Question

A geostationary satellite is orbiting the earth at a height of $5\,R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2\,R$ from the surface of the earth is

• A. 5
• B. 10
• C. $6\sqrt2$
• D. $\frac {6}{\sqrt2}$

Answer 1

The Correct Option is C

 To solve the problem of finding the time period of a satellite orbiting the Earth, we need to use Kepler's third law of planetary motion, which states that the square of the orbital period \(T\) of a satellite is directly proportional to the cube of the semi-major axis \(a\) of its orbit. Mathematically, it is expressed as:

\(T^2 \propto a^3\)

For a satellite orbiting the Earth, the semi-major axis \(a\) is the sum of Earth's radius \(R\) and the height of the satellite above Earth's surface.

Let's denote:

• \(R\) = Radius of Earth
• \(h\) = Height of the satellite above Earth's surface
• \(T_g\) = Time period of the geostationary satellite
• \(T_s\) = Time period of the second satellite

For the geostationary satellite, which orbits the Earth at a height of \(5R\), the total distance from the center of the Earth \(= R + 5R = 6R\).

For the geostationary satellite, the time period \(T_g = 24\) hours.

For the second satellite, the orbital height is \(2R\). Therefore, the total distance from the center of the Earth \(= R + 2R = 3R\).

According to Kepler's third law, we get the ratio:

\(\frac{T_s^2}{T_g^2} = \frac{(3R)^3}{(6R)^3}\)

Plugging in the given values, we have:

\(\frac{T_s^2}{24^2} = \frac{27}{216}\)

Simplify the fraction:

\(\frac{T_s^2}{576} = \frac{1}{8}\)

Solving for \(T_s\) :

\(T_s^2 = 72\) \(T_s = \sqrt{72} = 6\sqrt{2}\)

Thus, the time period of the second satellite is \(6\sqrt{2}\) hours.

Therefore, the correct answer is:

\(6\sqrt2\)