The asteroid Icarus has a perihelion of 0.19 AU and an orbital eccentricity of 0.83. How do you calculate the asteroid's orbital semimajor axis and aphelion distance from the Sun?

Answer 1

Semi-Major Axis Distance:
#a = d_p/(1-e) = (0.19\quad AU)/(1-0.83) = 1.12\quad AU#
Aphelion Distance:
#d_a = a+c=a+ea=(1+e)a#
#\qquad=(1+0.83)\times1.12\quad AU = 2.04\quad AU#

#a# - Semi-major distance; #\quad b# - Semi-minor distance; #c# - Centre-to-Focal Point distance; #\quad e# - Eccentricity of the ellipse;
#d_p# - Perihelion distance; #\quad d_a # - Aphelion distance;
Eccentricity - Focal Point Relation: #e = c/a# ...... (1) Perihelion distance - Eccentricity Relation: #d_p = a-c = a-ea = (1-e)a# ...... (2)

Determine the semi-major axis distance using Equation 2.

Semi-Major Axis Distance: #a = d_p/(1-e) = (0.19\quad AU)/(1-0.83) = 1.12\quad AU# Aphelion Distance: #d_a = a+c=a+ea=(1+e)a# #\qquad=(1+0.83)\times1.12\quad AU = 2.04\quad AU#
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Answer 2

The formula for calculating the semimajor axis of an asteroid's orbit is a = (2 * q) / (1 - e), where 'a' is the semimajor axis, 'q' is the perihelion distance, and 'e' is the eccentricity. For example, for Icarus, the formula is a = (2 * 0.19) / (1 - 0.83). The formula for calculating the aphelion distance is Q = a * (1 + e). Substituting 'a' and 'e' into the formula yields Q = a * (1 + 0.83). Solve for 'a' and 'Q' to determine the semimajor axis and aphelion distance.

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Answer 3

To calculate the asteroid's orbital semimajor axis ((a)) and aphelion distance from the Sun, you can use the following formulas:

  1. Orbital semimajor axis ((a)) can be calculated using the formula: [a = \frac{{\text{{Perihelion}}}}{{1 - \text{{Eccentricity}}}}]

  2. Aphelion distance ((Q)) can be calculated using the formula: [Q = a \times (1 + \text{{Eccentricity}})]

Using the given values:

  • Perihelion ((r)) = 0.19 AU
  • Eccentricity ((e)) = 0.83

Substituting these values into the formulas: [a = \frac{{0.19}}{{1 - 0.83}}]

[Q = a \times (1 + 0.83)]

Calculating: [a = \frac{{0.19}}{{1 - 0.83}} = \frac{{0.19}}{{0.17}} ≈ 1.11765 , \text{AU}]

[Q = 1.11765 \times (1 + 0.83) ≈ 2.04758 , \text{AU}]

So, the asteroid's orbital semimajor axis ((a)) is approximately 1.11765 AU, and its aphelion distance from the Sun ((Q)) is approximately 2.04758 AU.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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