Question

Consider a planet moving around a star in an elliptical orbit with period $T$. The area of the elliptical orbit is proportional to,

• A. $T^{\frac{4}{3}}$
• B. $T$
• C. $T^{\frac{2}{3}}$
• D. $T^{\frac{1}{2}}$

Answer 1

Answer: (A)

Area of ellipse
$A=\pi r_{1} r_{2} $
$\because r_{1}=a-a e=a(1-e)$
and $ r_{2}=a+a e=a(1+e) $
$A =\pi\{a(1-e)\}\{a(1+e)\}$
$=\pi a^{2}(1-e)(1+e)$
$=\pi a^{2}\left(1^{2}-e^{2}\right) $
$\because e^{2} < < a^{2} $ then
$A=\pi a^{2}$
So, $a \propto A^{\sqrt{2}}$
According to Keplar's III law
$T^{2} \propto a^{3} $
$T^{2} \propto\left[(A)^{1 / 2}\right]^{3} $
$T^{2} \propto A^{3 / 2} $
$A \propto\left(T^{2}\right)^{2 / 3} $
$A \propto T^{4 / 3}$