Question

According to Kepler's second law of planetary motion, the magnitude of areal velocity \(\left(\frac{\Delta A}{\Delta t}\right)\) of a planet of mass '\(m\)' revolving around the sun is constant and is equal to ('\(p\)' is the linear momentum and '\(L\)' is the angular momentum)

• A. \(\frac{p^{2}}{2m}\)
• B. \(2mL\)
• C. \(\frac{p}{m}\)
• D. \(\frac{L}{2m}\)

Hint:

The area swept per unit time equals half of the angular momentum divided by the mass.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In a small time \(\Delta t\), the radius vector sweeps a thin triangle of area \(\Delta A=\dfrac{1}{2}r\,(v\,\Delta t)\sin\theta\).

Step 2: Calculation
\(\dfrac{\Delta A}{\Delta t}=\dfrac{1}{2}rv\sin\theta\). Angular momentum \(L=mrv\sin\theta\), so \(rv\sin\theta=\dfrac{L}{m}\).

Step 3: Result
\(\dfrac{\Delta A}{\Delta t}=\dfrac{L}{2m}\). It is constant because gravity exerts no torque about the sun, so \(L\) is conserved.

Final Answer:
The areal velocity equals L divided by 2m. \[ \boxed{\text{(D) }\dfrac{L}{2m}} \]