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

Area covered $=2\times \frac{1}{2}ae\times b+\frac{\pi a b}{2}$
$=ab\left(e + \frac{\pi }{2}\right)$
$\pi ab \rightarrow T$
Then $ab\left(e + \frac{\pi }{2}\right) \rightarrow T^{'}$
Areal velocity will be same at both instant,
$\Rightarrow \frac{\pi a b}{T}=\frac{a b \left(e + \frac{\pi }{2}\right)}{T '}$
$\Rightarrow T'=\frac{T}{\pi }\left(e + \frac{\pi }{2}\right)$
$\Rightarrow T'=\frac{365}{\pi }\left(\frac{1}{2} + \frac{\pi }{2}\right)=240.59\simeq 241 days$