A planet revolves about the sun in elliptical orbit. The arial velocity `((dA)/(dt))` of the planet is `4.0xx10^(16) m^(2)//s`. The least distance between planet and the sun is `2xx10^(12) m`. Then the maximum speed of the planet in `km//s` is -

Options

A. 10
B. 20
C. 40
D. 80

Answer

To find the maximum speed of the planet in its elliptical orbit around the sun, we will use the information provided about the aerial velocity and the least distance from the sun.

### Step-by-Step Solution:

1. **Understanding Aerial Velocity**:
The aerial velocity \( \frac{dA}{dt} \) is given as \( 4.0 \times 10^{16} \, \text{m}^2/\text{s} \). According to Kepler's second law, this value remains constant for a planet in an elliptical orbit.

2. **Using the Formula for Aerial Velocity**:
The aerial velocity can be expressed as:
\[ \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} \]
where \( r \) is the distance from the sun (which is the least distance in this case) and \( \frac{d\theta}{dt} \) is the angular velocity \( \omega \).

3. **Substituting Known Values**:
The least distance \( r \) is given as \( 2.0 \times 10^{12} \, \text{m} \). We substitute the values into the aerial velocity formula:
\[ 4.0 \times 10^{16} = \frac{1}{2} (2.0 \times 10^{12})^2 \omega \]

4. **Calculating \( \omega \)**:
First, calculate \( (2.0 \times 10^{12})^2 \):
\[ (2.0 \times 10^{12})^2 = 4.0 \times 10^{24} \, \text{m}^2 \]
Now substituting back into the equation:
\[ 4.0 \times 10^{16} = \frac{1}{2} (4.0 \times 10^{24}) \omega \]
Simplifying gives:
\[ 4.0 \times 10^{16} = 2.0 \times 10^{24} \omega \]
Now, solve for \( \omega \):
\[ \omega = \frac{4.0 \times 10^{16}}{2.0 \times 10^{24}} = 2.0 \times 10^{-8} \, \text{rad/s} \]

5. **Finding Maximum Speed**:
The maximum speed \( v \) of the planet is given by:
\[ v = r \omega \]
Substituting the values:
\[ v = (2.0 \times 10^{12}) (2.0 \times 10^{-8}) = 4.0 \times 10^{4} \, \text{m/s} \]

6. **Converting to km/s**:
To convert \( v \) from m/s to km/s:
\[ v = \frac{4.0 \times 10^{4}}{1000} = 40 \, \text{km/s} \]

### Final Answer:
The maximum speed of the planet is \( 40 \, \text{km/s} \).

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