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Ch. 36 - The Special Theory of Relativity
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 36 - The Special Theory of RelativityProblem 79
Chapter 35, Problem 79

The Sun radiates energy at a rate of about 4 x 10²⁶ W.
(a) At what rate is the Sun’s mass decreasing?
(b) How long does it take for the Sun to lose a mass equal to that of Earth?
(c) Estimate how long the Sun could last if it radiated constantly at this rate.

Verified step by step guidance
1
Step 1: Use Einstein's mass-energy equivalence formula, \( E = mc^2 \), to relate the energy radiated by the Sun to the decrease in its mass. Rearrange the formula to solve for the rate of mass loss: \( \frac{dm}{dt} = \frac{P}{c^2} \), where \( P \) is the power radiated by the Sun (\( 4 \times 10^{26} \ \text{W} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \ \text{m/s} \)).
Step 2: For part (b), calculate the time it takes for the Sun to lose a mass equal to that of Earth. Use the formula \( t = \frac{m}{\frac{dm}{dt}} \), where \( m \) is the mass of Earth (\( 5.97 \times 10^{24} \ \text{kg} \)) and \( \frac{dm}{dt} \) is the rate of mass loss calculated in step 1.
Step 3: For part (c), estimate how long the Sun could last if it radiated constantly at this rate. Use the formula \( t = \frac{M_{\text{Sun}}}{\frac{dm}{dt}} \), where \( M_{\text{Sun}} \) is the mass of the Sun (\( 1.989 \times 10^{30} \ \text{kg} \)) and \( \frac{dm}{dt} \) is the rate of mass loss calculated in step 1.
Step 4: Substitute the given values into the formulas derived in steps 1, 2, and 3. Ensure that the units are consistent (e.g., power in watts, speed of light in meters per second, and mass in kilograms). Perform the necessary algebraic manipulations to simplify the expressions.
Step 5: Interpret the results. For part (a), the rate of mass loss will be a very small value due to the large value of \( c^2 \). For part (b), the time to lose Earth's mass will be relatively short compared to the Sun's lifetime. For part (c), the Sun's lifetime estimate will be significantly longer, as its total mass is much larger than the rate of mass loss.

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Key Concepts

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

Mass-Energy Equivalence

Mass-energy equivalence, expressed by Einstein's equation E=mc², states that mass can be converted into energy and vice versa. In the context of the Sun, the energy it radiates is a result of nuclear fusion processes that convert mass into energy, leading to a decrease in the Sun's mass over time.
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Luminosity

Luminosity is the total amount of energy emitted by a star per unit time, measured in watts. The Sun's luminosity, approximately 4 x 10²⁶ W, indicates how much energy it radiates into space, which can be used to calculate the rate at which it loses mass through the conversion of mass to energy.
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Lifetime of a Star

The lifetime of a star is determined by its mass and the rate at which it consumes its nuclear fuel. For the Sun, its estimated lifetime is around 10 billion years, and by calculating the total energy it can produce and its current luminosity, we can estimate how long it will continue to radiate energy at its current rate.
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