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Calculate the Schwarzschild radius of a black hole with the mass of a) the Sun. b) a proton. How does this result compare with the size scale \(10^{-15} \mathrm{~m}\) usually associated with a proton?

Short Answer

Expert verified
Answer: The Schwarzschild radius of a black hole with the mass of the Sun is approximately \(2.954 \times 10^3 \mathrm{~m}\), and the Schwarzschild radius of a black hole with the mass of a proton is approximately \(2.478 \times 10^{-54} \mathrm{~m}\). Comparing these values to the size scale usually associated with a proton (\(10^{-15} \mathrm{~m}\)), it shows that a black hole with the mass of a proton would have a Schwarzschild radius much smaller than the proton's size scale.

Step by step solution

01

Define the Given Values and Constants

First, let us define the given values for the mass of the Sun and a proton, and the constants: - Mass of the Sun, \(M_\odot = 1.989 × 10^{30} \mathrm{~kg}\) - Mass of a proton, \(M_p = 1.673 × 10^{-27} \mathrm{~kg}\) - Gravitational constant, \(G = 6.674 × 10^{-11} m^3 kg^{-1} s^{-2}\) - Speed of light, \(c = 2.998 × 10^8 \mathrm{~m/s}\)
02

Calculate the Schwarzschild Radius of the Sun

Now, we will calculate the Schwarzschild radius of a black hole with the mass of the Sun using the formula: \(R_{s_\odot} = \frac{2GM_\odot}{c^2}\) Plug in the values we defined earlier: \(R_{s_\odot} = \frac{2 × 6.674 × 10^{-11} \mathrm{m^3 kg^{-1} s^{-2}} × 1.989 × 10^{30} \mathrm{~kg}}{(2.998 × 10^8 \mathrm{~m/s})^2}\) After evaluating the expression, we get: \(R_{s_\odot} \approx 2.954 \times 10^3 \mathrm{~m}\) So, the Schwarzschild radius of a black hole with the mass of the Sun is approximately \(2.954 \times 10^3 \mathrm{~m}\).
03

Calculate the Schwarzschild Radius of a Proton

Now, we will calculate the Schwarzschild radius of a black hole with the mass of a proton using the formula: \(R_{s_p} = \frac{2GM_p}{c^2}\) Plug in the values we defined earlier: \(R_{s_p} = \frac{2 × 6.674 × 10^{-11} \mathrm{m^3 kg^{-1} s^{-2}} × 1.673 × 10^{-27} \mathrm{~kg}}{(2.998 × 10^8 \mathrm{~m/s})^2}\) After evaluating the expression, we get: \(R_{s_p} \approx 2.478 × 10^{-54} \mathrm{~m}\) So, the Schwarzschild radius of a black hole with the mass of a proton is approximately \(2.478 \times 10^{-54} \mathrm{~m}\).
04

Compare the Schwarzschild Radius with the Size Scale Associated with a Proton

Now, let's compare the calculated Schwarzschild radius of a proton-sized black hole with the size scale usually associated with a proton, which is \(10^{-15} \mathrm{~m}\). The Schwarzschild radius we calculated is much smaller than the size scale associated with a proton: \(R_{s_p} \approx 2.478 \times 10^{-54} \mathrm{~m} << 10^{-15} \mathrm{~m}\) This result shows that a black hole with the mass of a proton would have a Schwarzschild radius much smaller than the size scale usually associated with a proton.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Black Hole Physics
The concept of a black hole is one of the most fascinating subjects in the field of astrophysics. At its core, a black hole is a region in space where the pull of gravity is so strong that nothing, not even light, can escape from it. This intense gravitational attraction is due to the presence of an extremely dense mass concentrated in a small area.

A critical concept when discussing black holes is the Schwarzschild radius, also known as the event horizon. It represents the boundary around a black hole beyond which no information can return to the outside universe. The Schwarzschild radius is determined by the mass of the black hole and is directly proportional to it. Using the known values of constants like the speed of light and the gravitational constant, scientists can calculate this radius and gain insights into the properties of a black hole.

To truly understand the magnitude of a black hole's gravity, consider that the Schwarzschild radius for a black hole with the mass of the Sun is about 3 kilometers. If the Sun were to be compressed to within this radius, the escape velocity at its surface would be equal to the speed of light, thus transforming it into a black hole.
Gravitational Constant
The gravitational constant, denoted as \(G\), is a key value in the calculation of the Schwarzschild radius. It appears in Newton's law of universal gravitation as well as in Einstein’s theory of general relativity. The gravitational constant represents the force of attraction between two masses that are one meter apart.

The value of \(G\) is approximately \(6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}\). This constant is crucial for calculations related to gravity, including the dynamics of planetary orbits, the behavior of objects in free-fall on Earth, and the properties of black holes. When we plug this constant into the formula for the Schwarzschild radius, it helps us comprehend just how the mass of an object affects the extent of its gravitational pull, indicating the size the mass must be confined to for it to become a black hole.
Mass of the Sun
The Sun, with a mass of approximately \(1.989 \times 10^{30} kg\), is the most massive object in our solar system, accounting for about 99.86% of the system's total mass. This immense mass influences the orbit of planets, comets, and asteroids through its gravity.

Using the aforementioned Schwarzschild radius formula, if the Sun were to be compressed below approximately 3 kilometers, it would become a black hole. This radius is minuscule compared to the Sun's current radius of about 696,340 kilometers. The concept is theoretical and helps to illustrate the scale of forces at play within black holes of different masses. This comparison between the Schwarzschild radius and the actual size of the Sun conveys the extraordinary density that a black hole represents.
Mass of a Proton
The mass of a proton is about \(1.673 \times 10^{-27} kg\). In the context of the universe, protons are extremely small; however, they have a significant impact on the scale of atoms and molecules. A comparison between the Schwarzschild radius of a proton-sized black hole and the size of an actual proton illustrates fascinating aspects of scale in the universe.

As calculated, the Schwarzschild radius for a black hole with the mass of a proton is incredibly small, at about \(2.478 \times 10^{-54} m\). In contrast, the size of a proton is about \(10^{-15} m\), which is many orders of magnitude larger. This result is hypothetical and serves to emphasize that, in our current understanding of physics, black holes are expected to form only from significantly larger masses, as the conditions to compress a proton to within its Schwarzschild radius are not met in nature. These calculations allow students to appreciate the extent of densities and sizes we consider when exploring the cosmos.

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Most popular questions from this chapter

Consider motion in one spatial dimension. For any velocity \(v,\) define parameter \(\theta\) via the relation \(v=c \tanh \theta\) where \(c\) is the vacuum speed of light. This quantity is variously called the velocity parameter or the rapidity corresponding to velocity \(v\). a) Prove that for two velocities, which add according to the Lorentzian rule, the corresponding velocity parameters simply add algebraically, that is, like Galilean velocities. b) Consider two reference frames in motion at speed \(v\) in the \(x\) -direction relative to one another, with axes parallel and origins coinciding when clocks at the origin in both frames read zero. Write the Lorentz transformation between the two coordinate systems entirely in terms of the velocity parameter corresponding to \(v\), and the coordinates.

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, you see that the particle is, in fact, moving toward the wire.)

A spacecraft travels along a straight line from Earth to the Moon, a distance of \(3.84 \cdot 10^{8} \mathrm{~m}\). Its speed measured on Earth is \(0.50 c\). a) How long does the trip take, according to a clock on Earth? b) How long does the trip take, according to a clock on the spacecraft? c) Determine the distance between Earth and the Moon if it were measured by a person on the spacecraft.

In Jules Verne's classic Around the World in Eighty Days, Phileas Fogg travels around the world in, according to his calculation, 81 days. Due to crossing the International Date Line he actually made it only 80 days. How fast would he have to go in order to have time dilation make 80 days to seem like \(81 ?\) (Of course, at this speed, it would take a lot less than even 1 day to get around the world \(\ldots . .)\)

Michelson and Morley used an interferometer to show that the speed of light is constant, regardless of Earth's motion through any perceived luminiferous aether. An analogy can be understood from the different times it takes for a rowboat to travel two different round-trip paths in a river that flows at a constant velocity \((u)\) downstream. Let one path be for a distance \(D\) directly across the river, then back again; and let the other path be the same distance \(D\) directly upstream, then back again. Assume that the rowboat travels at constant speed, \(v\) (with respect to the water), for both trips. Neglect the time it takes for the rowboat to turn around. Find the ratio of the cross-stream time divided by the upstream-downstream time, as a function of the given constants.

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