Question

The planet mercury is revolving in an elliptical orbit around the sun as shown in the figure. The kinetic energy of mercury will be greater at

• A. $A$
• B. $B$
• C. $C$
• D. $D$

Answer 1

The correct option is A $A$

Kinetic energy ($K.E.$) of the planet
$K.E.=\frac{1}{2}m{v}^2$

using Kepler's secound law :
$mvr=const.$
so, $v\propto \frac{1}{r}$

From the above expression, it can be observed that when a planet having less $r$ (distance), velocity ($v$) of the planet will be maximum and so as the kinetic energy.
$v\propto \frac{1}{r}\propto K.E.$

Here, A is having the least distance. Hence at point A, the kinetic energy will be maximum.