A star spins one revolution every 10 days and its radius is 7.00*105 km. It then collapses and becomes a neutron star with a radius of 20.0 km. It also loses one half its mass during the collapse. ). What is the new revolutions /day? (The moment of inertia of a star is 2/5 mr²) O 3.13*108 rev/day O 3.20*108 rev/day O 3.32*108 rev/day 1.70*108 rev/day O 2.45*108 rev/day
A star spins one revolution every 10 days and its radius is 7.00*105 km. It then collapses and becomes a neutron star with a radius of 20.0 km. It also loses one half its mass during the collapse. ). What is the new revolutions /day? (The moment of inertia of a star is 2/5 mr²) O 3.13*108 rev/day O 3.20*108 rev/day O 3.32*108 rev/day 1.70*108 rev/day O 2.45*108 rev/day
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter34: Frontiers Of Physics
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Problem 21PE: Construct Your Own Problem Consider a star moving in a circular orbit at the edge at a galaxy....
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![### Angular Momentum and Neutron Star Formation
In this educational example, we explore the fascinating dynamics of a star's collapse into a neutron star and how its rotational characteristics change due to conservation of angular momentum.
#### Scenario:
- A star completes one revolution every 10 days.
- Initial radius: \(7.00 \times 10^5 \) km.
- The star collapses into a neutron star:
- New radius: 20.0 km.
- The star loses one half of its mass during the collapse.
#### Question:
**What is the new rate of revolutions per day?**
(The moment of inertia of a star is given by \( \frac{2}{5} mr^2 \).)
#### Multiple Choice Options:
- \(3.13 \times 10^8 \) rev/day
- \(3.20 \times 10^8 \) rev/day
- \(3.32 \times 10^8 \) rev/day
- \(1.70 \times 10^8 \) rev/day
- \(2.45 \times 10^8 \) rev/day
#### Explanation:
To find the new revolutions per day, consider the conservation of angular momentum. Angular momentum \(L\) is conserved and is defined as \(L = I \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
1. **Initial Moment of Inertia (\(I_{\text{initial}}\)):**
\[
I_{\text{initial}} = \frac{2}{5} m r_{\text{initial}}^2
\]
Given:
- Initial radius \( r_{\text{initial}} = 7.00 \times 10^5 \) km.
2. **New Moment of Inertia (\(I_{\text{final}}\)):**
\[
I_{\text{final}} = \frac{2}{5} \left( \frac{m}{2} \right) r_{\text{final}}^2
\]
Given:
- Final radius \( r_{\text{final}} = 20.0 \) km.
- The mass is halved: \( \frac{m}{2} \).
3. **Calculate Initial Angular Velocity (\(\omega_{\text{initial}}\)):**
- The period (\(T\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff63a57a-dcd1-4774-86da-c01ab69a7706%2F606578be-66f8-4e8a-a5f7-df321ba33e1d%2Fgs40f6u_processed.png&w=3840&q=75)
Transcribed Image Text:### Angular Momentum and Neutron Star Formation
In this educational example, we explore the fascinating dynamics of a star's collapse into a neutron star and how its rotational characteristics change due to conservation of angular momentum.
#### Scenario:
- A star completes one revolution every 10 days.
- Initial radius: \(7.00 \times 10^5 \) km.
- The star collapses into a neutron star:
- New radius: 20.0 km.
- The star loses one half of its mass during the collapse.
#### Question:
**What is the new rate of revolutions per day?**
(The moment of inertia of a star is given by \( \frac{2}{5} mr^2 \).)
#### Multiple Choice Options:
- \(3.13 \times 10^8 \) rev/day
- \(3.20 \times 10^8 \) rev/day
- \(3.32 \times 10^8 \) rev/day
- \(1.70 \times 10^8 \) rev/day
- \(2.45 \times 10^8 \) rev/day
#### Explanation:
To find the new revolutions per day, consider the conservation of angular momentum. Angular momentum \(L\) is conserved and is defined as \(L = I \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
1. **Initial Moment of Inertia (\(I_{\text{initial}}\)):**
\[
I_{\text{initial}} = \frac{2}{5} m r_{\text{initial}}^2
\]
Given:
- Initial radius \( r_{\text{initial}} = 7.00 \times 10^5 \) km.
2. **New Moment of Inertia (\(I_{\text{final}}\)):**
\[
I_{\text{final}} = \frac{2}{5} \left( \frac{m}{2} \right) r_{\text{final}}^2
\]
Given:
- Final radius \( r_{\text{final}} = 20.0 \) km.
- The mass is halved: \( \frac{m}{2} \).
3. **Calculate Initial Angular Velocity (\(\omega_{\text{initial}}\)):**
- The period (\(T\
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