Our sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit is 2.2 1020 m, and the angular speed of the sun is 1.2 10-15 rad/s.
(a) What is the tangential speed of the sun?
m/s
(b) How long (in years) does it take for the sun to make one revolution around the center?
y
a. 264000
a) V = (angular velocity) x (orbit radius)
b) Period = 2 pi/(angular velocity)
The answer will be in seconds, so you must convert that to years by dividing my the number of seconds in a year.
34.67
a. Vt= w*r
= (1.2*10^-15)*(2.2*10^20)
= 264000 m/s
b. w= period/time
1.2*10^-15= 2pi/t
t= 5.2*10^15 seconds
t= 1.7*10^8 yrs
To find the tangential speed of the sun, we can use the formula:
v = r * ω
where v is the tangential speed, r is the radius of the orbit, and ω is the angular speed.
(a) The radius of the sun's orbit is given as 2.2 * 10^20 m, and the angular speed is given as 1.2 * 10^-15 rad/s. Plugging these values into the formula:
v = (2.2 * 10^20 m) * (1.2 * 10^-15 rad/s)
Evaluating this expression, we get:
v ≈ 2.64 * 10^5 m/s
Therefore, the tangential speed of the sun is approximately 2.64 * 10^5 m/s.
Now let's move on to part (b). To find the time it takes for the sun to make one revolution around the center of the Milky Way galaxy, we can use the formula:
T = (2π) / ω
where T is the period of revolution and ω is the angular speed.
(b) Plugging in the given value for the angular speed, we have:
T = (2π) / (1.2 * 10^-15 rad/s)
Calculating this expression, we get:
T ≈ 5.24 * 10^15 seconds
To convert this time into years, we divide by the number of seconds in a year. Assuming there are approximately 3.15 * 10^7 seconds in a year, we have:
T ≈ (5.24 * 10^15 s) / (3.15 * 10^7 s/year)
Evaluating the expression, we get:
T ≈ 1.66 * 10^8 years
Therefore, it takes approximately 1.66 * 10^8 years for the sun to make one revolution around the center of the Milky Way galaxy.