Question 128: Gravitation

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.

Accepted Answer

$$\because {T^2} \propto {R^3}$$
$$	herefore$$ $${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$$
$${{{{24}^2}} \over {T_2^2}} = {{R_1^3} \over {{{\left( {{{{R_1}} \over 4}} ight)}^3}}}$$
$${{{{24}^2}} \over {T_2^2}} = {4^3}$$
$${T_2} = {{24} \over {{2^3}}} = 3$$ hours

Answer

12 hours

Answer

3 hours

Answer

6 hours

Answer

4 hours

Gravitation

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.

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