps the Moon in orbit around Earth, and the Sun keeps the Earth in orbit around it. The gravitational field strength \$g\$ at a distance \$r\$ from a mass \$M\$ is:
\$g = \frac{GM}{r^2}\$
Here, \$G\$ is the gravitational constant. The same formula that gives centripetal acceleration also tells us how strong gravity is at any point.
Tip: When solving for centripetal acceleration, always check units: speed in m/s, radius in m → acceleration in m/s².
Common Mistake: Mixing up centripetal (toward centre) with centrifugal (outward) forces. Remember: the force you calculate is always inward.
| Body | Mass (kg) | Radius from Surface (m) | Field Strength \$g\$ (m/s²) |
|---|---|---|---|
| Earth | \$5.97\times10^{24}\$ | 0 (surface) | 9.81 |
| Moon | \$7.35\times10^{22}\$ | 0 (surface) | 1.62 |
| Sun | \$1.99\times10^{30}\$ | 1.5×10¹¹ (1 AU) | 0.006 |
For an object in circular orbit, the gravitational force provides the necessary centripetal force:
\$\frac{GMm}{r^2} = \frac{mv^2}{r}\$
Simplifying gives the orbital speed:
\$v = \sqrt{\frac{GM}{r}}\$
Notice how the same constants appear in both centripetal and gravitational equations – they’re two sides of the same coin.
Tip: When given a problem about orbital speed or period, check if you can use \$v = \sqrt{GM/r}\$ or the period formula \$T = 2\pi\sqrt{r^3/GM}\$.
Remember: The gravitational constant \$G\$ is \$6.674\times10^{-11}\,\text{N·m}^2/\text{kg}^2\$. It’s tiny, but the huge masses make the product \$GM\$ large enough to produce noticeable gravity.