Question

The sun travels about the center of our galaxy in a nearly circular orbit of radius 2.50 x 10^17 km in a period of about 2.00 x 10^8 yr. Compute the magnitudes of the velocity and the acceleration of the sun relative to the center of the galaxy. (1 yr= 3.156 x 10^7 s) Please show formulas.

Answer 1

here radius of circular path R = 2.50 x 10¹⁷ Km = 2.50 x 10²⁰ m

time period T = 2 x 10⁸ yr = 2 x 10⁸ ( 3.154 x 10⁷ ) = 6.308 x 10¹⁵ s

as we know that the angula velocity

w = 2 x pie x f

w = ( 2 x pie ) / T

= ( 2 x 3.14) / (6.308 x 10¹⁵ ) = 0.99556 x 10^-15 rad/s

so the velocity is given by

V = wR

= 0.99556 x 10^-15 x 2.50 x 10²⁰

= 2.48890298 x 10⁵ m/s

= 248890.298 m/s Ans

and the acceleration is given by

a = w²R

= (0.99556 x 10^-15)² x 2.50 x 10²⁰

= 2.4778492 x 10^-10 m/s²   Ans