analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes

Gravitational Force Between Point Masses

Learning Objective

Analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes.

1. Newton’s Law of Universal Gravitation

The attractive force between two point masses \$m1\$ and \$m2\$ separated by a distance \$r\$ is

\$Fg = \frac{G\,m1\,m_2}{r^{2}}\$

where \$G = 6.674\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\$ is the universal gravitational constant.

2. Centripetal Force for Uniform Circular Motion

An object of mass \$m\$ moving with speed \$v\$ in a circle of radius \$r\$ requires a centripetal force

\$F_c = \frac{m v^{2}}{r}\$

3. Equating Gravitational and Centripetal Forces

For a satellite of mass \$m\$ orbiting a much larger body of mass \$M\$ (e.g., a planet or the Sun), the gravitational attraction provides the necessary centripetal force:

\$\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}\$

Cancel \$m\$ and one factor of \$r\$ to obtain the orbital speed:

\$v = \sqrt{\frac{G M}{r}}\$

4. Orbital Period

The period \$T\$ is the time to complete one full orbit. Since the circumference of the orbit is \$2\pi r\$,

\$T = \frac{2\pi r}{v} = 2\pi r \sqrt{\frac{r}{G M}} = 2\pi\sqrt{\frac{r^{3}}{G M}}\$

This is Kepler’s third law for circular orbits.

5. Example Calculation – Satellite Around Earth

1. Given: \$M_{\text{Earth}} = 5.97\times10^{24}\ \text{kg}\$, orbital radius \$r = 6.78\times10^{6}\ \text{m}\$ (≈400 km above the surface).
2. Orbital speed:

   \$\$v = \sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.78\times10^{6}}}

   \approx 7.67\times10^{3}\ \text{m s}^{-1}\$\$

3. Orbital period:

   \$\$T = 2\pi\sqrt{\frac{(6.78\times10^{6})^{3}}{(6.674\times10^{-11})(5.97\times10^{24})}}

   \approx 5.5\times10^{3}\ \text{s} \approx 92\ \text{min}\$\$

6. Key Relationships Summary

7. Common Misconceptions

• The gravitational force does not depend on the speed of the orbiting body; it depends only on the masses and separation.
• For a given central mass \$M\$, a larger orbital radius \$r\$ always means a slower orbital speed and a longer period.
• Even though the satellite’s mass \$m\$ cancels in the derivation of \$v\$, \$m\$ still determines the kinetic energy \$ \tfrac12 m v^{2}\$ and the required launch energy.

8. Extending to Non‑Circular (Elliptical) Orbits

For elliptical orbits, the instantaneous centripetal acceleration is still provided by gravity, but \$r\$ varies with true anomaly. The vis‑viva equation generalises the speed:

\$v = \sqrt{G M\!\left(\frac{2}{r} - \frac{1}{a}\right)}\$

where \$a\$ is the semi‑major axis. When \$r = a\$, the expression reduces to the circular‑orbit speed derived above.

9. Quick Check Questions

1. Derive the expression for the orbital period of a moon orbiting Jupiter, given \$M_{\text{Jupiter}} = 1.90\times10^{27}\ \text{kg}\$ and \$r = 4.22\times10^{8}\ \text{m}\$.
2. If the orbital radius of a satellite is doubled, by what factor does its orbital speed change? By what factor does its period change?
3. Explain why a low‑Earth orbiting satellite must travel faster than a geostationary satellite.

10. Summary

By equating Newton’s law of universal gravitation with the requirement for centripetal acceleration, we obtain simple formulae that describe the speed and period of objects in circular orbits. These relationships form the basis for analysing more complex orbital motions and for solving typical A‑Level physics problems.