A partical of mass m moves along a straight line with constant speed in the x - direction , a distance b from the x - axis (Fig. 4.17). Show that Kelper's second law is satisfied by demonstrating that the two shaded triangles in the figure have the same area when $t_{4} - t_{3} = t_{2} - t_{1} .$

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Solution

Area of triangle between $t_{1}$ and $t_{2}$ is:
$A_{1} = \frac{1}{2} (v t_{2} - v t_{1}) \times b = \frac{1}{2} v b (t_{2} - t_{1})$

Area of triangle between $t_{3}$ and $t_{4}$ is:
$A_{2} = \frac{1}{2} (v t_{4} - v t_{3}) \times b = \frac{1}{2} v b (t_{4} - t_{3})$

Now both the areas will be equal if: $t_{4} - t_{3} = t_{2} - t_{1}$

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