According to Kepler's second law of planetary motion, the magnitude of areal velocity ( Δ t Δ A ) of a planet of mass ' m ' revolving around the sun is constant and is equal to (&#039

Question

According to Kepler's second law of planetary motion, the magnitude of areal velocity ( Δ t Δ A ​ ) 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)

TestHub · Accepted Answer

The area swept by the radius vector in a small time interval $Δ t$ is given by $Δ A = \frac{1}{2} ∣ r \times Δ r ∣$ .

Dividing by $Δ t$ , the areal velocity is $\frac{Δ A}{Δ t} = \frac{1}{2} r \times \frac{Δ r}{Δ t} = \frac{1}{2} ∣ r \times v ∣$ .

The angular momentum of the planet is given by $L = m (r \times v)$ , which implies $∣ r \times v ∣ = \frac{L}{m}$ .

Substituting this into the expression for areal velocity, we get $\frac{Δ A}{Δ t} = \frac{L}{2 m}$ .

Answer: $\frac{L}{2 m}$

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