Finding Maximum Distance from Sun

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Homework Help Overview

The discussion revolves around finding the maximum distance (aphelion) from the Sun for an extreme elliptical orbit, given a semi-major axis of 26. Participants explore the relationship between eccentricity, perihelion, and aphelion in the context of elliptical orbits.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of eccentricity and its implications for extreme elliptical orbits, questioning what value should be used. There are inquiries about the smallest possible perihelion and how it relates to the distance from the Sun.

Discussion Status

Some participants have provided hints and guidance regarding the relationship between perihelion and aphelion, while others are exploring the implications of eccentricity on the shape of the orbit. The conversation reflects a mix of interpretations and attempts to clarify concepts without reaching a definitive conclusion.

Contextual Notes

Participants note that the Sun is positioned at one of the foci of the ellipse, which influences the calculations of perihelion and aphelion. There is also mention of constraints related to the distance of comets from the Sun.

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Homework Statement


Find the maximum (aphelion) distance from the Sun for extreme elliptical orbit. (Semi-Major axis = 26)

Homework Equations


Eccentricity = Distance between foci/Length of major axis
Major Axis = 2 * Semi-Major Axis
Major Axis = Perihelion + Aphelion

The Attempt at a Solution


Eccentricity = Distance / (26*2) = Distance / 52
Eccentricity of an elliptical orbit is between 0 and 1, 0 being a circle.
What eccentricity should I be using for an extreme elliptical orbit?
My best guess would be right under 1 (ex: 0.99) but I'm not sure if that makes sense. (Would it be too close to the sun?)

Then once I have the distance between the foci, I would do semi-major axis - 1/2 distance between foci, in order to get the perihelion. From there I can get the aphelion.
 
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What is the smallest possible value for perihelion?
 
Hm I'm not sure how to find this, it's not anywhere in my textbook and I've tried Googling this too. What can I do to find the smallest perihelion value?
 
Think about it for a bit. If I gave you any better hint than I did I would have given the answer away.
 
Remember that the sun is at a focus of the ellipse.
How does the distance between foci (and shape and position of the ellipse) change as the eccentricity becomes larger?
 
D H said:
Think about it for a bit. If I gave you any better hint than I did I would have given the answer away.

Hm.. The closest a comet can get without going through the sun would be r(comet)+r(sun), but that would mean it's touching the sun. It would have to be a little bigger than this? Am I on the right track?

mnova said:
Remember that the sun is at a focus of the ellipse.
How does the distance between foci (and shape and position of the ellipse) change as the eccentricity becomes larger?

As the eccentricity increases, the orbit goes from being a circle to a line, so the distance of the foci approaches the major axis of the whole orbit.. I'm not sure where to go from there.
 
colourpalette said:
Hm.. The closest a comet can get without going through the sun would be r(comet)+r(sun), but that would mean it's touching the sun. It would have to be a little bigger than this? Am I on the right track?
Correct. So what does this tell you about perihelion, and thus about apihelion? (Hint: You know what the semi-major axis is.)
 
D H said:
Correct. So what does this tell you about perihelion, and thus about apihelion? (Hint: You know what the semi-major axis is.)

This means the perihelion is r(comet) + r(sun)? And therefore the aphelion would be 2(semi-major axis) - perihelion.
Aphelion = 2(26AU * 149 598 000 km/AU) - r(comet) - r(sun)
= 7 779 096 000 km - r(comet) - 695 500 km
I don't know the r(comet) but I'm assuming it would be negligible at this point? If so I'd get *huge number* and converting it back to AU would make it 51.995 AU! Is this correct?
 
Correct.
 
  • #10
Thank you!
 

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