Gravitational Circular Motion
Have you ever wondered how planets and satellites remain in orbit? This phenomenon occurs when the gravitational force between masses provides the centripetal force needed for orbital motion. Natural satellites like planets, moons, and stars, as well as artificial satellites are kept in their orbital paths by gravity. Although gravity is one of the weakest fundamental forces, in the vacuum of space it acts unimpeded by other forces, and its effects can be precisely calculated using gravitational field equations.
Use this page to revise the following concepts within gravitational circular motion:
Gravitational forces
Gravity provides the centripetal force needed for objects to keep satellite s in orbit . This acts the same as horizontal circular motion but at a much larger scale. While actual orbits are elliptical and can occur at various angles, we'll analyse them as circular motion in a flat plane for simplicity. When analysing orbital paths, we measure the radius from the centre of the orbited body (like Earth) to the satellite, rather than just the distance above the surface.
Worked example
A satellite orbits Earth in a circular path at an altitude of \(400 \mathrm{~km}\). The gravitational force acting on the satellite is \(15000 \mathrm{~N}\) and its mass is \(2000 \mathrm{~kg}\). Calculate its orbital velocity. Earth's radius is \(6370 \mathrm{~km}\).
Identify known variables:
\(F = 15000 \mathrm{~N}\)
\(m = 2000 \mathrm{~kg}\)
\(r = (400 + 6370) \times 10^3 \mathrm{~m} = 6.77 \times 10^6 \mathrm{~m}\)
Use the centripetal force equation:
\[ \begin{aligned} F &= \frac{mv^2}{r} \\ v &= \sqrt{\frac{Fr}{m}} \end{aligned} \]
Substitute values:
\[ \begin{aligned} v &= \sqrt{\frac{(15000)(6.77 \times 10^6)}{2000}} \\ &\approx 7125.66 \mathrm{~ms}^{-1} \end{aligned} \]
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Gravitational fields
Gravity is a relatively weak force that depends on the mass of both objects and the distance between their centres:
- The greater the mass, the stronger the force
- The greater the distance, the weaker the force
This can be expressed by Newton’s law of universal gravitation:
\[ F = G\frac{m_1m_2}{r^2} \]
Where:
- \(F\) = magnitude of the gravitational force between the two masses \((\mathrm{N})\)
- \(G\) = universal gravitational constant \((6.67 \times 10^{-11} \mathrm{~Nm}^2\mathrm{~kg}^{-2})\)
- \(m_1\) = mass of object \(1\) \((\mathrm{kg})\)
- \(m_2\) = mass of object \(2\) \((\mathrm{kg})\)
- \(r\) = distance between the centres of the masses \((\mathrm{m})\)
More advanced gravitational field concepts are covered in the Fields and interactions topic.
Since gravity provides the centripetal force needed for orbital motion, the gravitational force and centripetal force can be equated:
\[ F_c = F_g \]
Therefore:
\[ \frac{m_2v^2}{r} = G\frac{m_1m_2}{r^2} \]
Solving for \(v\), we obtain the orbital speed of the smaller mass \(m_2\) as it orbits the larger mass \(m_1\):
\[ v = \sqrt{\frac{Gm_1}{r}} \]
Where:
- \(v\) = orbital speed \((\mathrm{ms}^{-1})\)
While we often treat the larger body (like the Sun in our solar system) as stationary for simplicity, both bodies orbit the centre of mass of the system. The location of this point depends on the relative masses of the two bodies. For example, while the Earth appears to orbit the Sun, both bodies technically orbit a point that lies within the Sun due to its much greater mass.
Worked example
Calculate the gravitational force between the Earth and the Sun. The mass of the Sun is \(1.989 \times 10^{30} \mathrm{~kg}\), and the mass of the Earth is \(5.972 \times 10^{24} \mathrm{~kg}\). The distance between the Earth and Sun is \(1.496 \times 10^{11} \mathrm{~m}\). Use the gravitational constant \(G = 6.67 \times 10^{-11} \mathrm{~Nm}^2\mathrm{~kg}^{-2}\).
Substitute values into Newton’s law of universal gravitation:
\[ \begin{aligned} F &= G\frac{m_1m_2}{r^2} \\ F &= \frac{(6.67 \times 10^{-11})(1.989 \times 10^{30})(5.972 \times 10^{24})}{(1.496 \times 10^{11})^2} \\ F &= \frac{7.92283 \times 10^{44}}{2.23802 \times 10^{22}} \\ F &= 3.54 \times 10^{22} \mathrm{~N} \end{aligned} \]