Relating Newton's Force to Black Holes (Calculation?)

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SUMMARY

The discussion centers on calculating the distance from a black hole's center at which an average-sized student would be torn apart by gravitational forces. The relevant equation used is the Schwarzschild radius formula, RSchwarzschild = (2MG)/c², where M is the mass of the black hole (1 solar mass), G is the gravitational constant, and c is the speed of light. The initial calculation yielded an incorrect Schwarzschild radius of 1.48x10-27 m, indicating a misunderstanding of unit compatibility. The correct approach requires incorporating the mass of the sun to ensure accurate unit conversion and calculation.

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  • Understanding of gravitational forces and their effects on objects
  • Familiarity with the Schwarzschild radius formula
  • Basic knowledge of unit conversions in physics
  • Concept of body surface area (BSA) and its relevance in physical calculations
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  • Research the concept of gravitational tidal forces and their effects near black holes
  • Learn about the mass of the sun and its role in astrophysical calculations
  • Study unit conversion techniques in physics to avoid calculation errors
  • Explore the implications of the event horizon in black hole physics
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alphaj
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Homework Statement


Muscle can be torn apart by a force of 100,000 N applied across an area of 1 m2. A 10 cm2 muscle therefore will be torn by a force of 100 N.

If a student of average size were being lowered into a black hole of 1 solar mass, at about what distance from the hole's center will he be torn apart?

2. (What I thought was a) Relevant equation

RSchwarzschild= (2MG)/c2

Where M is the mass of the black hole; G is the gravity constant; c is the speed of light.
(seen here: http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkhol.html#c2)

The Attempt at a Solution



I did some background research on black holes and I thought that if I calculated the Schwarzschild radius, I would be able to identify the event horizon (defined as the last distance at which light can escape the pull of a black hole) of the black hole, and thus, that would tell me the distance from which the student would be torn apart.

RSchwarzschild= (2MG)/c2
= [2(1 solar mass)(6.67 x 10-11 Nm2/kg2]/(2.998 x 108 m/s)2

R = 1.48x10-27 m

I think this would be the correct answer, but it doesn't use any of the numerical givens other than the mass of the black hole.

When I consider the givens regarding surface area and force; I can tell that I am supposed to use an estimate of an 'average student's size' to attain my answer. If I estimate that the average person is about 1.75 m2 (estimate using numbers from BSA https://en.wikipedia.org/wiki/Body_surface_area), then I believe I can say that 100000 N is to 1 m2, as x N is to 1.75 m2 and cross multiply to find that it would take x=175000 N of force to tear apart the student's muscles. But, I am unsure of how to relate this force to black holes and distance.
Any help or hints would be appreciated! Thank you.
 
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alphaj said:

Homework Statement


Muscle can be torn apart by a force of 100,000 N applied across an area of 1 m2. A 10 cm2 muscle therefore will be torn by a force of 100 N.

If a student of average size were being lowered into a black hole of 1 solar mass, at about what distance from the hole's center will he be torn apart?

2. (What I thought was a) Relevant equation

RSchwarzschild= (2MG)/c2

Where M is the mass of the black hole; G is the gravity constant; c is the speed of light.
(seen here: http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkhol.html#c2)

The Attempt at a Solution



I did some background research on black holes and I thought that if I calculated the Schwarzschild radius, I would be able to identify the event horizon (defined as the last distance at which light can escape the pull of a black hole) of the black hole, and thus, that would tell me the distance from which the student would be torn apart.

RSchwarzschild= (2MG)/c2
= [2(1 solar mass)(6.67 x 10-11 Nm2/kg2]/(2.998 x 108 m/s)2

I don't see units of "solar mass" in the rest of this equation. How can the result for the Schwarzschild radius be in meters?

R = 1.48x10-27 m

Always check the units of your calculations. The resulting units should be the same on both sides of the equation.

I think this would be the correct answer, but it doesn't use any of the numerical givens other than the mass of the black hole.

When I consider the givens regarding surface area and force; I can tell that I am supposed to use an estimate of an 'average student's size' to attain my answer. If I estimate that the average person is about 1.75 m2 (estimate using numbers from BSA https://en.wikipedia.org/wiki/Body_surface_area), then I believe I can say that 100000 N is to 1 m2, as x N is to 1.75 m2 and cross multiply to find that it would take x=175000 N of force to tear apart the student's muscles. But, I am unsure of how to relate this force to black holes and distance.

Any help or hints would be appreciated! Thank you.

You looked up some other numbers related to this problem on the internet. Why didn't you do the same for the mass of the sun?
 
Thank you for the input!
I'm not sure why I would need the mass of the Sun to evaluate this problem?
 
alphaj said:
Thank you for the input!
I'm not sure why I would need the mass of the Sun to evaluate this problem?
I thought I explained it very clearly. Units of "solar mass" in the formula for the Schwarzschild radius are not compatible with the units for the constant G.

Your calculation does not result in units of meters for the Schwarzschild radius.

10-27 m is to the size of a hydrogen atom as a hydrogen atom is to the size of a beach ball.
 
I see, thank you for clarifying. Sorry to make you repeat yourself! I'm going on my third week ever in a Physics class, so thanks again, your input is invaluable.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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