Question

The minimum and maximum distances of a planet revolving around the Sun are $x_1$ and $x_2$. If the minimum speed of the planet on its trajectory is $v_0$ then its maximum speed will be :

• A. $\frac{v_0 x_2^2}{x_1^2}$
• B. $\frac{v_0 x_1}{x_2}$
• C. $\frac{v_0 x_2}{x_1}$
• D. $\frac{v_0 x_1^2}{x_2^2}$

Hint:

Remember the simple inverse relation: \(v \propto \frac{1}{r}\) for the extremal points of an orbit. If distance is minimum, speed is maximum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Angular momentum of a planet revolving around the Sun is conserved because the gravitational force is a central force.
The planet moves in an elliptical orbit. The minimum speed occurs at the furthest point (aphelion) and the maximum speed occurs at the nearest point (perihelion).
Step 2: Key Formula or Approach:
Conservation of Angular Momentum: \(L = mvr \sin \phi\).
At perihelion and aphelion, the velocity is perpendicular to the position vector (\(\phi = 90^\circ\)), so:
\[ m v_{max} r_{min} = m v_{min} r_{max} \]
Step 3: Detailed Explanation:
Given:
Minimum distance (Perihelion) = \(x_1\).
Maximum distance (Aphelion) = \(x_2\).
Minimum speed (at \(x_2\)) = \(v_0\).
Let the maximum speed (at \(x_1\)) be \(v\).
By Conservation of Angular Momentum:
\[ m v x_1 = m v_0 x_2 \]
\[ v = \frac{v_0 x_2}{x_1} \]
Step 4: Final Answer:
The maximum speed will be \(\frac{v_0 x_2}{x_1}\).