Question

A planet revolves about the sun in elliptical orbit. The arial velocity $\left(\frac{dA}{dt}\right)$ of the planet is $4.0\times {10}^{16}{m}^2/s$. The least distance between planet and the sun is $2\times {10}^{12}$m. Then the maximum speed of the planet in km/s is

• A. 10
• B. 20
• C. 40
• D. none of these

Answer 1

Answer: (C)

40

We have: $\frac{dA}{dt}=\frac{1}{2}{r}^2\frac{d\theta }{dt}$
$4\times {10}^{16}=\frac{1}{2}(2\times {10}^{12}{)}^2\frac{d\theta }{dt}$
$\omega =\frac{d\theta }{dt}=2\times {10}^{-8}$

Using, $v=r\omega$

We get: $v=2\times {10}^{12}\times 2\times {10}^{-8}=40km/s$