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For a planet orbiting the sun, rp is the distance from the sun to the perihelion and ra is the distance from the sun to the aph...Asked by Jack_from_9th_grade
For a planet orbiting the sun, rp is the distance from the sun to the perihelion and ra is the distance from the sun to the aphelion. What is a planet’s orbital eccentricity if rp is equal to 0.45ra ?(1 point) Responses 0.55 0.55 2.64 2.64 0.38 0.38 1.45
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Answered by
Jack_from_9th_grade
Which statement summarizes Kepler's First Law of Planetary Motion?(1 point) Responses Planets move fastest while Earth is at perihelion. Planets move fastest while Earth is at perihelion. Planets in the solar system revolve around the sun. Planets in the solar system revolve around the sun. Planets closest to the sun move faster in their orbits than those farther away. Planets closest to the sun move faster in their orbits than those farther away. Planets revolve around the sun in elliptical orbits.
Answered by
Jack_from_9th_grade
Which of these can happen to objects in the solar system that do not have the momentum to maintain a stable orbit around the sun? Select the two correct answers.(1 point) Responses They can spiral into the sun. They can spiral into the sun. They can follow erratic orbits that take on different shapes. They can follow erratic orbits that take on different shapes. They can fall into circular orbits. They can fall into circular orbits. They can swing around the sun and be ejected outward into space.
Answered by
Jack_from_9th_grade
A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity if the satellite is moved to an orbital radius of 4r?(1 point) Responses 14v 1 fourth v 2v 2 v 12v 1 half v 4v
Answered by
Jack_from_9th_grade
What is a context clue?(1 point) Responses the feeling or impression a word gives readers the feeling or impression a word gives readers a dictionary definition for a word a dictionary definition for a word text near a word that helps clarify its meaning text near a word that helps clarify its meaning a detail that helps readers guess what might happen
Answered by
Jack_from_9th_grade
If generate means “to produce,” what does regenerate mean?(1 point) Responses to produce too many to produce too many to produce again to produce again to produce the opposite to produce the opposite to produce the most to produce the most
Answered by
GPT-4o mini
To find the orbital eccentricity \( e \) of a planet, we can use the following relationships for perihelion distance (\( r_p \)) and aphelion distance (\( r_a \)):
\[
r_p = a(1 - e)
\]
\[
r_a = a(1 + e)
\]
where \( a \) is the semi-major axis of the orbit.
From the problem, we know that \( r_p = 0.45 r_a \). Substituting the formulas for \( r_p \) and \( r_a \):
\[
a(1 - e) = 0.45 \cdot a(1 + e)
\]
Since \( a \) is common on both sides, we can divide both sides by \( a \) (assuming \( a \neq 0 \)):
\[
1 - e = 0.45(1 + e)
\]
Now, distribute the 0.45 on the right side:
\[
1 - e = 0.45 + 0.45e
\]
Next, combine the terms involving \( e \):
\[
1 - 0.45 = e + 0.45e
\]
\[
0.55 = e + 0.45e
\]
\[
0.55 = 1.45e
\]
Now, solve for \( e \):
\[
e = \frac{0.55}{1.45}
\]
\[
e \approx 0.3793
\]
Which can be rounded to \( e \approx 0.38 \).
So, the planet's orbital eccentricity is approximately **0.38**.
\[
r_p = a(1 - e)
\]
\[
r_a = a(1 + e)
\]
where \( a \) is the semi-major axis of the orbit.
From the problem, we know that \( r_p = 0.45 r_a \). Substituting the formulas for \( r_p \) and \( r_a \):
\[
a(1 - e) = 0.45 \cdot a(1 + e)
\]
Since \( a \) is common on both sides, we can divide both sides by \( a \) (assuming \( a \neq 0 \)):
\[
1 - e = 0.45(1 + e)
\]
Now, distribute the 0.45 on the right side:
\[
1 - e = 0.45 + 0.45e
\]
Next, combine the terms involving \( e \):
\[
1 - 0.45 = e + 0.45e
\]
\[
0.55 = e + 0.45e
\]
\[
0.55 = 1.45e
\]
Now, solve for \( e \):
\[
e = \frac{0.55}{1.45}
\]
\[
e \approx 0.3793
\]
Which can be rounded to \( e \approx 0.38 \).
So, the planet's orbital eccentricity is approximately **0.38**.
Answered by
GPT-4o mini
The statement that best summarizes Kepler's First Law of Planetary Motion is:
**Planets revolve around the sun in elliptical orbits.**
Kepler's First Law states that the orbits of planets around the Sun are elliptical, with the Sun at one of the foci of the ellipse.
**Planets revolve around the sun in elliptical orbits.**
Kepler's First Law states that the orbits of planets around the Sun are elliptical, with the Sun at one of the foci of the ellipse.
Answered by
GPT-4o mini
The two correct answers regarding what can happen to objects in the solar system that do not have the momentum to maintain a stable orbit around the Sun are:
1. **They can spiral into the sun.**
2. **They can swing around the sun and be ejected outward into space.**
Objects that do not have sufficient momentum may lose energy due to gravitational interactions or other forces, leading them to spiral into the Sun. Alternatively, they may have a close encounter with a massive object, gaining enough velocity to be ejected from the solar system entirely.
1. **They can spiral into the sun.**
2. **They can swing around the sun and be ejected outward into space.**
Objects that do not have sufficient momentum may lose energy due to gravitational interactions or other forces, leading them to spiral into the Sun. Alternatively, they may have a close encounter with a massive object, gaining enough velocity to be ejected from the solar system entirely.
Answered by
GPT-4o mini
To determine the velocity of a satellite in orbit at a different radius, we can use the formula for the orbital velocity of a satellite:
\[
v = \sqrt{\frac{GM}{r}}
\]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the orbital radius.
If the satellite moves to a new orbital radius of \( 4r \), the new velocity \( v' \) can be calculated as follows:
\[
v' = \sqrt{\frac{GM}{4r}} = \sqrt{\frac{1}{4}} \sqrt{\frac{GM}{r}} = \frac{1}{2} \sqrt{\frac{GM}{r}} = \frac{1}{2} v
\]
Thus, the velocity of the satellite when moved to an orbital radius of \( 4r \) is:
**1 half v**.
\[
v = \sqrt{\frac{GM}{r}}
\]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the orbital radius.
If the satellite moves to a new orbital radius of \( 4r \), the new velocity \( v' \) can be calculated as follows:
\[
v' = \sqrt{\frac{GM}{4r}} = \sqrt{\frac{1}{4}} \sqrt{\frac{GM}{r}} = \frac{1}{2} \sqrt{\frac{GM}{r}} = \frac{1}{2} v
\]
Thus, the velocity of the satellite when moved to an orbital radius of \( 4r \) is:
**1 half v**.
Answered by
GPT-4o mini
A context clue is:
**text near a word that helps clarify its meaning.**
Context clues provide hints or information surrounding a word within a sentence or passage that help readers understand the word's meaning.
**text near a word that helps clarify its meaning.**
Context clues provide hints or information surrounding a word within a sentence or passage that help readers understand the word's meaning.
Answered by
GPT-4o mini
Regenerate means:
**to produce again.**
The prefix "re-" typically implies "again" or "back," so regenerate refers to the process of producing something anew or restoring it.
**to produce again.**
The prefix "re-" typically implies "again" or "back," so regenerate refers to the process of producing something anew or restoring it.
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