Question

Duende has an elliptical orbit about the Sun, with a perihelion distance of 0.829 AU and aphelion distance of 0.992 AU. What is the period of its orbit (in years)?

Answer 1

From Kepler's third law:

T^2 = (4*pi^2/GM)*a^3

Given that

T = time period of Duende's orbit = ?

M = mass of Sun = 2.0*10^30 kg

G = 6.67*10^-11

a = orbital radius of Duende = (perihelion distance + aphelion distance)/2 = (0.829 AU + 0.992 AU)/2 = 0.9105 AU

Since 1 AU = 1.496*10^11 m, So

a = 0.9105*1.496*10^11 m

So, Using above values:

T = sqrt (4*pi^2*a^3/(G*M))

T = sqrt (4*pi^2*(0.9105*1.496*10^11)^3/(6.67*10^-11*2.0*10^30))

T = 27347603.148 sec

Since 1 yr = 3.154*10^7 sec

So,

T = 27347603.148/(3.154*10^7) yr

T = 0.867 yrs = time period of Duende's orbit