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ns209hw3sol

# ns209hw3sol - 2 H o = 70 km/s/Mpc What would be the...

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Homework 3 Due April 10th 1. According to Hubble's law how long will it take for two galaxies to double their distances. (Hint, the speed of galaxies with respect to each other is not constant. as distance between galaxies increases, so does the speed. So you need to construct a differential equation. The solution to this differential equation dx/dt - Cx = 0 is x=A e Ct . A is arbitrary constant. ) Note: This problem has far reaching meaning as you will see later in the class. Solution: HUBBLE LAW! v = H o x v=dx/dt then dx/dt = H o x dx/dt - H o x = 0 !!! I have already given you the solution to this eq. x = A e Hot . This is instantenous distance as a function of time. x i = A e Hoti (initial) x f = A e Hotf (final). Take the ratio x f / x i = e Ho(tf-ti) But x f is 2x i , and t f -t i = Δt we want. Taking ln ln 2 = Ho Δt, Δt=ln2 x (1/H o ) so Δt is Hubble time, multiplied with ln 2. This does not take into account density reduction. .. sorry. Not counted emrah Please use maximum of three significant digits in your answers. Do not just copy what the calculator results in.

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Unformatted text preview: 2. H o = 70 km/s/Mpc. What would be the apparent diameter (in arcseconds) of a galaxy which has an actual diameter of 60 kpc if its 656.3 nm H line is observed at 740 nm. Solution: redshift=Δλ/λ=(740-656.3)nm / 656.3nm = 0.128 = v/c v = 300000 km/s * 0.128 = 38400 km/s d = v/Ho = 38400 km/s / 70 km/s/Mpc = 549 Mpc. 60 kpc = 549 000 kpc x apparent d. / 57.3 o . app d = 0.006 o = 22.5'' 3. A quasar with a luminosity of 10 39 W has 5x10 8 M  of material in its accretion disk. Assuming constant luminosity and 10% efficiency, how long would this quasar consume its fuel? Solution: Total mass-energy of the accretion disk, E tot = mc 2 = 5 x 10 8 x 2 x 10 30 kg x 9 x10 16 m/s = 9 x 10 55 J efficiency is 10%, then the total energy that can be radiated away is E rad = E tot * 0.1 = 9x10 54 J power radiated away per second by the quasar is P rad = 10 39 J/s Then time = E rad /P rad = 9 x 10 15 s = 300 million years...
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ns209hw3sol - 2 H o = 70 km/s/Mpc What would be the...

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