Question

The maximum and minimum distances of a planet from sun are $r_1$ and $r_2$ respectively. If minimum velocity is $V_1$ and minimum velocity is $V_2$, then the ration $V_1 : V_2$ is:

Hint:

Remember the see-saw relationship in elliptical orbits due to angular momentum conservation: Largest radius goes with smallest velocity, and smallest radius goes with largest velocity ($v \propto 1/r$).

Answer 1

Step 1: Understanding the Concept:
According to Kepler's Second Law (Law of Areas), a line joining a planet and the sun sweeps out equal areas during equal intervals of time. A direct consequence of this is the conservation of angular momentum for the planet as it orbits the sun.

Step 2: Key Formula or Approach:
The angular momentum $L$ of a planet of mass $m$ moving with velocity $v$ at a distance $r$ from the sun is given by $L = mvr\sin(\theta)$.
At the maximum distance (aphelion) and minimum distance (perihelion), the velocity vector is strictly perpendicular to the position vector, so $\sin(90^\circ) = 1$.
Therefore, conservation of angular momentum simplifies to:
\[ m v_{max} r_{min} = m v_{min} r_{max} \]

Step 3: Detailed Explanation:
The problem text contains an obvious typographical error: "minimum velocity is $V_1$ and minimum velocity is $V_2$".
Based on the physical laws, velocity is inversely proportional to distance. Therefore:
- At maximum distance $r_1$, the planet moves slowest. Let this minimum velocity be $V_1$.
- At minimum distance $r_2$, the planet moves fastest. Let this maximum velocity be $V_2$.
Applying the conservation of angular momentum:
\[ m \cdot V_1 \cdot r_1 = m \cdot V_2 \cdot r_2 \]
Cancel the mass $m$:
\[ V_1 \cdot r_1 = V_2 \cdot r_2 \]
We need to find the ratio $V_1 : V_2$, which is $V_1 / V_2$:
\[ \frac{V_1}{V_2} = \frac{r_2}{r_1} \]

Step 4: Final Answer:
The ratio $V_1 : V_2$ is $r_2 : r_1$.