A planet revolves about the sun in elliptical orbit of semimajor axis $2 imes 10^{12} \mathrm{~m}$

Question

A planet revolves about the sun in elliptical orbit of semimajor axis $2 	imes 10^{12} \mathrm{~m}$. The areal velocity of the planet when it is nearest to the sun is $4.4 	imes 10^{16} \mathrm{~m}^2 / \mathrm{s}$. The least distance between planet and the sun is $1.8 	imes 10^{12} \mathrm{~m}$. The minimum speed of the planet is $(5 n) \mathrm{km} / \mathrm{s}$. Find ' $n$ '.

Professor PC Thomas Classes · Answer

Area covered by line joining planet and sun in time dt is $\boldsymbol{d} \boldsymbol{S}=\frac{1}{2} \boldsymbol{x}^\iota \boldsymbol{d} \boldsymbol{	heta}$; Areal velocity $=\mathrm{dS} / \mathrm{dt}=\frac{1}{2} \boldsymbol{x}^\iota \boldsymbol{d} \boldsymbol{	heta} / \boldsymbol{d} \boldsymbol{t}=\frac{1}{2} \boldsymbol{x}^\iota \omega$ where $\mathrm{x}=$ distance between planet and sun and $\omega=$ angular speed of planet about sun. From Keplers second law Areal velocity of planet is constant.
At farthest position
or (least speed). (Using values)
$$
A=d S / d t=\frac{1}{2}(2 R-r)^2 \omega=\frac{1}{2}(2 R-r)[(2 R-r) \omega]=\frac{1}{2}(2 R-r) V_B
$$or $V_B=\frac{2 A}{2 R-r}$ (least speed). (Using values)
$$
\mathrm{V}_{\mathrm{B}}=40 \mathrm{~km} / \mathrm{s}
$$