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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.9.109c
Chapter 8, Problem 8.9.109c

109. Escape velocity and black holes The work required to launch an object from the surface of Earth to outer space is given by W = ∫ from R to ∞ of F(x) dx, where R = 6370 km is the approximate radius of Earth, F(x) = (GMm)/x² is the gravitational force between Earth and the object, G is the gravitational constant, M is the mass of Earth, m is the mass of the object, and GM = 4 × 10¹⁴ m³/s².
c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, c = 300,000 km/s, then light cannot escape the body and it cannot be seen. Show that such a body has a radius R ≤ 2GM/c². For Earth to be a black hole, what would its radius need to be?

Verified step by step guidance
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Start with the expression for the work required to escape Earth's gravity, which is the integral of the gravitational force from the radius R to infinity: \[ W = \int_{R}^{\infty} F(x) \, dx = \int_{R}^{\infty} \frac{GMm}{x^2} \, dx \]
Evaluate the integral to find the work done (or energy required) to escape Earth's gravitational field: \[ W = GMm \int_{R}^{\infty} x^{-2} \, dx = GMm \left[-\frac{1}{x}\right]_{R}^{\infty} = \frac{GMm}{R} \]
Recall that the escape velocity \(v_e\) is the velocity needed so that the kinetic energy equals the work done against gravity: \[ \frac{1}{2} m v_e^2 = \frac{GMm}{R} \]
Solve for the escape velocity \(v_e\): \[ v_e = \sqrt{\frac{2GM}{R}} \]
To find the radius \(R\) for which the escape velocity equals the speed of light \(c\), set \(v_e = c\) and solve for \(R\): \[ c = \sqrt{\frac{2GM}{R}} \implies R = \frac{2GM}{c^2} \] This shows that if a body's radius is less than or equal to \(\frac{2GM}{c^2}\), its escape velocity is at least \(c\), making it a black hole. For Earth, substitute the given values of \(GM\) and \(c\) to find the radius it would need to become a black hole.

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Key Concepts

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

Gravitational Force and Work Done by a Variable Force

Gravitational force between two masses varies with distance as F(x) = GMm/x². Calculating work done to move an object against this force involves integrating F(x) over the distance from the Earth's surface to infinity. This integral represents the energy required to escape Earth's gravity, linking force, distance, and work in a variable force field.
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Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from a celestial body's gravitational pull without further propulsion. It is derived by equating kinetic energy to gravitational potential energy, resulting in v = √(2GM/R). This concept connects gravitational parameters with motion and energy conservation.
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Schwarzschild Radius and Black Holes

The Schwarzschild radius (R ≤ 2GM/c²) defines the critical radius at which an object's escape velocity equals the speed of light, making it a black hole. If a mass is compressed within this radius, not even light can escape its gravity, rendering it invisible. This concept bridges classical gravity with relativistic limits on information escape.
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Related Practice
Textbook Question

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Textbook Question

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1
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