HINT:#
The question asks us to determine the energy released by a mass swallowed by a black hole.
HINT:#
The problem tells us that the mass of the black hole is equal to 109 solar masses, the mass fed into the black hole is equal to one solar mass and that the energy released is approximately equal to the loss of gravitational potential energy of the object, given by
E = Gm1m2/R, with m1 equal to the mass fed into the black hole, and m2 equal to the mass of the black hole. We also know that R = Rs, the Schwarzschild radius, calculated in the previous problem. G is the gravitational constant. The problem also tells us to calculate the number of ergs released per second if the given mass is fed into the black hole over one year.
HINT:#
Energy Released = Gm1m2/Rs
Energy Released/second = (Gm1m2/RS)/t
HINT:#
G = 6.7 x 10-8 cm3/g*s2
m1 = mblack hole = 109 solar masses
m2 = mmass eaten = 1 solar mass
R = Rs = 3 x 1014 cm
t = 1 year
HINT:#
Let's use unit analysis:
Energy Released = Gm1m2/R
Energy Released = ergs = g*cm2/s2 = (cm3/g*s2)(solar masses)(solar masses)/(cm)

Since we have the masses of the object eaten and the black hole in terms of Solar Masses, and we are working with CGS (centimeter, grams, seconds) units we will have to convert the masses into grams.
m1 = mblack hole = 109 solar masses * 2 x 1033 grams/solar mass = 2 x 1042 g
m2 = mmass eaten = 1 solar mass * 2 x 1033 grams/solar mass = 2 x 1033 g


Energy Released/second = Energy Released/t
Since we have time in years, we will have to convert time from years to seconds t = 1 yr * 3.16 x 107 = 3.16 x 107 s
HINT:#
Energy Released = Gm1m2/R
= (6.7 x 10-8 cm3/g*s2)(2 x 1042 g)(2 x 1033 g)/(3 x 1014 cm)
= 9 x 1053 ergs

Energy Released/s = 9 x 1053 ergs/3.16 x 107 s = 2.8 x 1046
This is more than enough to power 3C273 as observed!

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