Question

A comet has a very elliptical orbit with a period of 124.3 y. If the closest approach of the comet to the Sun is 0.09 AU, what is its greatest distance from the Sun?

AU

Answer 1

R_min = 0.09 AU = 0.09*1.5*10^11 = 1.35*10^10 m

T = 124.3 = 124.3*365*24*60*60 = 3.9*10^9 s

By Kepler’s third law,

T² = [(4π²)/(GM)]*a³

T² = [(4π²a³)/(GM)]

Where T = period

G=gravitational constant,

M= mass of the Sun = 1.99*10^30 kg

a= (R_max+R_min)

T² = [(4π²a³)/(GM)]

a³ = (T²GM)/(4π²)

(R_max+R_min)³ = (T²GM)/(4π²)

Thus

Plugging values,

(R_max+R_min)³ = [(3.9*10^9)^2*6.67*10^-11*1.99*10^30]/(4*3.14^2)

(R_max+R_min)³ = 5.12*10^37

Taking cube root,

R_max+R_min = 3.7*10^10 m

R_max= 3.7*10^10 m + R_min

Plug value of R_min

R_max= 3.7*10^10 + 1.35*10^10

R_max= 5.05*10^`10 m

Convert it in AU

R_max= (5.05*10^`10 m)/(1.5*10^11m )

R_max= 0.337 AU