A mercury like planet has a semi major axis of 0.371 AU has and an eccentricity of 0.22 what is the perihelion distance in AU

Answer 1

Hello Rose!

Knowns:

a = 0.371 [AU]

e = 0.22 [unitless]

Useful Formulas:

r_p = a(1-e)

r_a = a(1+e)

Note: a = semi-major axis

e = eccentricity

r_p = radius at periapsis

r_a = radius at apoapsis

Solution:

r_p = (0.371 [AU]) * (1 - 0.22)

= 0.29 [AU] ~ 0.3 [AU]