Question 226: Gravitation

Question

The time period of satellite of earth is $$5$$ hours. If the separation between the earth and the satellite is increased to $$4$$ times the previous value, the new time period will become

Accepted Answer

According to kepler's law,
$${T^2} \propto {R^3}$$
$$	herefore$$ $${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$$
$$ \Rightarrow $$ $${T_2} = {T_1}{\left( {{{{R_2}} \over {{R_1}}}} ight)^{{aise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = 5 	imes {\left[ {{{4R} \over R}} ight]^{{aise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
$$ = 5 	imes {2^3} = 40\,\,$$ hour

Answer

$$10$$ hours

Answer

$$80$$ hours

Answer

$$40$$ hours

Answer

$$20$$ hours

Gravitation

The time period of satellite of earth is $5 hours. If the separation between the earth and the satellite is increased to 4$ times the previous value, the new time period will become

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