An asteroid has a perihelion distance of 2.0 AU, an aphelion distance of 4.0 AU, and an orbital semimajor axis of 3.0 A.U. Calculate its eccentricity
An asteroid has a perihelion distance of 2.0 AU, an aphelion distance of 4.0 AU, and an orbital semimajor axis of 3.0 A.U. Calculate its eccentricity
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![### Calculating the Eccentricity of an Asteroid's Orbit
An asteroid has a perihelion distance (the closest point to the Sun) of 2.0 AU (Astronomical Units), an aphelion distance (the farthest point from the Sun) of 4.0 AU, and an orbital semimajor axis of 3.0 AU. To find the orbital eccentricity of the asteroid, we use the following formula:
\[ e = \frac{a - p}{a + p} \]
where:
- \( e \) = eccentricity
- \( a \) = aphelion distance
- \( p \) = perihelion distance
Given:
- \( p = 2.0 \) AU
- \( a = 4.0 \) AU
Substitute these values into the formula:
\[ e = \frac{4.0 \, \text{AU} - 2.0 \, \text{AU}}{4.0 \, \text{AU} + 2.0 \, \text{AU}} \]
\[ e = \frac{2.0 \, \text{AU}}{6.0 \, \text{AU}} \]
\[ e = \frac{1}{3} \]
\[ e = 0.33 \]
Therefore, the eccentricity of the asteroid's orbit is 0.33.
This eccentricity value indicates that the orbit is slightly elliptical, since a perfectly circular orbit would have an eccentricity of 0. An eccentricity closer to 1 would indicate a much more elongated orbit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fe8677b-d2b4-4cf1-b1ed-08820154fcb5%2F83a6c8d6-86fb-438a-8848-56cdf4518358%2Frmjkgv2_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Eccentricity of an Asteroid's Orbit
An asteroid has a perihelion distance (the closest point to the Sun) of 2.0 AU (Astronomical Units), an aphelion distance (the farthest point from the Sun) of 4.0 AU, and an orbital semimajor axis of 3.0 AU. To find the orbital eccentricity of the asteroid, we use the following formula:
\[ e = \frac{a - p}{a + p} \]
where:
- \( e \) = eccentricity
- \( a \) = aphelion distance
- \( p \) = perihelion distance
Given:
- \( p = 2.0 \) AU
- \( a = 4.0 \) AU
Substitute these values into the formula:
\[ e = \frac{4.0 \, \text{AU} - 2.0 \, \text{AU}}{4.0 \, \text{AU} + 2.0 \, \text{AU}} \]
\[ e = \frac{2.0 \, \text{AU}}{6.0 \, \text{AU}} \]
\[ e = \frac{1}{3} \]
\[ e = 0.33 \]
Therefore, the eccentricity of the asteroid's orbit is 0.33.
This eccentricity value indicates that the orbit is slightly elliptical, since a perfectly circular orbit would have an eccentricity of 0. An eccentricity closer to 1 would indicate a much more elongated orbit.
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