Question
The figure shows elliptical orbit of a planet $$m$$ about the sun $$S.$$ The shaded area $$SCD$$ is twice the shaded area $$SAB.$$ If $${t_1}$$ is the time for the planet to move from $$C$$ to $$D$$ and $${t_2}$$ is the time to move from $$A$$ to $$B,$$ then
The figure shows elliptical orbit of a planet $$m$$ about the sun $$S.$$ The shaded area $$SCD$$ is twice the shaded area $$SAB.$$ If $${t_1}$$ is the time for the planet to move from $$C$$ to $$D$$ and $${t_2}$$ is the time to move from $$A$$ to $$B,$$ then
A.
$${t_1} > {t_2}$$
B.
$${t_1} = 4{t_2}$$
C.
$${t_1} = 2{t_2}$$
D.
$${t_1} = {t_2}$$
Answer :
$${t_1} = 2{t_2}$$
Solution :
Apply Kepler's second law. The line joining the sun to the planet sweeps out equal areas in equal time interval i.e. areal velocity is constant.
$$\eqalign{ & \frac{{dA}}{{dt}} = {\text{constant}} \cr & {\text{or}}\,\,\frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} \cr} $$
where, $${{A_1}} =$$ area under $$SCD$$
$${{A_2}} =$$ area under $$ABS$$
$$\eqalign{ & \Rightarrow {t_1} = \frac{{{A_1}}}{{{A_2}}}{t_2} \cr & {\text{Given,}}\,\,{A_1} = 2{A_2} \cr & \therefore {t_1} = 2{t_2} \cr} $$
Apply Kepler's second law. The line joining the sun to the planet sweeps out equal areas in equal time interval i.e. areal velocity is constant.
$$\eqalign{ & \frac{{dA}}{{dt}} = {\text{constant}} \cr & {\text{or}}\,\,\frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} \cr} $$
where, $${{A_1}} =$$ area under $$SCD$$
$${{A_2}} =$$ area under $$ABS$$
$$\eqalign{ & \Rightarrow {t_1} = \frac{{{A_1}}}{{{A_2}}}{t_2} \cr & {\text{Given,}}\,\,{A_1} = 2{A_2} \cr & \therefore {t_1} = 2{t_2} \cr} $$