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:

\$ac = g = \frac{GME}{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{GME/r}\$ with \$GME = 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

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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