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

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SUMMARY

The discussion centers on calculating the mass gain of a black hole from X-ray emissions in a binary system, specifically using Cygnus X-1 as a case study. The power output of Cygnus X-1 is estimated at 4.00x10^31 W, with 0.84% of the in-falling mass escaping as X-ray energy. The correct calculation shows that the black hole gains mass at a rate of approximately 5.247x10^16 kg/s, derived from the energy-mass equivalence formula E=mc². Participants clarified the interpretation of power absorption and the correct application of the equations involved.

PREREQUISITES
  • Understanding of black hole physics and binary systems
  • Familiarity with the electromagnetic spectrum, particularly X-ray emissions
  • Knowledge of the energy-mass equivalence principle (E=mc²)
  • Basic proficiency in algebra and unit conversions
NEXT STEPS
  • Research the properties and characteristics of Cygnus X-1 as a binary system
  • Learn about the mechanisms of X-ray emissions in astrophysical contexts
  • Study the implications of mass-energy conversion in black holes
  • Explore advanced calculations involving gravitational effects on mass gain in black holes
USEFUL FOR

Astronomers, astrophysicists, students studying black hole dynamics, and anyone interested in the mechanics of mass gain in binary star 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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