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SOLUTION TO EXERCISE 33
Let's imagine an object orbiting around another - a comet around the Sun, for example. At one moment in time, its distance to the Sun is given by r1 and its orbital velocity is given by v1. If it travels for a period of time t, then the distance it travels is given by d1 = v1 * t. That means the area it sweeps out during that time is a triangle, with base length d1 and height r1. The area of this triangle is thus given by A1 = (d1 * r1) / 2 = (v1 * r1 * t) / 2. Now, at a different moment in time, the comet's distance to the sun is now given by r2, and its orbital velocity is now given as v2. If the comet again travels for a period of time t, then by the same reasoning, the area of this triangle is A2 = (d2 * r2) / 2 = (v2 * r2 * t) / 2. Now, we know that the angular momentum of the comet is given by L = mvr. The Law of Conservation of Angular Momentum states that L will stay constant as long as the orbit is stable and not torqued. So: At the first moment in time, L = m * v1 * r1 and At the second moment in time, L = m * v2 * r2 and they're equal. This means that m * v1 * r1 = m * v2 * r2, and thus v1 * r1 = v2 * r2. That, in turn, means that (v1 * r1 * t) / 2 = (v2 * r2 * t) / 2, which means that A1 = A2. So the two swept-out areas are equal! This is Kepler's Second Law, that an orbit sweeps out equal areas in equal times, and we've now proven that it is true as long as the Conservation of Angular Momentum is true. (For an animated demonstration of Kepler's Second Law, see http://physics.bu.edu/~duffy/semester1/c17_kepler2.html) |