Question

The Sun rotates with the angular rate Ω₀=2.97×10⁻⁶ rad/s at the equator.

Calculate the sidereal rotation period of the Sun (in days) P₀ at the equator (that is the number of days it takes a sunspot at the equator to complete one full 360-degree rotation.)

Answer 1

To calculate the sidereal rotation period of the Sun at the equator, we use the given angular rate of 2.97×10−6 rad/s and the formula T = (2π) / Ω. The resulting sidereal rotation period is approximately 24.48 days.

Step-by-step explanation:

To calculate the sidereal rotation period of the Sun at the equator (labeled P0), given its angular rate (Ω0) of 2.97×10−6 rad/s, we can use the relationship between angular velocity and the period of rotation. The sidereal rotation period is the time it takes for an observer to see a sunspot complete one full 360-degree rotation relative to the distant stars, which differs slightly from the solar day due to Earth's orbit.

The formula to calculate the period (T) of rotation from angular velocity (Ω) is:

T = Ω−1

Since 2π radians is equivalent to one full rotation (360 degrees), the calculation is:

T = (2π) / Ω0

Using the given angular rate:

T = (2π) / (2.97×10−6 rad/s)

T = (6.283185307179586) / (2.97×10−6) s

T ≈ 2.115×106 s

Converting seconds to days (since there are 86,400 seconds in a day):

T = 2.115×106 s / 86,400 s/day

T ≈ 24.48 days

Therefore, the sidereal rotation period of the Sun at the equator is approximately 24.48 days.