Universe
Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 21, Problem 60Q

(a)

To determine

The Schwarzschild radius of a black hole having mass equal to that of Earth. Also calculate the density if matter is uniformly spread throughout the volume of the event horizon.

(a)

Expert Solution
Check Mark

Answer to Problem 60Q

Solution:

RSch=8.89×103 m; ρ=2.03×1030 kg/m3.

Explanation of Solution

Given data:

A black hole having mass equal to that of Earth.

Formula used:

The expression for Schwarzschild radius of an object is given as,

RSch=2GMc2

Here, RSch is the Schwarzschild radius, G is the universal gravitational constant, M is the mass of the object and c is the speed of light.

The expression for density of an object is given as,

ρ=MV

Here, ρ is the density, M is the mass and V is the volume.

Explanation:

The mass of black hole is equal to that of earth, which is equal to 5.976×1024 kg.

Schwarzschild radius of the black hole is calculated as follows:

Recall the expression for Schwarzschild radius of an object.

RSch=2GMc2

Upon substituting 6.67×1011 Nm2/kg2 for G, 5.976×1024 kg for M and 3×108 m/s2 for c,

RSch=2(6.67×1011 Nm2/kg2)(5.976×1024 kg)(3×108 m/s2)2=8.89×103 m

Density of the black hole is calculated by considering it to be a sphere.

Recall the expression for density of an object.

ρ=MV

Upon substituting 43πR3 for V and 5.976×1024 kg for M,

ρ=(5.976×1024 kg)(43πR3)

Substitute 8.89×103 m for R,

ρ=5.976×1024 kg43π(8.89×103 m)3=2.03×1030 kg/m3

Conclusion:

Hence, Schwarzschild radius of the black hole is 8.89×103 m and its density is 2.03×1030 kg/m3.

(b)

To determine

The Schwarzschild radius of a black hole having mass equal to that of Sun. Also calculate the density if matter is uniformly spread throughout the volume of the event horizon.

(b)

Expert Solution
Check Mark

Answer to Problem 60Q

Solution:

RSch=2.93×103 m; ρ=1.87×1019 kg/m3.

Explanation of Solution

Given data:

A black hole having mass equal to that of Sun.

Formula used:

The expression for Schwarzschild radius of an object is given as,

RSch=2GMc2

Here, RSch is the Schwarzschild radius, G is the universal gravitational constant, M is the mass of the object and c is the speed of light.

The expression for density of an object is given as,

ρ=MV

Here, ρ is the density, M is the mass and V is the volume.

Explanation:

The mass of black hole is equal to that of Sun, which is equal to 1.989×1030 kg.

Schwarzschild radius of the black hole is calculated as follows:

Recall the expression for Schwarzschild radius of an object.

RSch=2GMc2

Upon substituting 6.67×1011 Nm2/kg2 for G, 1.989×1030 kg for M and 3×108 m/s2 for c,

RSch=2(6.67×1011 Nm2/kg2)(1.989×1030 kg)(3×108 m/s2)2=2.94×103 m

Density of the black hole is calculated by considering it to be a sphere.

Recall the expression for density of an object.

ρ=MV

Upon substituting 43πR3 for V and 1.989×1030 kg for M,

ρ=(1.989×1030 kg)(43πR3)

Substitute 2.94×103 m for R,

ρ=1.989×1030 kg43π(2.94×103 m)3=1.87×1019 kg/m3

Conclusion:

Hence, Schwarzschild radius of the black hole is 2.94×103 m and its density is 1.87×1019 kg/m3.

(c)

To determine

The Schwarzschild radius of a black hole having mass equal to that of a supermassive black hole in NGC4261. Also calculate the density if matter is uniformly spread throughout the volume of the event horizon.

(c)

Expert Solution
Check Mark

Answer to Problem 60Q

Solution:

RSch=3.54×1012 m; ρ=12.9 kg/m3.

Explanation of Solution

Given data:

A black hole having mass equal to that of a supermassive black hole in NGC4261.

Formula used:

The expression for Schwarzschild radius of an object is given as,

RSch=2GMc2

Here, RSch is the Schwarzschild radius, G is the universal gravitational constant, M is the mass of the object and c is the speed of light.

The expression for density of an object is given as,

ρ=MV

Here, ρ is the density, M is the mass and V is the volume.

Explanation:

The mass of black hole is equal to that of a supermassive black hole in NGC4261, which is equal to 2.39×1039 kg.

Calculate Schwarzschild radius of a black hole.

Recall the expression of Schwarzschild radius of an object.

RSch=2GMc2

Upon substituting 6.67×1011 Nm2/kg2 for G, 2.39×1039 kg for M and 3×108 m/s2 for c,

RSch=2(6.67×1011 Nm2/kg2)(2.39×1039 kg)(3×108 m/s2)2=3.54×1012 m

Density of the black hole is calculated by considering to be a sphere.

Recall the expression for density of an object.

ρ=MV

Upon substituting 43πR3 for V and 5.976×1024 kg for M,

ρ=(2.39×1039 kg)(43πR3)

Substitute 3.54×1012 m for R,

ρ=2.39×1039 kg43π(3.54×1012 m)3=12.9 kg/m3

Conclusion:

Hence, Schwarzschild radius of the black hole is 3.54×1012 m and its density is 12.9 kg/m3.

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Chapter 21 Solutions

Universe

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