Mass of a black hole - given only the diameter

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Homework Help Overview

The discussion revolves around calculating the mass of a mini black hole based on its diameter, specifically using the diameter of a proton as a reference. The problem is situated within the context of cosmology and general relativity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the Schwarzschild radius and mass, with some attempting to derive mass from the given diameter using gravitational equations. Questions arise regarding the correct formulation and factors involved in the equations.

Discussion Status

Participants are actively discussing the correct application of the Schwarzschild radius formula and its rearrangement to find mass. Some have pointed out potential errors in earlier calculations, while others are confirming their understanding of the relationships involved.

Contextual Notes

There is mention of confusion regarding the inclusion of factors and the square root in the equations, indicating that participants are working through the mathematical details and assumptions related to black hole physics.

PirateFan308
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Homework Statement


Cosmologists have speculated that black holes the size of a proton could have formed during the early days of the Big Bang when the universe began. If we take the diameter of a proton to be [itex]1.0*10^{-15}[/itex], what would be the mass of a mini black hole?

Homework Equations



[itex]v=\sqrt{\frac{Gm}{r}}[/itex]

The Attempt at a Solution



[itex]v=\sqrt{\frac{Gm}{r}}[/itex]

[itex]m=\frac{v^{2}r}{G}[/itex]

[itex]m=\frac{(3.0*10^{8})^{2}(0.5*10^{-15})}{(6.67*10^{-11})}[/itex]

[itex]m=6.75*10^{11} kg[/itex]

It says that this is wrong, but I can't find my mistake. Thanks!
 
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The Schwarzschild radius is given by:
[tex]r_s = \frac{2 G M}{c^2}[/tex]
I think you forgot the factor of 2.

EDIT: D'Oh. No square root! Fixed it.
 
Last edited:
gneill said:
The Schwarzschild radius is given by:
[tex]r_s = \sqrt{\frac{2 G M}{c^2}}[/tex]
I think you forgot the factor of 2.

Isn't the schwarzchild radius simply:

[itex]R_s=\frac{2GM}{c^2}[/itex]

So this would rearrange to

[itex]m=\frac{rc^2}{2G}[/itex]

and plugging in the values, I would get [itex]m=3.37*10^{11}kg[/itex]

Is this now correct?
 
PirateFan308 said:
Isn't the schwarzchild radius simply:

[itex]R_s=\frac{2GM}{c^2}[/itex]

So this would rearrange to

[itex]m=\frac{rc^2}{2G}[/itex]

and plugging in the values, I would get [itex]m=3.37*10^{11}kg[/itex]

Is this now correct?

Yes, and yes. Sorry about the square root distraction, I don't know where my head was at!
 
Thanks for the help!
 

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