Calculate the speed with which the moon orbits the earth given the distance from earth to moon as R = 3.84 · 108 m. (Astronomers note that the true orbital period of the moon, is 27.3 Earth days. Interestingly, this would mean that there are approximately 13 months in a year. Use the 27.3 days/month for T - the time required for one revolution in your calculation.)

A. 2.78x10^4m/s

B. 2.41x10^4m/s

C. 2.45x10^4m/s

D. 1.02x10^3m/s

Question

Clay Mcdowell

Physics

6 April, 20:27

Calculate the speed with which the moon orbits the earth given the distance from earth to moon as R = 3.84 · 108 m. (Astronomers note that the true orbital period of the moon, is 27.3 Earth days. Interestingly, this would mean that there are approximately 13 months in a year. Use the 27.3 days/month for T - the time required for one revolution in your calculation.)

A. 2.78x10^4m/s

B. 2.41x10^4m/s

C. 2.45x10^4m/s

D. 1.02x10^3m/s

+2

Answers (1)

Know the Answer?

Matilda Mcclure · Answer 1

D. 1.02*10³ m/s

Explanation:

The radius of the moon's orbit is 3.84*10⁸ m. So the circumference is:

C = 2πR

C = 2π (3.84*10⁸ m)

C = 2.41*10⁹ m

The period is 27.3 days. Converting to seconds:

T = 27.3 days * (24 hrs / day)

T = 655.2 hrs * (3600 s / hr)

T = 2.36*10⁶ s

So the speed is:

v = C / T

v = (2.41*10⁹ m) / (2.36*10⁶ s)

v = 1.02*10³ m/s