The Moon orbits the Earth in an approximately circular path. The position of the moon as a function of time is given by: x (t) = r cos (ωt) y (t) = r sin (ωt) where r = 3.84 108 m and ω = 2.46 10-6 radians/s. What is the average velocity of the Moon measured over the interval from t = 0 to t = 3.92 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis.

Answer: Average Velocity = - 643.42 i + 512.66 j m/s

Magnitude = 822.7 m/s

Direction = 141.45°

Explanation:

r = 3.84 x 10^8 m

w = 2.46 x 10^-6 rad/s

Formula for Average velocity = displacement / time

at t = 0

x (0) = r

y (0) = 0

at t = 8.45 days

= 8.45 x 24 x 3600 s = 730080 sec

w t = 2.46 x 10^-6 x 730080 = 1.80 rad Or 102.90°

xf = r cos (w t) = - 0.2233r

yf = r sin (w t) = 0.9747r

Displacement = (xf - x0) i + (yf - y0) j = - 1.2233r i + 0.9747r j

= dispalcement / t = (-1.2233r i + 0.9747r j) / (730080 s)

= - 643.42 i + 512.66 j m/s

Magnitude

= sqrt (643.42^2 + 512.66^2)

= 822.7 m/s

Direction

= 180 - tan^-1 (512.66 / 643.42)

= 141.45°

Given

x (t) = r cos (ωt)

y (t) = r sin (ωt)

where r = 3.84*10⁸ m and ω = 2.46*10-⁶radians/s

Average velocity = Δx / Δt

At t = 0s

x (0) = 3.84*10⁸*cos (2.46*10-⁶*0)

x (0) = 3.84*10⁸*1 = 3.84*10⁸m

At t = 3.92days = 3.92*86400s = 3.39*10⁵s

x (3.39*10⁵) = 3.84*10⁸*cos (2.46*10-⁶*3.39*10⁵) = - 2.58*10⁸m

Δx = x (3.39*10⁵) - x (0) = - 2.58*10⁸-3.84*10⁸m = - 6.42*10⁸m

Δt = 3.39*10⁵ - 0 = 3.39*10⁵

vx = - 6.42*10⁸/3.39*10⁵ = - 1894m/s

At t = 0s

y (0) = 3.84*10⁸*sin (2.46*10-⁶*0)

y (0) = 3.84*10⁸*0 = 0m

At t = 3.92days = 3.92*86400s = 3.39*10⁵s

y (3.39*10⁵) = 3.84*10⁸*sin (2.46*10-⁶*3.39*10⁵) = 2.84*10⁸m

Δy = y (3.39*10⁵) - y (0) = 2.84*10⁸ - 0 = 2.84*10⁸m

Δt = 3.39*10⁵ - 0 = 3.39*10⁵

vy = 2.84*10⁸/3.39*10⁵ = 838m/s

vy = 838m/s

v = √ (vx² + vy²) = √ ((-1894) ² + 838²)

v = 2071m/s

θ = tan-¹ (vy/vx) = tan-¹ (838/1894) = - 23.9°