Gravitational fields

Centripetal Acceleration and Gravitational Fields

Learning Objective

Understand how centripetal acceleration arises in uniform circular motion and how it relates to the gravitational field of a massive body.

Key Concepts

• Uniform circular motion
• Centripetal force and acceleration
• Gravitational field strength $g$
• Equivalence of gravitational and centripetal forces for orbital motion

1. Centripetal Acceleration

For an object moving in a circle of radius $r$ with constant speed $v$, the direction of the velocity vector changes continuously. This change requires a net inward (centripetal) acceleration given by

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

where $\omega = \dfrac{v}{r}$ is the angular speed (rad s$^{-1}$).

Derivation

1. Consider two velocity vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ separated by a small angle $\Delta\theta$.
2. The change in velocity $\Delta\mathbf{v}$ is directed towards the centre of the circle.
3. Taking the limit $\Delta t \to 0$, the magnitude of the acceleration is $a_c = \dfrac{v\,\Delta\theta}{\Delta t} = v\omega = \dfrac{v^2}{r}$.

2. Gravitational Field Strength

The gravitational field $g$ at a distance $r$ from a point mass $M$ is defined as the force per unit mass:

$$g = \frac{F_g}{m} = \frac{GM}{r^2}$$

where $G = 6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ is the universal gravitational constant.

3. Connecting Centripetal Acceleration to Gravitational Fields

For a satellite in a circular orbit of radius $r$ around Earth (mass $M_E$), the required centripetal force is supplied by gravity:

$$\frac{mv^2}{r} = \frac{GM_E m}{r^2}$$

Canceling $m$ gives the orbital speed:

$$v = \sqrt{\frac{GM_E}{r}}$$

Dividing both sides of the force equation by $m$ shows that the centripetal acceleration equals the gravitational field strength at that altitude:

$$a_c = g = \frac{GM_E}{r^2}$$

4. Example Calculation

Find the orbital speed of a satellite 300 km above the Earth's surface.

1. Earth's radius $R_E = 6.37\times10^6\,$m.
2. Total orbital radius $r = R_E + 300\,000\,$m = $6.67\times10^6\,$m.
3. Use $v = \sqrt{GM_E/r}$ with $GM_E = 3.986\times10^{14}\,$m$^3\,$s$^{-2}$.
4. $$v = \sqrt{\frac{3.986\times10^{14}}{6.67\times10^6}} \approx 7.73\times10^3\ \text{m s}^{-1}$$

5. Comparison Table

Quantity	Symbol	Expression	Units
Centripetal acceleration	$a_c$	$\displaystyle \frac{v^2}{r} = \omega^2 r$	m s$^{-2}$
Gravitational field strength	$g$	$\displaystyle \frac{GM}{r^2}$	m s$^{-2}$
Force providing centripetal acceleration	$F_c$	$\displaystyle m\frac{v^2}{r}$	N
Gravitational force	$F_g$	$\displaystyle \frac{GMm}{r^2}$	N

6. Common Misconceptions

• Confusing centripetal force with a separate “force”; it is the net inward force acting on the object.
• Assuming the gravitational field is constant at all heights; it decreases with $1/r^2$.
• Thinking that a larger speed always means a larger centripetal force; the radius also matters.

7. Practice Questions

1. Calculate the centripetal acceleration of a car travelling at 20 m s$^{-1}$ around a curve of radius 50 m.
2. Determine the gravitational field strength at an altitude of 400 km above Earth’s surface.
3. A satellite orbits Mars (mass $M_M = 6.42\times10^{23}$ kg) at a radius of $1.0\times10^7$ m. Find its orbital speed and the centripetal acceleration.

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