Question

Earth revolves around Sun with time period of $T=365$ days. In the figure shown below, find the time taken (in days approximate to nearest integer) by the Earth to move from $A$ to $B$ in the path $ACB$ if the eccentricity of ellipse is $0.5$ .

Answer 1

Answer: 241

Area covered $=2\times \frac{1}{2}ae\times b+\frac{\pi ab}{2}$
$=ab\left(e+\frac{\pi }{2}\right)$
$\pi ab\to T$
Then $ab\left(e+\frac{\pi }{2}\right)\to T^{{}^{\prime }}$
Areal velocity will be same at both instant,
$\Rightarrow \frac{\pi ab}{T}=\frac{ab\left(e+\frac{\pi }{2}\right)}{T^{\prime }}$
$\Rightarrow T^{\prime }=\frac{T}{\pi }\left(e+\frac{\pi }{2}\right)$
$\Rightarrow T^{\prime }=\frac{365}{\pi }\left(\frac{1}{2}+\frac{\pi }{2}\right)=240.59\simeq 241days$