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SOLUTIONS TO EXERCISE 15
(a) We can adapt Isaac Newton's formulation of the expansion of the universe for this class: the universe will expand forever if any object in that universe is expanding at a velocity at or above the escape velocity of all the matter in the universe holding it back. The formula for escape velocity is something we've used often: v = sqrt(2GM/r) In this case, M = (4/3) * pi * r^3 * (rho), where r is the radius of the universe and (rho) is the density of the universe at which v equals the escape velocity -- known as the "critical density." Furthermore, v = Hr as explained by the Hubble law of universal expansion, where H is the Hubble constant. Thus, the equation can be rewritten 2 * G * (4/3) * pi * r^3 * (rho) Hr = sqrt(--------------------------------). Square both sides, and the r "r" terms cancel completely! So we don't even need to know how big the universe is to compute the critical density. We get H^2 = 2 * G * (4/3) * pi * (rho) Which, when rearranged, gives (rho) = (3 * H^2)/(8 * pi * G). Now, if H = 70 km/s/Mpc, we can rewrite H in units of "per second" like this: (70 km/Mpc)/s H = ------------------ = 2.26 * 10^-18 s^-1 3.1 * 10^19 km/Mpc So (rho) = 3 * (2.26 * 10^-18)^2 / (8 * 3.14 * 6.67 * 10^-11) = 9.1 * 10^-27 kg/m^3 or, in hydrogen atoms, (rho) = (9.1 * 10^-27)/(1.67 * 10^-27) = 5.4 h-atoms/m^3. (c) If the average mass density of the universe is about 2 hydrogen atoms per cubic meter, then this value is less than half the critical density as calculated in Part (b) above. Thus, the expansion velocity of the universe is greater than the escape velocity. Based on our calculations, the universe apparently will expand forever! |