Question

Determine the time it takes for a satellite to orbit the Earth in a circular near-Earth orbit. A “near-Earth” orbit is at a height above the surface of the Earth that is very small compared to the radius of the Earth. [Hint. You may take the acceleration due to gravity as essentially the same as that on the surface.] Does your result depend on the mass of the satellite?

Short answer

The period of a satellite is $5070.6\mathrm{s}$.

Step-by-step solution

Step 1. Understanding the relation of speed of the satellite

In this problem, for evaluating the time period of the satellite utilize the relation of the speed of the satellite in the orbit of Earth having radius r and mass m.

Step 2. Estimating the time period of the satellite

The relation of speed of a satelliteis given by,

$\begin{array}{c}v=\sqrt{\frac{Gm}{r}}\\ \frac{2\pi r}{T}=\sqrt{\frac{Gm}{r}}\\ T=\frac{2\pi r}{\sqrt{\frac{Gm}{r}}}\end{array}$

Here, m is the mass of the Earth, G is the gravitational constant, r is the radius of the Earth and T is the time period.

On plugging the values in the above relation.

$\begin{array}{l}T=\frac{2\pi \left(6.38\times {10}^6\mathrm{m}\right)}{\sqrt{\frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{\left(6.38\times {10}^6\mathrm{m}\right)}}}\\ T=5070.6\mathrm{s}\end{array}$

Thus, $T=5070.6\mathrm{s}$is the required period of a satellite.