Question

A geo-stationary satellite is orbiting the earth at height of 6\(R\) above the surface of the earth, \(R\) being the radius of the earth. The time period of another satellite at a height of 2.5 \(R\) from the surface of the earth is

• A. 10 h
• B. \(6\sqrt{2}\) h
• C. 6 h
• D. \(6\sqrt{2}\) h

Hint:

\(T \propto r^{3/2}\). Always measure orbital radius from Earth's centre: \(r = R_{Earth} + h\).

Answer 1

The Correct Option is B

Step 1: Basic Principle
By Kepler's third law: \(T^2 \propto r^3\), where \(r\) is the orbital radius from the Earth's centre.
Step 2: Solution Procedure:
Geostationary: \(r_1 = R + 6R = 7R\), \(T_1 = 24\) h
New satellite: \(r_2 = R + 2.5R = 3.5R\)
\[ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} = \left(\frac{3.5R}{7R}\right)^{3/2} = \left(\frac{1}{2}\right)^{3/2} = \frac{1}{2\sqrt{2}} \] \[ T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \text{ h} \]
Step 3: Required Answer:
Time period \(= \mathbf{6\sqrt{2}}\) h.