Question

If you know the mass of a star and an object

Answer 1

Let the mass of the comet = m

radius of sun = 7.0 x 10⁸ m

closest distance = R_c = 8.3 x 10⁸ m + 7.0 x 10⁸ m = 15.3 x 10⁸ m

speed at closest distance = V_c

farthest distance = R_f = 2.4 x 10¹³ m.

speed at farthest distance = V_f

Using conservation of angular momentum ::

m V_f R_f = m V_c R_c

V_f R_f = V_c R_c                     

V_f = V_c R_c/R_f                                          Eq-1                 ('m'' cancel out)

gravitational Potential energy at farthest distance = -GM_s m/R_f

gravitational Potential energy at closest distance = -GM_s m/R_c

Kinetic energy at farthest distance = 1/2 m V_f²

Kinetic energy at closest distance = 1/2 m V_c²

Total energy at closest = -GM_s m/R_c + 1/2 m V_c²

Total energy at farthest = -GM_s m/R_f + 1/2 m V_f²

Using conservation of energy ::

-GM_s m/R_c + 1/2 m V_c²    = -GM_s m/R_f + 1/2 m V_f²

-GM_s /R_c + 1/2 V_c²    = -GM_s/R_f + 1/2 V_f²

GM_s/R_f - GM_s /R_c = 1/2 V_f² - 1/2 V_c²

2 GM_s ( 1/R_f - 1/R_c) = V_f² - V_c²

2 (6.67 x 10^-11) (2 x 10³⁰ ) ( 1/(2.4 x 10¹³ m.) - 1/ (15.3 x 10⁸ m)) =V_f² - V_c²

-1.74 x 10¹¹ = (V_c R_c/R_f)² - V_c²

-1.74 x 10¹¹ = ((R_c/R_f)² - 1 ) V_c²

-1.74 x 10¹¹ = (((15.3 x 10⁸)/(2.4 x 10¹³))² - 1 ) V_c²

V_c = 4.2 x 10⁵ m/s

V_f = V_c R_c/R_f = ( 4.2 x 10⁵)(15.3 x 10⁸)/(2.4 x 10¹³) = 26.78 m/s

b)

distance of earth from sun = R_e = 1.5 x 10¹¹

distance of comet from sun = R_c = 2.4 x 10¹³

time period of eartha around sun = 1 yr

Using kepler's law

(T_e/T_c)² = (R_e/R_c)³

1 / T_c² = (1.5 x 10¹¹ /2.4 x 10¹³ )³

T_c = 2023 yrs