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Chapter 39: Problem 68

Imagine that the entire Sun collapses to a sphere of radius \(R_{g}\) such that the work required to remove a small mass \(m\) from the surface would be equal to its rest energy \(m c^{2} .\) This radius is called the gravitational radius for the Sun. Find \(R_{g}\) (It is believed that the ultimate fate of very massive stars is to collapse beyond their gravitational radii into black holes.)

Short Answer

Expert verified
The gravitational radius of the Sun, if it were to collapse into a black hole, would be approximately 2.96 kilometers.

Step by step solution

01

Understanding given variables

For the exercise, some known constants are: the speed of light \( c = 3 \times 10^{8} m/s \), the gravitation constant \( G = 6.67 \times 10^{-11} m^{3} kg^{-1} s^{-2} \), and the mass of the sun \( M_\odot = 1.99 \times 10^{30} kg \). The exercise asks to find the gravitational radius \( R_{g} \) where the work required to remove a small mass \( m \) equals its rest energy.
02

Setting up the equation

According to Einstein鈥檚 mass-energy equivalence, the rest energy \( E \) of a body with mass \( m \) is given by the formula \( E = mc^{2} \). The Work-Energy Theorem tells us this energy will be equal to the work done to remove a small mass \( m \) from the surface of the Sun, which corresponds to the gravitational potential energy (negative because it's bound), given by the formula \( W = - \frac{GM_\odot m}{R_{g}} \). Equating these, we have \( - \frac{GM_\odot m}{R_{g}} = mc^{2} \).
03

Solving for \( R_{g} \)

To solve for \( R_{g} \), clear the equation of \( m \) by dividing each side by \( m \) (since \( m \neq 0 \)). This gives: \( - \frac{GM_\odot}{R_{g}} = c^{2} \). Now isolate \( R_{g} \) by multiplying each side by \( -R_{g} \) and divide each side by \( c^{2} \): \( R_{g} = \frac{GM_\odot }{c^{2}} \).
04

Substituting in values

Substitute the known values into the equation to find \( R_{g} \): \( R_{g} = \frac{(6.67 \times 10^{-11} m^{3} kg^{-1} s^{-2}) (1.99 \times 10^{30} kg)}{(3 \times 10^{8} m/s)^{2}} \).
05

Calculating the result

Perform the calculation to find that \( R_{g} = 2.96 \times 10^{3} \) meters, or approximately 2.96 kilometers.

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Key Concepts

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

Mass-Energy Equivalence
One of the most famous equations in physics is Albert Einstein's mass-energy equivalence formula, expressed as \( E = mc^2 \). This equation describes how mass (\

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Most popular questions from this chapter

A Klingon spacecraft moves away from the Earth at a speed of \(0.800 c\) (Fig. \(\mathrm{P} 39.26\) ). The starship Enterprise pursues at a speed of \(0.900 c\) relative to the Earth. Observers on the Earth see the Enterprise overtaking the Klingon craft at a relative speed of \(0.100 c .\) With what speed is the Enterprise overtaking the Klingon craft as seen by the crew of the Enterprise?

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