give me proof of 2nd law of Kepler
Answers
Newton visualized the motion of an object acted on by a gravitational force as a succession of small kicks or impulses which in the limit become a continuously applied influence. Newton imagined an object travelling along part of an orbit AB which then receives an impulse directed towards the point S. As a result, it then travels along the line BC instead of Bc. Similar impulses carry it to D, E and F. Newton visalized the displacement BC as being, in effect, the combination of the displacement Bc, equal to AB, that the object would have undergone if it had continued for an equal length of time with its original velocity, together with the displacement cC parallel to the line BS along which the impulse was applied. This at once yields Kepler's second law by a simple argument: The triangles SAB and SBc are equal, having equal bases (AB and Bc) and the same altitude. The triangles SBc and SBC are equal, having a common base (SB) and lying between the same parallels. Hence triangle SAB = triangle SBC.
A modern Newtonian derivation of Kepler's second law requires the concept of an orbiting body's angular momentum
L = r X p = m (r X v)
where m is the body's mass, r is its position vector and p its linear momentum (= mv, where v is its velocity). Note that for the first time in this course we distinguish between vector quantities and scalar quantities by writing vector quantities in a bold face. The vector cross product (denoted by X) is an operation that yields the product of the perpendicular components of two vectors; hence if r and p are parallel, then r X p = 0. Angular momentum is a vector quantity L with the units kgm2s-1. Differentiating L, we have
dL/dt = d(r X p)/dt = v X p + r X (dp/dt) = r X F
since v is parallel to p and dp/dt is the definition of force according to Newton's second law. We call dL/dt the torque (with units kgm2s-2) and see that when F and r are co-linear, due to a central force such as gravitation, the torque vanishes. Hence L is constant in time and so angular momentum is conserved for all central forces. The conservation of angular momentum is a very powerful tool in celestial mechanics and can be used to derive Kepler's second law as follows.
Deduction of Kepler's 2nd law is given in the pic below ...
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