Hi! I really need help with the following please:
The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical. The Sun is located at a focus of the ellipse.
The perihelion is the point in a planet’s orbit that is closest to the Sun. So, it is the endpoint of the major axis that is closest to the Sun.
The aphelion is the point in the planet’s orbit that is furthest from the Sun. So, it is the endpoint of the major axis that is furthest from the Sun.
The closest Mercury comes to the Sun is about 46 million miles. The farthest Mercury travels from the Sun is about 70 million miles.
(Score for Question 1: ___ of 3 points)
What is the distance between the perihelion and the aphelion?
Answer:
(Score for Question 2: ___ of 5 points)
What is the distance from the center of Mercury’s elliptical orbit and the Sun?
Answer:
(Score for Question 3: ___ of 8 points)
Write the equation of the elliptical orbit of Mercury, where the major axis runs horizontally. Allow a and b to be measured in millions of miles. Use the origin as the center of the ellipse.
Answer:
(Score for Question 4: ___ of 2 points)
What is the eccentricity of the ellipse? Round your answer to the nearest thousandth.
Answer:
(Score for Question 5: ___ of 2 points)
What does the value of the eccentricity tell you about the relative shape of the ellipse?
Answer:
Thank you very much in advance!
Daniel.
+118687 Looks like you are doing a test right now.
Chris if you want to answer these could you please leave it for a couple of hours.
+118687 I am not good with conics so I probably will not attempt to help anyway .
But if I was to help I would want you to spell out what you had tried in order to help yourself first.
+1713 Okay, nvm, it's conics...
And this very suspiciously looks like homework...
+1713 conics and math
idk the points are just staring at me in the face...
+129899 1) What is the distance between the perihelion and the aphelion?
70 + 46 = 116 million miles
2) What is the distance from the center of Mercury’s elliptical orbit and the Sun?
The major axis is 116 million miles
This = 2a
So
2a = 116
a = 58
This is the distance from the center of the ellipse to one endpoint on the major axis
And c = the distance from the center of its elliptical orbit to the Sun = 58 - 46 = 12 million miles
3) Write the equation of the elliptical orbit of Mercury, where the major axis runs horizontally. Allow a and b to be measured in millions of miles. Use the origin as the center of the ellipse.
b = sqrt [ 58^2 - 12^2] = sqrt (3220)
b^2 = 3220
a^2 = 58^2 = 3364
The equation is
x^2 y^2
____ + ____ = 1
3364 3220
4) Eccentricity = c/a = 12 / 58 ≈ .207
5) The closer to 0 the eccentrcity is, the more circular the ellipse....the nearer to 1, the more elongated
Thus.....the elliptical path here is somewhat circular
Here's a graph : https://www.desmos.com/calculator/cj4ey4e2gx
+129899 No
Imagine the focus is the Sun
The nearest that Mercury is to the Sun is 46 million miles....this is the distance a - c
The farthest that Mercury is to the Sun is 70 million miles....this is the distance a + c
So adding these distances we have (a - c) + (a + c) = 2a = the length of the major axis =116 million miles
