By Neb Duric
Astronomy describes the mechanics of the universe within the extra easy language of physics, in a different way referred to as "astrophysics". wisdom of black holes, quasars and extrasolar planets calls for realizing of the physics underlying astrophysics. This publication clarifies the basic ideas of the sector in addition to the real astronomical phenomena it describes. Readers will achieve a better appreciation of the relationship among physics and astronomy.
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For a stationary system (one that is relaxed) the moment of inertia does not change (on average) with time. Thus m i r¨i ri = −2E k . i (66) 36 Galaxy dynamics Repeating the same operation on the right-hand side of (64) we get − mi m j G ij =− (ri − r j ) · ri |ri − r j |3 mi m j G ij 1 =− 2 ij (r j − ri ) · r j |ri − r j |3 mi m j G = EG . |ri − r j | (67) Equating (66) and (67) yields the relation that defines the theorem 1 1 E k = − E G = |E G |. 2 2 ⇒ the Virial theorem (68) According to the Virial theorem the total energy of a stationary system is E T = Ek + E G = 1 EG .
As we will see this affects the ability to measure the masses of the stars. 6 The mass function From page 3 and the discussion of the center of gravity (center of mass), we have a1 a1 m2 = = a1 + a 2 a0 m1 + m2 so that (a1 sin i)3 = a0 m 2 sin i m1 + m2 3 = m 32 a03 sin3 i . (m 1 + m 2 )3 (40) Re-examining Kepler’s Third Law, m1 + m2 = 42 a03 . 05 Fig. 8 The apparent orbit of 70 Tauri on the sky. P. and Latham, D. , 1997, ApJ, 479, pp. 268–278). Combining (40) and (41), yields m 32 sin3 i 42 (a1 sin i)3 = .
The kinetic energy of the expansion can be written as Ek = ρv 2 3 4ρ H 2 dr= 2 2 R r 4 dr = 0 2 ρ H 2 R5. 5 The gravitational potential energy can be expressed as E G = −G M(r )ρ 3 4 3 d r = −G r ρ r 3 (4)2 2 ⇒ EG = ρ G R5. 15 We see that MR ∝ ρ R3 M R 1/3 R∝ ρ 1/3 5/3 EG ∝ ρ M → 0 as ρ → 0. 1 The expansion of the Universe 41 r Fig. 1 Cosmic expansion according to Newtonian mechanics. The expansion is assumed to be uniform and isotropic. The total energy in the expansion can now be obtained E T = Ek 1 + EG Ek = Ek 1 − 8ρG 3H 2 ≡ Ek 1 − ρ ρc ≡ E k (1 − Ω) where ρc = 3H02 = 10−29 g cm−3 8G = 6 × 10−6 m p cm−3 H0 70 km s−1 Mpc−1 H0 70 km s−1 Mpc−1 2 2 , where H0 is known as the Hubble constant and represents the present rate of expansion.