A plenet moving along an elliptical orbit is closest to the sun at a distance `r_(1)` and farthest away at a distance of `r_(2)`. If `v_(1)` and `v_(2)` are the linear velocities at these points respectively, then the ratio `(v_(1))/(v_(2))` is

Options

A. `r_(2)//r_(1)`
B. `(r_(2)//r_(1))^(2)`
C. `r_(1)//r_(2)`
D. `(r_(1)//r_(2))^(2)`

Answer

To solve the problem, we need to find the ratio of the linear velocities \( \frac{v_1}{v_2} \) of a planet moving in an elliptical orbit at its closest distance \( r_1 \) and farthest distance \( r_2 \) from the Sun.

### Step-by-Step Solution:

1. **Understanding Angular Momentum Conservation**:
The angular momentum \( L \) of a planet moving in an orbit is conserved because the net external torque acting on the system is zero. This is due to the gravitational force acting along the line connecting the planet and the Sun, which does not create any torque.

\[ L = m \cdot r \cdot v \]

where \( m \) is the mass of the planet, \( r \) is the distance from the Sun, and \( v \) is the linear velocity.

2. **Setting Up the Angular Momentum Equation**:
For the two positions of the planet (closest and farthest from the Sun), we can write the angular momentum as follows:

- At the closest point (distance \( r_1 \)):
\[ L_1 = m \cdot r_1 \cdot v_1 \]

- At the farthest point (distance \( r_2 \)):
\[ L_2 = m \cdot r_2 \cdot v_2 \]

3. **Equating Angular Momenta**:
Since angular momentum is conserved, we have:
\[ L_1 = L_2 \]
Therefore:
\[ m \cdot r_1 \cdot v_1 = m \cdot r_2 \cdot v_2 \]

4. **Canceling Mass**:
The mass \( m \) of the planet cancels out from both sides of the equation:
\[ r_1 \cdot v_1 = r_2 \cdot v_2 \]

5. **Finding the Ratio**:
Rearranging the equation gives us:
\[ \frac{v_1}{v_2} = \frac{r_2}{r_1} \]

### Final Result:
Thus, the ratio of the linear velocities at the closest and farthest points from the Sun is:
\[ \frac{v_1}{v_2} = \frac{r_2}{r_1} \]