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Question

"Mercury has a perihelion (closest distance) of 46 000 000km from the sun and an aphelion (furthest distance) of 69 820 000km. The sun has a radius of 695 500km. Find the eccentricity of Mercury's orbit and an equation for the orbit"

I'm thinking maybe the formule e=PF/PM comes into this but I dont know how to work out where the focus is :S

Answer 1

Let the Sun be at a focus at (c,0) and the right vertex at (a,0)
Then (-c,0) is the other focus and (-a,0) is the left vertex.
The distance from (c,0) to (a,0) is a-c = 4.6 X 10^7 km. The distance fron (c,0) to ( -a,0) is c+a = 6.982 X 10^7km.
Thus 2a = 11.582X10^7, which is the major axis of the ellipse, and and a = 5.791 X10^7 km, so (5.791X10^7,0) is the right vertex of the ellipse.

c = 6.982X10^7 - 5.791X 10^7 = 1.191X10^7 km.

b = (a^2-c^2)^.5 = 5.666X10^7. So 2b = minor axis =
2*5.666X10^7 = 11.332X107 km.

eccentricity = c/a = 1.191X10^7/5.791X10^7= 0.206

Equation of orbit is x^2/a^2 +y^2/b^2 = 1

Answer 2

the path is a simple ellipse. Perihelion plus the radius of the sun and aphelion plus the radius of the sun give the a and b values used to calculate eccentricity.
then assume the sun to be at 0,0 on the cartesian plane and aphelion on the x axis, perihelion on the y, then the equation is
x^2 / a^2 + y^2/b^2 = 1