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# 9.2: Kepler's Second Law from Conservation of Angular Momentum

• • Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria $$\text{FIGURE IX.1}$$

In figure $$\text{IX.1}$$, a particle of mass $$m$$ is moving in some sort of trajectory in which the only force on it is directed towards or away from the point $$\text{O}$$. At some time, its polar coordinates are $$(r, θ)$$. At a time $$δt$$ later these coordinates have increased by $$δr$$ and $$δθ$$.

Using the formula one half base times height for the area of a triangle, we see that the area swept out by the radius vector is approximately

$δA = \frac{1}{2} r^2 δ θ+ \frac{1}{2} r δ θδ r. \label{9.3.1} \tag{9.3.1}$

On dividing both sides by $$δt$$ and taking the limit as $$δt → 0$$, we see that the rate at which the radius vector sweeps out area is

$\dot{A} = \frac{1}{2} r^2 \dot{θ}. \label{9.3.2} \tag{9.3.2}$

But the angular momentum is , $$mr^2\dot{θ}$$ and since this is constant, the areal speed is also constant. The areal speed, in fact, is half the angular momentum per unit mass.