Calculating Black Hole Mass Gain from X-ray Emissions of Binary Systems

AI Thread Summary
In a binary system where a black hole and a normal star interact, gases from the star heat up and emit X-rays due to friction. Cygnus X-1, a prominent X-ray source, radiates at approximately 4.00x10^31 W, with 0.84% of the in-falling mass escaping as X-ray energy. The discussion revolves around calculating the mass gain of the black hole based on the energy absorbed, leading to the conclusion that the black hole gains mass at a rate of approximately 5.247x10^16 kg/s. The calculations involve determining the energy absorbed by the black hole and applying Einstein's equation, E=mc^2. This process highlights the relationship between X-ray emissions and black hole mass gain in binary systems.
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Homework Statement


If a black hole and a "normal" star orbit each other, gases from the normal star falling into the black hole can have their temperature increased by millions of degrees due to frictional heating. When the gases are heated that much, they begin to radiate light in the X-ray region of the electromagnetic spectrum (high-energy light photons). Cygnus X-1, the second strongest known X-ray source in the sky, is thought to be one such binary system; it radiates at an estimated power of 4.00x10^31 W. If we assume that 0.84 percent of the in-falling mass escapes as X ray energy, at what rate is the black hole gaining mass?

Homework Equations


E=m*c^2

The Attempt at a Solution


Wondering if someone could verify this because this question confuses me.

I'm assuming that the black hole absorbs 100-.84=99.16% of the power.

I take 4.00x10^31 * .9916 = 3.9664x10^31 J

Using E=m*c^2 => m = 3.9664x10^31/((3x10^8)^2) = 4.407x10^4 kg/s

Does this seem a logical conclusion, am I interpreting the question correctly? Thanks in advance.
 
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That's not right. The 4.00x10^31 W isn't the total power but only 0.84% of it. The other 99.16% goes into the hole.
 
Right, which is why I take 4x10^31*.9916 to get the J absorbed by the black hole per second. Am I right in doing so, or am I doing something wrong?
 
Ok I see now,

Total*.0084 = 4x10^31

Mass entering per second = 4x10^31*.9916/.0084 = 4.72x10^33



4.72x10^33 / ((3x10^8)^2) = 5.247x10^16 kg/s is the right answer, thanks for the help!
 
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