Question

A satellite is placed in a circular orbit very close to the Earth's surface, so that its orbital radius is almost equal to the Earth's radius. What is the approximate time period of this satellite?

Hint:

Use T equals 2 pi times the square root of R over g, with R as the Earth's radius.

Answer 1

The time period of a satellite orbiting close to the Earth's surface is about 84.6 minutes.

Step 1: Formula.
For a satellite in a circular orbit of radius r, the time period is \(T = 2\pi \sqrt{\dfrac{r^3}{GM}}\). Since \(GM = gR^2\) near the Earth's surface, this becomes \(T = 2\pi \sqrt{\dfrac{R}{g}}\) when the orbit is close to the surface, so r is approximately equal to R, the Earth's radius.

Step 2: Substituting values.
Take \(R = 6400\) km \(= 6.4 \times 10^6\) m and \(g = 9.8\) m/s^2.
\(T = 2\pi \sqrt{\dfrac{6.4 \times 10^6}{9.8}} = 2\pi \sqrt{6.53 \times 10^5}\).
\(\sqrt{6.53 \times 10^5} \approx 808\) seconds, so \(T \approx 2\pi \times 808 \approx 5078\) seconds.

Step 3: Converting to minutes.
\(5078 / 60 \approx 84.6\) minutes.

Step 4: Answer.
So a satellite orbiting just above the Earth's surface takes about 84.6 minutes to complete one revolution. This is close to the shortest possible orbital period for any Earth satellite.