Question

A geostationary satellite is orbiting the earth at a height of \(4R\) above the surface of the earth, where \(R\) is the radius of the earth. Another satellite is orbiting the earth at a height \(1.5R\) from the surface of the earth with periodic time 'T' in hour. The value of 'T' in hour is

• A. \(\frac{6}{\sqrt{2}}\)
• B. \(6\sqrt{2}\)
• C. \(4\)
• D. \(8\)

Hint:

Use Kepler's third law T^2 proportional to r^3 with r measured from the centre of the earth.

Answer 1

The Correct Option is B

Step 1: Orbital Period Formula:
$T=2\pi\sqrt{\dfrac{r^3}{GM}}$, so $T\propto r^{3/2}$.

Step 2: Ratio:
$\dfrac{T_2}{T_1}=\left(\dfrac{2.5R}{5R}\right)^{3/2}=\left(\dfrac12\right)^{3/2}=\dfrac1{2\sqrt2}$.

Step 3: Compute:
$T_2=\dfrac{24}{2\sqrt2}=\dfrac{12}{\sqrt2}=6\sqrt2$ h. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } 6\sqrt{2}\ \text{h}} \]