Gravitational fields
Centripetal Acceleration 🚀
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.
Key Formula
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.
Analogy: The Merry‑Go‑Round 🎠
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.
Real‑World Example: Car Turning 🏎️
- Car travels at 20 m/s around a curve with radius 50 m.
- Compute $a_c$:
$$a_c = \frac{(20)^2}{50} = 8 \text{ m/s}^2$$
That 8 m/s² is the acceleration pulling the car toward the centre of the curve.
Connecting to Gravitational Fields 🌍
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.
Exam Tip Box 📚
Common Mistake: Mixing up centripetal (toward centre) with centrifugal (outward) forces. Remember: the force you calculate is always inward.
Gravitational Fields & Orbital Motion 🌌
Gravitational Field Strength Table
| 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 |
Orbiting a Planet: The Balance of Forces
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.
Exam Tip Box 📚
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.
Revision
Log in to practice.