Question

A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's centre. Their time periods are in the ratio of

• A. 1:2
• B. 1:16
• C. 1:32
• D. 1:2√2

Hint:

According to Kepler's Law, the time period does not depend upon the mass of the satellite it only depends upon the orbital radius.

Answer 1

The Correct Option is D

To solve this problem, we need to find the ratio of the time periods of two satellites, A and B, orbiting the Earth. This requires us to use Kepler's Third Law, which states that the square of the time period T of a satellite is proportional to the cube of the semi-major axis a of its orbit:

T^2 \propto a^3

For a circular orbit, the semi-major axis a is simply the distance from the center of the Earth to the satellite.

Let's calculate the time period for each satellite:

1. Satellite A:
   • Distance from the center of the Earth = R + r (where R is the radius of the Earth).
   • Using Kepler's Third Law, T_A^2 \propto (R + r)^3.
2. Satellite B:
   • Distance from the center of the Earth = 2r.
   • Using Kepler's Third Law, T_B^2 \propto (2r)^3 = 8r^3.

Now, we need to find the ratio of their time periods:

\left( \frac{T_A}{T_B} \right)^2 = \frac{(R + r)^3}{8r^3}

Taking the square root on both sides, we obtain:

\frac{T_A}{T_B} = \sqrt{\frac{(R + r)^3}{8r^3}}

Under the assumption that the orbit radius for both satellites is large compared to the Earth’s radius, we can simplify this to:

\frac{T_A}{T_B} = \frac{(R + r)^{3/2}}{(2r)^{3/2}} = \frac{1}{2\sqrt{2}}

Thus, the ratio of the periods T_A\) to T_B\) is:

1:2\sqrt{2}

The correct answer is 1:2√2.