as the planet goes around. Hence, ∆ A / ∆ t is a constant according to the last equation. This is the law of areas. Gravitation is a central force and hence the law of areas follows.
Example . Let the speed of the planet at the perihelion P in Fig. . (a) be v P and the Sun-planet distance SP be r P .
Relate { r P , v P } to the corresponding quantities at the aphelion { r A, v A }. Will the planet take equal times to traverse BAC and CPB ? Answer The magnitude of the angular momentum at P is L p = m p r p v p , since inspection tells us that r p and v p are mutually perpendicular. Similarly, L A = m p r A v A .
From angular momentum conservation m p r p v p = m p r A v A or p A A p Since r A > r p , v p > v A . The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in Fig. . .
From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse BAC than CPB . . UNIVERSAL LAW OF GRAVITATION Legend has it that observing an apple falling from a tree, Newton was inspired to arrive at an universal law of gravitation that led to an explanation of terrestrial gravitation as well as of Kepler’s laws.
Newton’s reasoning was that the moon revolving in an orbit of radius R m was subject to a centripetal acceleration due to earth’s gravity of magnitude ( . ) where V is the speed of the moon related to the time period T by the relation . The time period T is about . days and R m was already known then to be about .
× m. If we substitute these numbers in Eq. ( . ), we get a value of a m much smaller than the value of acceleration due to gravity g on the surface of the earth, arising also due to earth’s gravitational attraction.
This clearly shows that the force due to earth’s gravity decreases with distance. If one assumes that the gravitational force due to the earth decreases in proportion to the inverse square of the distance from the centre of the earth, we will have a m α R − ; g α R − and we get ( . )