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Ch 13: Newton's Theory of Gravity
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 13: Newton's Theory of GravityProblem 21
Chapter 13, Problem 21

Nothing can escape the event horizon of a black hole, not even light. You can think of the event horizon as being the distance from a black hole at which the escape speed is the speed of light, 3.00 ✕ 10⁸ m/s, making all escape impossible. What is the radius of the event horizon for a black hole with a mass 5.0 times the mass of the sun? This distance is called the Schwarzschild radius.

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1
The Schwarzschild radius \( r_s \) is given by the formula \( r_s = \frac{2GM}{c^2} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \), \( M \) is the mass of the black hole, and \( c \) is the speed of light \( 3.00 \times 10^8 \, \text{m/s} \).
The mass of the black hole is given as \( 5.0 \) times the mass of the Sun. The mass of the Sun \( M_\odot \) is approximately \( 1.989 \times 10^{30} \, \text{kg} \). Therefore, the mass of the black hole \( M \) is \( M = 5.0 \times M_\odot = 5.0 \times 1.989 \times 10^{30} \, \text{kg} \).
Substitute the values of \( G \), \( M \), and \( c \) into the Schwarzschild radius formula: \( r_s = \frac{2 \times (6.674 \times 10^{-11}) \times (5.0 \times 1.989 \times 10^{30})}{(3.00 \times 10^8)^2} \).
Simplify the numerator by multiplying \( 2 \), \( G \), and \( M \), and simplify the denominator by squaring \( c \).
Divide the simplified numerator by the simplified denominator to find the Schwarzschild radius \( r_s \). This will give you the radius of the event horizon for the black hole.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Event Horizon

The event horizon is the boundary surrounding a black hole beyond which no information or matter can escape. It represents the point at which the escape velocity equals the speed of light, making it impossible for anything, including light, to break free from the gravitational pull of the black hole.
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Schwarzschild Radius

The Schwarzschild radius is a specific radius that defines the size of the event horizon for a non-rotating black hole. It is directly proportional to the mass of the black hole and can be calculated using the formula r_s = 2GM/c², where G is the gravitational constant, M is the mass of the black hole, and c is the speed of light.
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Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without any additional propulsion. For a black hole, the escape velocity at the event horizon reaches the speed of light, which is why nothing can escape once it crosses this threshold.
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Related Practice
Textbook Question

You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of earth but the free-fall acceleration at the surface is only one-fourth as large. What is the planet's radius?

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Textbook Question

The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 5.0 earth years. What are the asteroid's orbital radius and speed?

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Textbook Question

Two stars, one twice as massive as the other, are 1.0 light year (ly) apart. One light year is the distance light travels in one year at the speed of light, 3.00 ✕ 108 m/s . The gravitational potential energy of this double-star system is - 8.0 ✕ 1034 J. What is the mass of the lighter star?

Textbook Question

Three satellites orbit a planet of radius R, as shown in FIGURE EX13.24. Satellites S1 and S3 have mass m. Satellite S2 has mass 2m. Satellite S1 orbits in 250 minutes and the force on S1 is 10,000 N. What is the kinetic-energy ratio for K1 / K3 for S1 and S3?

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Textbook Question

A binary star system has two stars, each with the same mass as our sun, separated by 1.0 ✕ 1012 m. A comet is very far away and essentially at rest. Slowly but surely, gravity pulls the comet toward the stars. Suppose the comet travels along a trajectory that passes through the midpoint between the two stars. What is the comet's speed at the midpoint?

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Textbook Question

You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of earth but the free-fall acceleration at the surface is only one-fourth as large. To get back to earth, you need to escape the planet. What minimum speed does your rocket need?