Question

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=\frac{G M m}{x^{2}}\) 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 \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}\) a. Find the work required to launch an object in terms of \(m\). b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18 th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

Short answer

Under what conditions would Earth be considered a black hole? Answer: The work required to launch an object into space is given by \(W = G M m \frac{1}{R}\), where \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(R\) is the radius of Earth. The escape velocity is given by \(v_{e} = \sqrt{\frac{2 G M}{R}}\). For Earth to be considered a black hole, its radius would have to be compressed to approximately \(2.96 \times 10^{-3}\) meters.

Step-by-step solution

Identify the integral to be evaluated

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(F(x)=\frac{G M m}{x^{2}}\). We shall evaluate this integral to find the work in terms of \(m\).

Evaluate the integral

To find the work in terms of \(m\), we will evaluate the integral: $$ W = \int_{R}^{\infty} \frac{G M m}{x^{2}} d x $$ To evaluate, we will use the antiderivative of \(\frac{1}{x^2}\) which is \(-\frac{1}{x}\): $$ W = -G M m \left[ \frac{1}{x} \right]_{R}^{\infty} $$ Evaluating the limits, we get: $$ W = -G M m \left( \frac{1}{\infty} - \frac{1}{R} \right) $$ The term \(\frac{1}{\infty}\) approaches zero, thus: $$ W = G M m \frac{1}{R} $$ b. Finding the escape velocity

Equate work and kinetic energy

The escape velocity \(v_{e}\) is the velocity required for an object to have kinetic energy equal to the work done on the object. We are given the expression for kinetic energy: $$ \frac{1}{2} m v_{e}^{2} = W $$ We will now plug in the expression for work found in part a.

Solve for escape velocity

Equating work and kinetic energy, we get: $$ \frac{1}{2} m v_{e}^{2} = G M m \frac{1}{R} $$ Now solve for \(v_{e}\): $$ v_{e}^{2} = \frac{2 G M}{R} $$ $$ v_{e} = \sqrt{\frac{2 G M}{R}} $$ c. Conditions for a black hole

Set escape velocity equal to the speed of light

For a body to be considered a black hole, the escape velocity must be equal to or greater than the speed of light (\(c\)). We will now set the escape velocity equal to the speed of light: $$ c = \sqrt{\frac{2 G M}{R}} $$

Solve for the radius R

We will now solve for the radius \(R\): $$ R = \frac{2 G M}{c^{2}} $$ For Earth to be a black hole, we need to plug in the values \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}\) and \(c=300,000 \mathrm{km} / \mathrm{s}\): $$ R = \frac{2 (4 \times 10^{14} \mathrm{m^{3}/s^{2}})}{(300,000 \mathrm{km/s})^{2}} $$ $$ R \approx 2.96 \times 10^{-3} \mathrm{m} $$ So, for Earth to be considered a black hole, its radius would have to be compressed to approximately \(2.96 \times 10^{-3}\) meters.

Key concepts

Gravitational Force

Understanding gravitational force is crucial when exploring concepts like escape velocity and black holes. It's the force that attracts any two objects with mass. The strength of this force is determined by the universal law of gravitation, which can be expressed with the equation:

\[ F(x) = \frac{G M m}{x^{2}} \]
Where:

• \( F(x) \) is the gravitational force between two objects.
• \( G \) represents the gravitational constant, approximately \( 6.674 \times 10^{-11} \) m^3 kg^-1 s^-2.
• \( M \) and \( m \) are the masses of the two objects interacting gravitationally.
• \( x \) is the distance between the centers of the two masses.

This force decreases with the square of the distance, meaning gravity weakens rapidly as objects move farther apart, but it never becomes zero. Moreover, this gravitational interaction is mutual; Earth pulls the object as much as the object pulls on Earth, experiencing equal and opposite forces as per Newton's third law.

Work-Energy Principle

The work-energy principle connects force, displacement, and the energy transferred as work. In a gravitational context, work done by or against gravitational force is integral in calculating the escape velocity of an object. The work \( W \) done on an object to move it from Earth's surface to a point at infinity is represented by the integral:

\[ W = \int_{R}^{\infty} F(x) dx \]
The principle states that the work done equals the change in kinetic energy; hence, to escape Earth's gravity, an object requires energy to overcome gravitational attraction.

When calculating escape velocity, the energy imparted to the object becomes its kinetic energy, defined as \( \frac{1}{2} m v_e^2 \), where \( v_e \) is the escape velocity and \( m \) is the object's mass. Setting this equal to the work done and solving for velocity reveals the minimum speed needed for the object to break free from Earth's gravitational hold without further propulsion.

Black Holes

A black hole is an astronomical entity with a gravitational pull so intense that not even light, the fastest entity in the known universe, can escape from it. The concept was so radical that it took centuries from prediction to confirmation. The escape velocity formula we use for celestial bodies also applies to black holes:

\[ v_{e} = \sqrt{\frac{2GM}{R}} \]
When the escape velocity is equal to or exceeds the speed of light \( c \), the object could theoretically become a black hole. This implies a radius threshold of:

\[ R \leq \frac{2GM}{c^2} \]
This equation, deduced from general relativity, defines a black hole's event horizon, beyond which nothing can return. In our example calculation for Earth, the radius needed for it to be a black hole is minuscule and physically unfeasible, illustrating the extraordinary density and compactness of black holes.