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Chapter 16: Problem 48

What will the escape velocity be when the Sun becomes an AGB star with a radius 200 times greater and a mass only 0.7 times that of today? How will these changes in escape velocity affect mass loss from the surface of the Sun as an AGB star?

Short Answer

Expert verified
The escape velocity will be approximately 51.7 km/s. The reduced escape velocity will likely cause higher mass loss from the Sun's surface.

Step by step solution

01

Understanding the escape velocity formula

The escape velocity from the surface of a star is given by the formula: \[ v_e = \sqrt{ \frac{2GM}{R} } \] where \(v_e\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the star, and \(R\) is the radius of the star.
02

Sun's mass and radius as an AGB star

Given data: - New radius \(R_{AGB} = 200R_{Sun}\) - New mass \(M_{AGB} = 0.7M_{Sun}\) where \(R_{Sun}\) and \(M_{Sun}\) are the current radius and mass of the Sun.
03

Substitute values into the escape velocity formula

Substitute the given values: \[ v_{e,AGB} = \sqrt{ \frac{2G (0.7M_{Sun})}{(200R_{Sun})} } \]
04

Simplify the expression

Simplify the fraction inside the square root: \[ v_{e,AGB} = \sqrt{ \frac{2G (0.7M_{Sun})}{200R_{Sun}} } = \sqrt{ \frac{0.7 \times 2GM_{Sun}}{200R_{Sun}} } = \sqrt{ \frac{1.4GM_{Sun}}{200R_{Sun}} } \]
05

Factor out constants

Factor out constants and simplify further: \[ v_{e,AGB} = \sqrt{ \frac{GM_{Sun}}{R_{Sun}} \times \frac{1.4}{200} } = v_{e,Sun} \sqrt{ \frac{1.4}{200} } \] where \(v_{e,Sun} = \sqrt{ \frac{2GM_{Sun}}{R_{Sun}} }\) is the current escape velocity of the Sun.
06

Calculate the new escape velocity

Calculate the numerical value: \[ v_{e,AGB} = v_{e,Sun} \sqrt{ \frac{1.4}{200} } = v_{e,Sun} \sqrt{0.007} \] Given that \(v_{e,Sun} \approx 618 km/s \), \[ v_{e,AGB} \approx 618 \times \sqrt{0.007} \approx 51.7 km/s \]
07

Effects on mass loss

The decrease in escape velocity means particles on the surface require less energy to leave the star. Thus, the lower escape velocity during the AGB phase will likely lead to an increased rate of mass loss from the surface.

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Key Concepts

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

Escape Velocity
Escape velocity is the speed at which an object must travel to break free from a celestial body's gravitational pull. This critical speed ensures that the object doesn't fall back due to gravity. The formula to calculate escape velocity is given by:
\( v_e = \sqrt{ \frac{2GM}{R} } \)
Here, \( v_e \) is the escape velocity, \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.
When you plug in the mass and radius of the body, you get the escape velocity. For the Sun, as it transitions into an Asymptotic Giant Branch (AGB) star, we notice significant changes in these parameters, directly affecting the escape velocity.
Asymptotic Giant Branch (AGB) Star
An AGB star is a late-stage star undergoing significant changes. Here are the key points:
  • Characteristics: The star becomes much larger in size and its envelope expands.
  • Sun's Transformation: In the case of the Sun, it will reach a radius 200 times its current size but will only retain 70% of its mass.

These changes have profound effects on various properties.
As the Sun's radius increases drastically, the escape velocity decreases. This is because the formula \( v_e = \sqrt{ \frac{2GM}{R} } \) shows that a larger \( R \) results in a smaller \( v_e \), given the gravitational constant \( G \) and mass \( M \).
Mass Loss
Mass loss in stars, particularly AGB stars, is a critical process in stellar evolution.
With a lower escape velocity, particles on the star's surface need less energy to escape its gravitational pull. Here's how it works:
  • Reduced Escape Velocity: In our exercise, the Sun's escape velocity drops to approximately 51.7 km/s from 618 km/s.
  • Increased Particle Escape: Because of this reduction, more particles can leave the Sun's surface.
  • Enhanced Mass Loss: Consequently, during the AGB phase, the Sun will experience a higher rate of mass loss.

Understanding mass loss helps in studying how stars evolve and distribute elements in their surroundings. This phenomenon contributes to the enrichment of the interstellar medium and the future formation of new stars and planets.

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Most popular questions from this chapter

A white dwarf will become a nova if a. the original star was more than \(1.4 \mathrm{M}_{\odot}\) b. it accretes an additional \(1.4 M_{\circ}\) from a companion. c. even a small amount of mass falls on it from a companion. d. enough mass accretes from a companion to give the white dwarf a total mass of \(1.4 M_{\odot}.\)

Why are the paths along the H-R diagram that a star follows as it forms (from a protostar) and as it leaves the main sequence (climbing the red giant branch) so similar?

Stars begin burning helium to carbon when the temperature rises in the core. This temperature increase is caused by a. gravitational collapse. b. fusing hydrogen into helium in the core. c. fusing hydrogen into helium in a shell around the core. d. degenerate-electron pressure.

Each form of energy generation in stars depends on temperature. a. The rate of hydrogen fusion (proton-proton chain) near \(10^{7} \mathrm{K}\) increases with temperature as \(T^{4} .\) If the temperature of the hydrogen- burning core is raised by 10 percent, how much does the hydrogen fusion energy increase? b. Helium fusion (the triple-alpha process) at \(10^{8} \mathrm{K}\) increases with an increase in temperature at a rate of \(T^{40}\). If the temperature of the helium-burning core is raised by 10 percent, how much does the helium fusion energy increase?

Giant stars lose mass because a. the radiation pressure from fusion becomes so high. b. the mass of the star drops because of mass loss from fusion. c. the magnetic field causes increasing numbers of coronal mass ejections. d. the star swells until the surface gravity is too low to hold material.
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