Astroparticle Physics

364 17 SolutionsWith the ansatzR = At p ,Ṙ = pAt p−1one getsṘ 2R 2 = p2 A 2 t 2p t −2A 2 t 2p = p2t 2 = 8π ( ) 33 Gϱ R00 = 8π R 3 Gϱ 0R0 3 A−3 t −3p .Comparing the t dependence on the right- and left-hand sides givesp = 2 3⇒ R = At 2/3 .In this procedure the constant of proportionality A is automatically fixed asA = 3√ 6πGϱ 0 R 0 .3. For the early universe one can approximate the Friedmann equation byṘ 2R 2 = 8π 3Gϱ ; with ϱ =π2Since ṘR = H and G = 1m 2 one hasPl√8πH =3 g ∗ T 2.90 m Pl30 g ∗T 4 ⇒ Ṙ2R 2 = 8π 3 Gπ 230 g ∗T 4 = 8π 390 Gg ∗T 4 .4. [G] = m 3 s −2 kg −1 , [c] = ms −1 , [¯h] = Js= kg m 2 s −1 .[ ]¯hGTryc 3 = kg m2 s −1 m 3 s −2 kg −1m 3 s −3 = m 2 .√¯hGTherefore, has the dimension of a length, and this is the Planck length.c 35. The escape velocity from a massive object can be worked out from12 mv2 = G mMR ,where m is the mass of the escaping particle, and M and R are the mass and radius of themassive object. The mass of the escaping object does not enter into the escape velocity.If the escape velocity is equal to the velocity of light, even light cannot escape from theobject. If v is replaced by c this leads toc 2 = 2 GMRorR = 2 GMc 2 ,