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Chapter 12: Problem 55

The radius of a black hole is the distance from the black hole's center at which the escape speed is the speed of light. a) What is the radius of a black hole with a mass twice that of the Sun? b) At what radius from the center of the black hole in part (a) would the orbital speed be equal to the speed of light? c) What is the radius of a black hole with the same mass as that of the Earth?

Short Answer

Expert verified
Also, what is the radius of a black hole with the same mass as that of the Earth? Answer: The radius of a black hole with a mass twice that of the Sun is approximately 5.93 kilometers. At a radius of approximately 2.965 kilometers from the center of this black hole, the orbital speed would be equal to the speed of light. The radius of a black hole with the same mass as that of the Earth is approximately 8.87 millimeters.

Step by step solution

01

Setting up the escape velocity equation for a black hole

Since the escape velocity at the radius of a black hole is equal to the speed of light, we set the escape velocity formula equal to the speed of light: \(c = \sqrt{\frac{2GM}{r_s}}\), where \(c\) is the speed of light and \(r_s\) is the Schwarzschild radius.
02

Solve for the Schwarzschild radius

Now we can solve the equation for the Schwarzschild radius, \(r_s\), by squaring both sides and rearranging terms: \(r_s = \frac{2GM}{c^2}\).
03

Plug in given values

We are given a black hole with a mass twice the mass of the Sun. The mass of the Sun, \(M_\odot\), is approximately \(1.989 \times 10^{30}\) kg. Thus, the mass of the black hole is \(2M_\odot = 3.978 \times 10^{30}\) kg. Also, the speed of light \(c\) is approximately \(3 \times 10^8\) m/s and the gravitational constant \(G\) is approximately \(6.674 \times 10^{-11} \; \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\). Plugging these values into the formula for \(r_s\): \(r_s = \frac{2(6.674 \times 10^{-11})(3.978 \times 10^{30})}{(3 \times 10^8)^2}\).
04

Calculate the Schwarzschild radius

Finally, compute the value of the Schwarzschild radius: \(r_s \approx 5.93 \times 10^3 \mathrm{m}\). So the radius of a black hole with a mass twice that of the Sun is approximately \(5.93 \times 10^3\) meters (or \(5.93\) kilometers). #b) At what radius from the center of the black hole in part (a) would the orbital speed be equal to the speed of light? #
05

Setting up the orbital speed equation

We are given that the orbital speed is equal to the speed of light, so we set the orbital speed formula equal to the speed of light: \(c = \sqrt{\frac{GM}{r_o}}\), where \(r_o\) is the radius at which the orbital speed is equal to the speed of light.
06

Solve for the radius

Now we can solve the equation for the radius, \(r_o\), by squaring both sides and rearranging terms: \(r_o = \frac{GM}{c^2}\).
07

Plug in given values

We already have the mass of the black hole, the speed of light, and the gravitational constant from the previous part. Plugging these values into the formula for \(r_o\): \(r_o = \frac{(6.674 \times 10^{-11})(3.978 \times 10^{30})}{(3 \times 10^8)^2}\).
08

Calculate the radius

Finally, compute the value of the radius \(r_o\): \(r_o \approx 2.965 \times 10^3 \mathrm{m}\). So, at a radius of approximately \(2.965 \times 10^3\) meters (or \(2.965\) kilometers) from the center of the black hole, the orbital speed would be equal to the speed of light. #c) What is the radius of a black hole with the same mass as that of the Earth? # The process of finding the Schwarzschild radius for a black hole with the mass of the Earth is the same as in part a). We only need to substitute the mass of the Earth for the mass of the black hole.
09

Find the mass of the Earth

The mass of the Earth, \(M_\oplus\), is approximately \(5.972 \times 10^{24}\) kg.
10

Plug in values

Substitute the mass of the Earth into the formula for \(r_s\): \(r_s = \frac{2(6.674 \times 10^{-11})(5.972 \times 10^{24})}{(3 \times 10^8)^2}\).
11

Calculate the Schwarzschild radius

Finally, compute the Schwarzschild radius for a black hole with the mass of the Earth: \(r_s \approx 8.87 \times 10^{-3} \mathrm{m}\). So the radius of a black hole with the same mass as that of the Earth is approximately \(8.87 \times 10^{-3}\) meters (or \(8.87\) millimeters).

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Key Concepts

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

Escape Velocity
Escape velocity is a fundamental concept in astrophysics, particularly when studying celestial objects such as black holes. It is defined as the speed needed for an object to break free from the gravitational pull of a celestial body without further propulsion.

Imagine you’re throwing a ball upwards; the escape velocity is that initial throw speed needed for the ball to avoid falling back to Earth and instead escape into space. This speed depends on the mass of the celestial body and the distance from its center.

In the context of black holes, the escape velocity at the event horizon, known as the Schwarzschild radius, equals the speed of light. This characteristic makes black holes appear 'black' as no light can escape from within this radius. The exercise provided illustrates how to calculate the Schwarzschild radius using the escape velocity formula and the given black hole mass.
Black Hole Mass
The mass of a black hole is a critical factor in determining its gravitational pull and, consequently, the Schwarzschild radius. Black hole mass is typically expressed in multiples of the Sun's mass because the Sun is a well-known reference point.

The exercise showcased the calculation of the Schwarzschild radius for a black hole with a mass twice that of the Sun. This doubled mass greatly affects the Schwarzschild radius, showcasing the direct relationship between a black hole's mass and its gravitational effects.

To understand this further, remember that the heavier the black hole, the larger the radius at which its escape velocity would equal the speed of light. This concept is essential for comprehending the immense gravitational influence black holes have in their vicinity.
Gravitational Constant
The gravitational constant, denoted by the symbol 'G', is one of the fundamental constants in physics. It quantifies the strength of gravity between two objects and is used in the universal law of gravitation as well as in the equations governing black holes.

With a value of approximately 6.674 x 10-11 m3 kg-1 s-2, the gravitational constant G directly influences the escape velocity and Schwarzschild radius calculations. In the steps provided for the exercises, G is a vital coefficient in determining the exact size of the Schwarzschild radius for a given black hole mass. It’s interesting to note that G is the same everywhere in the universe, making it a constant not just in value, but also in its cosmic ubiquity.

The invariable nature of G allows us to make precise calculations about the properties of black holes and other astronomical phenomena regardless of where they are in the universe.

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Most popular questions from this chapter

An asteroid is discovered to have a tiny moon that orbits it in a circular path at a distance of \(100 . \mathrm{km}\) and with a period of \(40.0 \mathrm{~h}\). The asteroid is roughly spherical (unusual for such a small body) with a radius of \(20.0 \mathrm{~km} .\) a) Find the acceleration of gravity at the surface of the asteroid. b) Find the escape velocity from the asteroid.

Satellites in low orbit around the Earth lose energy from colliding with the gases of the upper atmosphere, causing them to slowly spiral inward. What happens to their kinetic energy as they fall inward?

A space shuttle is initially in a circular orbit at a radius of \(r=6.60 \cdot 10^{6} \mathrm{~m}\) from the center of the Earth. A retrorocket is fired forward, reducing the total energy of the space shuttle by \(10 \%\) (that is, increasing the magnitude of the negative total energy by \(10 \%\) ), and the space shuttle moves to a new circular orbit with a radius that is smaller than \(r\). Find the speed of the space shuttle (a) before and (b) after the retrorocket is fired.

Standing on the surface of a small spherical moon whose radius is \(6.30 \cdot 10^{4} \mathrm{~m}\) and whose mass is \(8.00 \cdot 10^{18} \mathrm{~kg}\) an astronaut throws a rock of mass 2.00 kg straight upward with an initial speed \(40.0 \mathrm{~m} / \mathrm{s}\). (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?

A scientist working for a space agency noticed that a Russian satellite of mass \(250 . \mathrm{kg}\) is on collision course with an American satellite of mass \(600 .\) kg orbiting at \(1000 . \mathrm{km}\) above the surface. Both satellites are moving in circular orbits but in opposite directions. If the two satellites collide and stick together, will they continue to orbit or crash to the Earth? Explain.
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