Question

A comet of mass 2.20 ×1014 kg orbits the sun of mass 1.99 × 1030 kg with a period of 365 days/year on an elliptical orbit of eccentricity 0.967.

a) Find the Perihelion and Aphelion, i.e. the closest and farthest distance between the sun and the comet.

b) Find the linear speed of the comet at the Perihelion.

Answer 1

A comet of mass m = 2.20 ×10¹⁴ kg

the sun of mass M = 1.99 × 10³⁰ kg

Time period of planet T = 365 days

elliptical orbit of eccentricity = 0.967

Universal gravity constant G = 6.67408 × 10^-11 m³ kg^-1 s^-2

the constant = GM = 1.99 × 10³⁰ x 6.67408 × 10^-11 = 1.32815 x 10²⁰

From III law, areal velocity (h/2) = 3.14 x ab / T

               h² = (2ab/T)² --------------------(1)

, where a is Perihelion and b is Aphelion

From keplar III law

h² = b²/a -------------------(2)

From (1) and (2)

(2ab/T)² = b²/a

a³ = T²/2 = (365x24x60x60)²(1.32815 x 10²⁰)/2

a = 276036.6x10⁶ m

But the ecentricity of the elliptical orbit is

² = 1 - (b/a)²

b² = (1 - ² )a² = ( 1 - 0.967²)(276036.6x10⁶)² = 4945971832.8x10¹²

b = 69613 x10⁶ m

b)

Average Velocity of planet

V = 22721.618 m/s