Question

A planet of mass m revolves around the sun of mass M in an elliptical orbit. The minimum and maximum distance of the planet from the sun are $r_1$ & $r_2$ respectively. If the minimum velocity of the planet is $\sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}}$ then it's maximum velocity will be :

• A. $\sqrt{\frac{2GMr_2}{(r_1+r_2)r_1}}$
• B. $g\sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}}$
• C. $\sqrt{\frac{2Gmr_2}{(r_1+r_2)r_1}}$
• D. $\sqrt{\frac{2GM}{r_1+r_2}}$

Answer 1

Answer: (A)

$\sqrt{\frac{2GMr_2}{(r_1+r_2)r_1}}$

Answer 2

The problem involves a planet revolving around the sun in an elliptical orbit. We are given the minimum and maximum distances from the sun ($r_1$ and $r_2$) and the minimum velocity ($v_{min}$). We need to find the maximum velocity ($v_{max}$).

Key Concepts:

1. Conservation of Angular Momentum: For a body in orbit under a central force (like gravity), its angular momentum ($L$) about the center of force is conserved. $L = mvr$, where $m$ is the mass of the orbiting body, $v$ is its speed, and $r$ is its distance from the center of force.

2. Perihelion and Aphelion: In an elliptical orbit, the point closest to the sun is called perihelion, and the point farthest from the sun is called aphelion.

   • At perihelion, the distance is minimum ($r_1$) and the velocity is maximum ($v_{max}$).
   • At aphelion, the distance is maximum ($r_2$) and the velocity is minimum ($v_{min}$).

Solution:

From the conservation of angular momentum, we have: Angular momentum at perihelion = Angular momentum at aphelion $m v_{max} r_1 = m v_{min} r_2$

Since the mass of the planet ($m$) is constant, we can cancel it out: $v_{max} r_1 = v_{min} r_2$

Now, we can express $v_{max}$ in terms of $v_{min}$, $r_1$, and $r_2$: $v_{max} = v_{min} \frac{r_2}{r_1}$

We are given the expression for the minimum velocity: $v_{min} = \sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}}$

Substitute this expression for $v_{min}$ into the equation for $v_{max}$: $v_{max} = \sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}} \times \frac{r_2}{r_1}$

To simplify, bring the term $\frac{r_2}{r_1}$ inside the square root. When brought inside, it becomes $\left(\frac{r_2}{r_1}\right)^2 = \frac{r_2^2}{r_1^2}$: $v_{max} = \sqrt{\frac{2GMr_1}{(r_1+r_2)r_2} \times \frac{r_2^2}{r_1^2}}$

Now, perform the multiplication and cancel out common terms: $v_{max} = \sqrt{\frac{2GM \cdot r_1 \cdot r_2^2}{(r_1+r_2) \cdot r_2 \cdot r_1^2}}$ $v_{max} = \sqrt{\frac{2GM \cdot r_2}{(r_1+r_2) \cdot r_1}}$

Comparing this result with the given options, our derived expression matches option (A).

Therefore, the maximum velocity is $\sqrt{\frac{2GMr_2}{(r_1+r_2)r_1}}$.

Explanation of the solution: The problem is solved using the principle of conservation of angular momentum for a planet orbiting the sun. At the closest point (perihelion, distance $r_1$), the velocity is maximum ($v_{max}$). At the farthest point (aphelion, distance $r_2$), the velocity is minimum ($v_{min}$). By conservation of angular momentum, $m v_{max} r_1 = m v_{min} r_2$, which simplifies to $v_{max} = v_{min} \frac{r_2}{r_1}$. Substituting the given expression for $v_{min}$ and simplifying the algebraic expression leads to the final result.