Physics - Gravitation Question with Solution | TestHub

Let the mass of the Sun be $M$ and the mass of the planet be $m$.

According to the law of conservation of mechanical energy for a planet in an elliptical orbit, the total energy $E$ remains constant.

$E=K+U=\text{constant}$

$K-\frac{GMm}{r}=-\frac{GMm}{2a}$

$K=\frac{GMm}{r}-\frac{GMm}{2a}$

From this relation, it is clear that the kinetic energy $K$ is inversely related to the distance $r$ from the Sun. As $r$ increases, $K$ decreases.

From the given figure, A is the perihelion (closest point) and C is the aphelion (farthest point).

Let $a$ be the semi-major axis and $e$ be the eccentricity of the ellipse.

The distances of points A, B, and C from the Sun (S) are:

$r_A=a(1-e)$

$r_C=a(1+e)$

Since SB is perpendicular to the major axis, B is at the end of the semi-latus rectum passing through the focus S. Thus, $r_B=a(1-e^2)$.

Since $0<e<1$, we have:

$1-e<1-e^2<1+e$

Therefore, $r_A<r_B<r_C$.

Since kinetic energy is maximum when the distance is minimum and vice-versa, we get:

$K_A>K_B>K_C$

Answer: $K_A>K_B>K_C$