A geostationary satellite is orbiting the earth at a height of 7R above the surface of the earth, R being the radius of the earth. The time period of another satellite at a height... A geostationary satellite is orbiting the earth at a height of 7R above the surface of the earth, R being the radius of the earth. The time period of another satellite at a height of 3R from the surface of the earth is: Read more

Steps to Solve

1. Understanding the Problem The problem involves two satellites orbiting the Earth, one at a height of ( 7R ) and the other at ( 3R ) above the Earth's surface (where ( R ) is the Earth's radius). We will use Kepler's Third Law to find the time period of the satellite at ( 3R ).

2. Kepler's Third Law Application Kepler’s Third Law states that the square of the period ( T ) of a satellite's orbit is directly proportional to the cube of the semi-major axis ( a ) of its orbit. Mathematically, this can be expressed as: $$ T^2 \propto a^3 $$

For our satellites, the semi-major axis ( a ) is given by the radius of the Earth plus the height of the satellite:

• For the first satellite at ( 7R ): $$ a_1 = R + 7R = 8R $$

• For the second satellite at ( 3R ): $$ a_2 = R + 3R = 4R $$

3. Calculating Periods From Kepler's Law, we can set up the ratios: $$ \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} $$

Substituting our values: $$ \frac{T_1^2}{T_2^2} = \frac{(8R)^3}{(4R)^3} $$

This simplifies to: $$ \frac{T_1^2}{T_2^2} = \frac{512R^3}{64R^3} = 8 $$

From here, taking the square root gives: $$ \frac{T_1}{T_2} = \sqrt{8} = 2\sqrt{2} $$

4. Finding the Time Period for Satellite at 3R Let’s assume the time period for the first satellite, ( T_1 ), is known (a geostationary satellite has a period of approximately 24 hours). Therefore: $$ T_2 = \frac{T_1}{2\sqrt{2}} $$

Calculating for ( T_1 = 24 ) hours: $$ T_2 = \frac{24}{2\sqrt{2}} = \frac{12\sqrt{2}}{2} = 12\sqrt{2} \text{ hours} $$

Given ( 12\sqrt{2} ) is approximately ( 16.97 ) hours, we need the closest match from the options given.