JEE MAIN - Physics (2003 - No. 31)
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
$$10$$ hours
$$80$$ hours
$$40$$ hours
$$20$$ hours
Explanation
According to kepler's law,
$${T^2} \propto {R^3}$$
$$\therefore$$ $${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$$
$$ \Rightarrow $$ $${T_2} = {T_1}{\left( {{{{R_2}} \over {{R_1}}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = 5 \times {\left[ {{{4R} \over R}} \right]^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
$$ = 5 \times {2^3} = 40\,\,$$ hour
$${T^2} \propto {R^3}$$
$$\therefore$$ $${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$$
$$ \Rightarrow $$ $${T_2} = {T_1}{\left( {{{{R_2}} \over {{R_1}}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = 5 \times {\left[ {{{4R} \over R}} \right]^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
$$ = 5 \times {2^3} = 40\,\,$$ hour
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