Question

A planet revolves about the sun in an elliptical orbit of semi-major axis $2\times 10^{12}m$ . The areal velocity of the planet when it is nearest to the sun is $4.4\times 10^{16}{m}^2{s}^{-1}$ . If the least distance between the planet and the sun is $1.8\times 10^{12}m$ , then the minimum speed of the planet is

• A. $v_{min}=40km{s}^{-1}$
• B. $v_{min}=30km{s}^{-1}$
• C. $v_{min}=10km{s}^{-1}$
• D. $v_{min}=20km{s}^{-1}$

Answer 1

Answer: (A)

The areal velocity of a planet around the sun is
$\frac{dA}{dt}=\frac{L}{2m}$ , where $L$ is the angular momentum of the planet around the sun and $m$ is its mass.
$4.4\times 10^{16}=\frac{m{v}_{min}{r}_{max}}{2m}$
$2\times 4.4\times 10^{16}=v_{\min }\left(4\times 10^{12}-1.8\times 10^{12}\right)$
$v_{min}=40km{s}^{-1}$