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: Radii From the Earth Centre:
Geostationary satellite: \(r_1=R+4R=5R\), period \(T_1=24\) h.
Second satellite: \(r_2=R+1.5R=2.5R\).

Step 2: Kepler's Third Law:
\[ \frac{T_2^2}{T_1^2}=\left(\frac{r_2}{r_1}\right)^3=\left(\frac{2.5}{5}\right)^3=\frac18 \]

Step 3: Solve:
\[ T_2=24\times\frac1{2\sqrt2}=\frac{12}{\sqrt2}=6\sqrt2\ \text{h} \]

Step 4: Check the Other Options:
\(6/\sqrt2\approx4.2\) h would need a radius ratio of about \(0.31\), and 4 h and 8 h correspond to \(T^2/T_1^2=1/36\) and \(1/9\), which do not match \(1/8\). So (B) is correct.

Final Answer:
The period is \(6\sqrt2\) hours, option (B). \[ \boxed{\text{(B) } 6\sqrt{2}\ \text{h}} \]