posted by silvia .
Mercury has a semi-major axis of 0.367AU and an eccentricity of e=0.206. Calculate its distance to the Sun at perihelion. Remember to use units of AU in your answer.
let c be the distance from the center to the focus
let a be the semi-major axis
e = c/a
c = e*a = .206*.367 = .0756
distance at perihelion = a-c = .367 - .0756 = 0.291
oh, yeah. AU (duh)
semi-major axis = a = 0.367 au
closest distance = (1-e)*a = 0.794 a
= 0.291 au