By Thanu Padmanabhan

ISBN-10: 9812566384

ISBN-13: 9789812566386

ISBN-10: 9812566872

ISBN-13: 9789812566874

This precise e-book presents a transparent and lucid description of a number of points of astrophysics and cosmology in a language comprehensible to a physicist or newbie in astrophysics. It provides the main issues in all branches of astrophysics and cosmology in an easy and concise language. The emphasis is on at the moment energetic learn parts and interesting new frontiers instead of on extra pedantic issues. Many advanced effects are brought with easy, novel derivations which improve the conceptual knowing of the topic. The booklet additionally includes over 100 routines so as to support scholars of their self learn. Undergraduate and graduate scholars in physics and astrophysics in addition to all physicists who're drawn to acquiring a brief snatch of astrophysical suggestions will locate this ebook worthwhile.

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**Sample text**

One of the simplest situations arises in the case of spacetime outside a spherically symmetric distribution of matter with total mass M. The metric for this case can be obtained in a heuristic manner, which we will now describe. Let us consider a point P a t a distance r from the origin where the gravitational potential is 4>N = —GM/r according to Newtonian theory. If we consider a small box around P which is freely falling towards the origin, then the metric in the coordinates used by a freely falling observer in the box will be just that of special relativity: ds2 = c2dt2n - dr?

23) An Invitation to Astrophysics 14 gives, ft x J = K where K is the external torque. 5 degrees to the normal to the plane containing the ring (see Fig. 2). A perfectly spherical earth, of course, will feel no torque from the ring; but since the earth has bulge around equator with equatorial and polar radii differing by 5R/R « 1/300, there will be a torque, as if earth is a small dumbbell with masses SM « M{8R/R) located at the ends of a rod of length 2R. This net force F is due to the difference in the gravitational force of the sun (treated as a ring around earth) on the two masses of the dumbbell; so F « {GMQSM/d3Q)(2Rcos9) and the corresponding torque is K « F(2Rsm6).

Next, taking the dot product of Eq. 18) with r, dividing by r 2 and re-arranging the terms gives G(mi+m 2 ) Gmi / d3 an Gmi ( ai\ M ' ^ +^O-T)- ™ Chapter 1: Gravitation 11 Since mia\ = m^a^, those two terms cancel out giving the result d = a. That is, we conclude | r - r 1 | = | r - r 2 | = a. }. This result is independent of the ratio (mi/m2). It can be verified that while Iq, L2, L3 are unstable equilibrium points, L4, L5 are indeed stable equilibrium points provided the mass ratio is less than about (1/25).

### An Invitation to Astrophysics by Thanu Padmanabhan

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