+91 The orbit of a comet around the sun is shaped like a ellipse in which the sun is at one focus. The two extrmeities of the major axis of a comet's orbit are called the comet's perihelion(closest distance to the sun) and the aphelion (furthest distance from the sun) These distances from the sun are usually expressed in astronomical units(AU), where 1AU is defined as the average distance between the sun and the earth, or about 93 million miles. The most famous comet in the solar system is Haley's comet, which has a perihelion of 0.59 AU and an aphelion of 35.09AU.
For the following assume that the orbit of Haleys comet is centered at the origin, and that the sun is located on the x-axis to the right of the origin. Give answers to two decimal places.
a) What is the length of the mjor axis of the orbit of Halley's Comet?
b) What are the coordinates of the sun?
c) Find the equation for the ellipse that describes the comet's orbit.
d) Find the eccentricity of the orbit to two decimal places from your equation...
+129899 Let the perhelion = .59 * 93,000,000 = 54,870,000 miles
Let the aphelion = 35.09 * 93,000,000 = 3,263,370,000 miles
(a) Then...the length of the major axis = 54,870,000 + 3,263,370,000 = 3,318,240,000 miles
Then the length of the semi-major axis = 3,318,240,000 / 2 = 1,659,120,000 miles = "c"
And the distance from the center of the ellipse to the Sun = "a" = 1,659,120,000 - 54,870,000 = 1,604,250,000 miles
(b) [ The co-ordinates of the Sun = (1,604,250,000 , 0 )
We need to find the length of the semi-minor axis = "b" thusly
b = √ [c^2 - a^2] = √ [ 1,659,120,000^2 - 1,604,250,000^2 ] ≈ 423,156,132 miles
So....our [approximate] equation is
x^2 /a^2 + y^2/b^2 = 1
(c) x^2 / 1,604,250,000^2 + y^2 / 423,156,132^2 = 1
The eccentricity = distance from the center to the foucus / length of the semi-major axis =
(d) 1,604,250,000 / 1,659,120,000 ≈ .9669 = .97
+129899 Let the perhelion = .59 * 93,000,000 = 54,870,000 miles
Let the aphelion = 35.09 * 93,000,000 = 3,263,370,000 miles
(a) Then...the length of the major axis = 54,870,000 + 3,263,370,000 = 3,318,240,000 miles
Then the length of the semi-major axis = 3,318,240,000 / 2 = 1,659,120,000 miles = "c"
And the distance from the center of the ellipse to the Sun = "a" = 1,659,120,000 - 54,870,000 = 1,604,250,000 miles
(b) [ The co-ordinates of the Sun = (1,604,250,000 , 0 )
We need to find the length of the semi-minor axis = "b" thusly
b = √ [c^2 - a^2] = √ [ 1,659,120,000^2 - 1,604,250,000^2 ] ≈ 423,156,132 miles
So....our [approximate] equation is
x^2 /a^2 + y^2/b^2 = 1
(c) x^2 / 1,604,250,000^2 + y^2 / 423,156,132^2 = 1
The eccentricity = distance from the center to the foucus / length of the semi-major axis =
(d) 1,604,250,000 / 1,659,120,000 ≈ .9669 = .97
