Time of Revolution in Circular Orbits

By /

In the case of a circular orbit, a body moves at constant speed along a path of fixed radius around a central mass under the influence of a gravitational force. The time of revolution (or orbital period) is the time it takes for the body to complete one full orbit.

Gravitational Force as Centripetal Force

For a body of mass m orbiting a central body of mass M in a circular orbit of radius r, the gravitational force provides the required centripetal force:

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

Canceling m and simplifying:

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

Taking the square root:

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

This is the orbital speed in a circular orbit.

Orbital Period

The orbital period T is the time taken to complete one full revolution:

T = \frac{\text{Distance}}{\text{Speed}} = \frac{2 \pi r}{v}

Substituting the expression for v:

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

Thus, the time of revolution is:

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

Key Observations

  • The period depends only on the radius of the orbit r and the mass of the central body M.
  • The orbiting body’s mass m does not affect the period.
  • The larger the radius, the longer the period.

Application: Satellites Around Earth

For a satellite orbiting Earth in a low circular orbit (e.g., 300 km above Earth’s surface), using:

G = 6.674 \times 10^{-11} , \text{Nm}^2/\text{kg}^2

M = 5.972 \times 10^{24} , \text{kg}

r = R_E + h = 6.371 \times 10^6 + 3 \times 10^5 , \text{m}

You can calculate the actual time of revolution in seconds or minutes.

Connection to Kepler’s Third Law

This result forms the basis of Kepler’s Third Law. It shows that for circular orbits:

T^2 \propto r^3

That is, the square of the orbital period is proportional to the cube of the orbital radius. This relationship holds for all small-body orbits around a much larger central mass under gravity.

Shopping Cart
Scroll to Top