Question: Stars and gas located 1 pc away from the center of our Galaxy are observed in a circular orbit with an average speed of 100 km/s. What is the total mass of the stars, gas, black hole, and perhaps dark matter located inside this radius?
Answer: Both the online lecture outline and the Unit 73 in the textbook show an example calculation for determining the mass inside the solar circle. Use the same equation but substitute r = 1 pc and v = 100 km/s. You will find that the answer is 2.3e6 solar masses (or 2.4 million Suns).
Question: The global energy consumption rate today is about 3.0e15 joules per day. If we can build a black hole engine that can extract 10% of the mass energy into a useful form, how long can we meet the entire energy need of the world by dumping 2 kg of trash into it?
Answer: Example calculation shown in section 50.1 of your textbook should gives an example calculation of turning mass into energy. You have to use the famous mass energy formula, E = mc2, except this BH engine is only 10% efficient. So, the energy you get out is 0.1 x m x c2. Here, m = 2 kg, and c = 3e8 m/s, and the energy output is 0.1 x 2 x 3e82 = 1.8e16 joules. We use up about 3.0e15 joules per day, so 2 kg of trash should generate enough energy to last 1.8e16/3.0e15 = 6 days. Not bad for about 3 lbs of trash!
Question: What is the mass-to-light ratio of a 0.1 solar mass brown dwarf with luminosity of 0.001 L(Sun)?
Answer: The mass-to-light ratio is literally mass divided by light (luminosity). So, mass-to-light ratio is (0.1)/(0.001) = 100 solar mass per solar luminosity. We did this as an inclass quiz problem already.
Question: The measured temperature of the cosmic microwave background (CMB) is about 3 Kelvin, and the peak of this thermal radiation is seen at the wavelength of 1,000,000 nm. If the temperature of the last scattering surface of the CMB was 3,000 Kelven, then the peak of the emitted thermal radiation was at a wavelength of 1000 nm. What is the redshift of this scattering surface?
Answer: As you already worked this out in HW5 (also see Unit 25), redshift z is given by [lambda(observed) - lambda(rest)]/lambda(rest). Here, lambda(observed) = 1,000,000 nm and lambda(rest) = 1000 nm. So, z = (1000000 - 1000)/1000 = 999.
Question: If the age of the universe is 15 billion (15e9) years, the diameter of the observable universe is __ cm.
Answer: The observable edge of the universe is at the distance traveled by light over the age of the universe. If the age of the universe is 15e9 years, then the radius of the observable universe is 15e9 light years. One light year is about 1.0e18 cm, so the diameter of the observable universe is 2 x 15e9 x 1.0e18 = 3.0e28 cm.
Question: During the Inflation Era, the universe increased its size by a factor of 1030. Suppose the age of the universe is 14 billion (14e9) years. Then the diameter of the current observable universe was only __ cm across before the Inflation. Is this bigger than the size of a pinhead?
Answer: Repeating the calculation in Question 5, the diameter of the observable universe is 2 x 14e9 x 1.0e18 = 2.8e28 cm. Before the inflation, this area was 1030 times smaller. Therefore, the diameter of the observable universe before the inflation was 2.8e28/1.0e30 = 0.028 cm. This is about 0.3 mm, which is about the thickness of your hair and is much smaller than the size of a pinhead.