2. Gravity from a Rod. Now, suppose mass M is uniformly distributed along a vertical line segment joining two points (0, -L) and (0, L ) So, a continuum of mass is distributed along this vertical line segment and the linear mass density is M . Suppose a test particle of mass m is located on the x-axis at the point (p, 0) . See the figure below. dM - M. dy 2 L dy HA [Figure 1. Various parts of the stick (yellow ) exert gravity to the test particle at C. The total gravity applied to the test particle is the sum of all these forces. Due to the symmetry of the configuration, the vertical components of the forces will B I cancel pairwise. There will remain only the sum of all the horizontal components. 2L P C The mass is distributed in continuum, so the summation of these forces is done by a definite integral. ] Show that the total gravity experienced by the test particle is F = GMm PVP2 + 12 [Hint: Consider a very short segment of the stick located near (0, y). The blue piece in the figure above. The length of this piece is dy. the horizontal component of gravity exerted by this part of the stick to the test particle is the cosine of the angle of the triangle shown in the figure times the magnitude of the gravity (pointing toward the blue section.) Thus, you can easily express the infinitesimal parts of the horizontal components of the gravity contributed by the tiny segments. dF = Gm.dM cos(ZACB)[Hint: Consider a very short segment of the stick located near (0, y). The blue piece in the figure above. The length of this piece is dy. the horizontal component of gravity exerted by this part of the stick to the test particle is the cosine of the angle of the triangle shown in the figure times the magnitude of the gravity (pointing toward the blue section.) Thus, you can easily express the infinitesimal parts of the horizontal components of the gravity contributed by the tiny segments. dF _ Gm.dM - cos(ZACB) where both |AC| and the "cos(ZACB) " are functions of y which you can find easily. The differential dM is the mass of the blue piece with length dy . By integrating dF from y = - L to y = L , you will have taken the "sum" of all the gravitational forces exerted on C caused by all the tiny vertical segments of the stick. ] Suggestion for this part: 1. Show detailed steps on how the integral is constructed from Newton's Law of Gravity. 2. Show how you performed the integral using a certain substitution that we learned in Chpater 3. Do not use the "Table of Integral" in order to receive full credit. 3. Taylor Series Expansion. [ Continuation of #2] Find the Taylor series expansion of F near = 0 ( i.e. a power series of .) You must display the first three non-zero terms. Then, use this Taylor series to show that F ~ GMm [ Linear Approximation] p2 when I