A planet revolves about the sun in an elliptical orbit of semi-major axis `2xx10^(12) m`. The areal velocity of the planet when it is nearest to the sun is `4.4xx10^(16) m//s`. The least distance between the planet and the sun is `1.8xx10^(12) m//s`. The minimum speed of the planet in `km//s` is `10 K`. determine the value of `K`.

Answer

4

Explanation

Area covered by line joining planet and the sun in time `dt` is `ds=1/2X^(2)d theta`, area velocity `=(dS)/(dt)=1/2X^(2)(d theta)/(dt)=1/2X^(2)omega` where `X=` distance between planet and sun and `omega=` angular speed of planet about sun. from kepler's second law areal velocity of planet is constant. at farthest position
`A=(dS)/(dt)=1/2(2R-r)^(2)=1/2(2R-r)[(2R-r)omega]`
`=1/2(2R-r)v_(B)` or `v_(B)=(2A)/(2R-r)` least speed, Using values `v_(B)=40 km//s `thus, `K=4`