HINT:#
The question asks us to determine the total energy emitted by 3C273.
HINT:#
The problem tells us that Chandra has observed 10-11 ergs/s of energy per square centimeter from 3C273. The previous problem tells us that the total energy emitted by the object is equal to the energy detected per square centimeter multiplied by the surface area of a sphere with a radius equal to the distance between the Earth and the 3C273, a value given in Activity 2. Since the emission of the 3C273 is given in terms of energy per second, we will infer that we want to determine total energy emitted per second.
HINT:#
ET = Ecm2 * 4Πd2
HINT:#
Ecm2 = 10-11 ergs/cm2*s
d = 800 mpc
HINT:#
Let's use unit analysis:
ET = Ecm2 * 4Πd2
ET/s = ergs/s = (ergs/cm2*s)(mpc2)

For all units to cancel except for ergs/s, we will have to convert our distance in mpc to cm.
d = 800 mpc * 1 x 106 pc/mpc * 3 x 1018 cm/pc = 2.4 x 1027 cm
HINT:#
ET = Ecm2 * 4Πd2
ET = 1 x 10-11 ergs/s * cm 2 * 4Π (2.4 x 1027 cm)2
ET = 7.3 x 1044 ergs/s

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