An asteroid has a 5:2 orbital resonance with Jupiter. What is its orbital period (in years)? How far is it from the Sun in AU? How does the distance between the orbits of Mars and the asteroid compare to the distance between the orbits of Mars and Jupiter? Part 1 of 3 Orbital resonances compare how many orbits a body, in this case an asteroid, makes in terms of the orbit of a much larger body, in this case Jupiter. We can write resonances as (number of orbits of smaller body): (number of orbits of larger body). Na: N₂ If we want to know the orbital period of the smaller body, we can express this as a relation. The second number tells us how many times the larger body orbits and the first number tells us how many times the smaller body orbits. If we want the orbital period, we just take the ratio of these and multiply by the orbital period of the larger body. N P₂= Na An asteroid has a 5:2 orbital resonance with Jupiter. What is its orbital period (in years)? Pa = years Submit Skip (you cannot come back)
An asteroid has a 5:2 orbital resonance with Jupiter. What is its orbital period (in years)? How far is it from the Sun in AU? How does the distance between the orbits of Mars and the asteroid compare to the distance between the orbits of Mars and Jupiter? Part 1 of 3 Orbital resonances compare how many orbits a body, in this case an asteroid, makes in terms of the orbit of a much larger body, in this case Jupiter. We can write resonances as (number of orbits of smaller body): (number of orbits of larger body). Na: N₂ If we want to know the orbital period of the smaller body, we can express this as a relation. The second number tells us how many times the larger body orbits and the first number tells us how many times the smaller body orbits. If we want the orbital period, we just take the ratio of these and multiply by the orbital period of the larger body. N P₂= Na An asteroid has a 5:2 orbital resonance with Jupiter. What is its orbital period (in years)? Pa = years Submit Skip (you cannot come back)
Applications and Investigations in Earth Science (9th Edition)
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ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
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An asteroid has a 5:2 orbital resonance with Jupiter. What is its orbital period (in years)?
How far is it from the Sun in AU?
How does the distance between the orbits of Mars and the asteroid compare to the distance between the orbits of Mars and Jupiter?
Part 1 of 3
Orbital resonances compare how many orbits a body, in this case an asteroid, makes in terms of the orbit of a much larger body, in this case Jupiter. We can write resonances as (number of orbits of smaller
body): (number of orbits of larger body).
Na: N₁
If we want to know the orbital period of the smaller body, we can express this as a relation. The second number tells us how many times the larger body orbits and the first number tells us how many times
the smaller body orbits. If we want the orbital period, we just take the ratio of these and multiply by the orbital period of the larger body.
Pa
=
Na
An asteroid has a 5:2 orbital resonance with Jupiter. What is its orbital period (in years)?
Pa
years
Submit Skip_(you cannot come back)
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![**Text Transcription:**
The region between Mars and Jupiter, where asteroids lie, extends from 1.52–5.20 AU from the Sun. To find the distance between Mars and this asteroid as a fraction of the total distance between Mars and Jupiter, we simply take their ratios:
\[ f = \frac{d_{Ma}}{d_{Mj}} \]
\[ f = \]
**Explanation:**
- **AU (Astronomical Unit):** A unit of measurement equivalent to the average distance from the Earth to the Sun, roughly 149.6 million kilometers (93 million miles).
- **Formula Explanation:**
- The formula \( f = \frac{d_{Ma}}{d_{Mj}} \) is used to calculate the fraction \( f \), which represents the distance between Mars and the asteroid as a fraction of the total distance between Mars and Jupiter.
- \( d_{Ma} \) is the distance from Mars to the asteroid.
- \( d_{Mj} \) is the distance from Mars to Jupiter.
- **Input Field:**
- Below the formula, there is an input field provided to enter the calculated fraction \( f \).](https://content.bartleby.com/qna-images/question/b60d532a-9aac-45a2-a466-49da1bb79676/794410e6-53b9-476f-b04e-2afda9740654/8wvgcgr_processed.png)
Transcribed Image Text:**Text Transcription:**
The region between Mars and Jupiter, where asteroids lie, extends from 1.52–5.20 AU from the Sun. To find the distance between Mars and this asteroid as a fraction of the total distance between Mars and Jupiter, we simply take their ratios:
\[ f = \frac{d_{Ma}}{d_{Mj}} \]
\[ f = \]
**Explanation:**
- **AU (Astronomical Unit):** A unit of measurement equivalent to the average distance from the Earth to the Sun, roughly 149.6 million kilometers (93 million miles).
- **Formula Explanation:**
- The formula \( f = \frac{d_{Ma}}{d_{Mj}} \) is used to calculate the fraction \( f \), which represents the distance between Mars and the asteroid as a fraction of the total distance between Mars and Jupiter.
- \( d_{Ma} \) is the distance from Mars to the asteroid.
- \( d_{Mj} \) is the distance from Mars to Jupiter.
- **Input Field:**
- Below the formula, there is an input field provided to enter the calculated fraction \( f \).
Solution
by Bartleby ExpertFollow-up Question
![Kepler's Third Law tells us how the distance from the Sun relates to the orbital period of a body:
\[
\left( \frac{a}{1 \text{ AU}} \right)^3 = \left( \frac{p}{1 \text{ yr}} \right)^2
\]
So the distance from the Sun in AU based on the period calculated in Step 1 is:
\[
a = \_\_\_ \text{ AU}
\]](https://content.bartleby.com/qna-images/question/b60d532a-9aac-45a2-a466-49da1bb79676/50a5d319-2920-4a6a-bf3c-7a689a62f268/4okh4qa_processed.png)
Transcribed Image Text:Kepler's Third Law tells us how the distance from the Sun relates to the orbital period of a body:
\[
\left( \frac{a}{1 \text{ AU}} \right)^3 = \left( \frac{p}{1 \text{ yr}} \right)^2
\]
So the distance from the Sun in AU based on the period calculated in Step 1 is:
\[
a = \_\_\_ \text{ AU}
\]
Solution
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