Question

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.

• A. 4 hours
• B. 6 hours
• C. 3 hours
• D. 12 hours

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between the time period of a satellite and its distance from the Earth. The time period of a satellite in circular orbit around the Earth is given by Kepler's Third Law:

T^2 = \frac{4\pi^2}{GM}a^3

Where:

• T is the time period of the satellite.
• G is the gravitational constant.
• M is the mass of the Earth.
• a is the semi-major axis or average distance from the Earth to the satellite.

Given that the satellite initially has a time period of 24 hours, we need to find the new time period when the distance is reduced to one fourth of its original value.

Let the initial distance be a and the initial time period be T = 24 hours. We use the formula:

T_1^2 \propto a^3

When the distance becomes \frac{a}{4}, the new time period T_2 is given by:

T_2^2 \propto \left(\frac{a}{4}\right)^3 = \frac{a^3}{64}

The ratio of the squares of the time periods is:

\frac{T_2^2}{T_1^2} = \frac{\left(\frac{a}{4}\right)^3}{a^3} = \frac{1}{64}

Solving for T_2, we have:

T_2 = T_1 \times \frac{1}{8}

Substitute T_1 = 24 hours:

T_2 = 24 \times \frac{1}{8} = 3 hours

Therefore, the new time period of the satellite is 3 hours, which matches the given correct answer.