Question

A planet revolves around the Sun in an elliptical orbit. The areal velocity of the planet is \(4 \times 10^{16}\,\text{m}^2\text{s}^{-1}\). If the maximum distance between the planet and the Sun is \(4 \times 10^{12}\,\text{m}\), then the minimum speed of the planet is:

• A. \( 1 \times 10^4\,\text{m s}^{-1} \)
• B. \( 2 \times 10^4\,\text{m s}^{-1} \)
• C. \( 4 \times 10^4\,\text{m s}^{-1} \)
• D. \( 8 \times 10^4\,\text{m s}^{-1} \)

Hint:

Shortcut: \[ v = \frac{2}{r}\frac{dA}{dt} \]

Answer 1

The Correct Option is B

Step 1: Recall Kepler's area rule.
A planet sweeps out equal areas in equal times. So the areal velocity, the area swept per second, stays the same all along the orbit.

Step 2: Write the areal velocity.
The rate of sweeping area is \[ \frac{dA}{dt} = \frac{1}{2}\,r\,v, \] where $r$ is the distance from the Sun and $v$ is the speed at that point.

Step 3: Find where the speed is smallest.
Because the product $r\,v$ is fixed, the speed is least when the distance is greatest. So the minimum speed happens at the maximum distance.

Step 4: Put in the known values.
Use the maximum distance $r_{\max} = 4 \times 10^{12}$ m and the given areal velocity $4 \times 10^{16}$: \[ 4 \times 10^{16} = \frac{1}{2} \times (4 \times 10^{12}) \times v_{\min}. \]

Step 5: Solve for the minimum speed.
The right side is $2 \times 10^{12}\,v_{\min}$, so \[ v_{\min} = \frac{4 \times 10^{16}}{2 \times 10^{12}} = 2 \times 10^{4}. \]

Step 6: State the answer.
The slowest speed of the planet is found at the farthest point. \[ \boxed{2 \times 10^{4} \ \text{m s}^{-1}} \]