Relative to the center of the Earth, the position of the Moon can be approximated by (t) = r [cos (ωt) + sin (ωt) ], where r = 3.84 108 m and ω = 2.46 10-6 radians/s.
(a) Find the magnitude and direction of the Moon's average velocity between t = 0 and t = 3.12 days. (This time is one-quarter of the 29.5 days it takes the Moon to complete one orbit.)
(b) Is the instantaneous speed of the Moon greater than, less than or the same as the average speed found in part (a)? Explain.
The position vector can be written as :
r= r{cos(wt)x + sin(wt)y}
whereas r=3.84x10^ 8 and w=2.46x10^-2
now the the average velocity is :
total displacement (divided by) time
if position vector at t=0 is r1 and at t=7.38 days is r2 then:
the resultant displacement is r2 -r1
putting values in the equation:
we get:
r1 = rx (because cos(0) = 1 and sin(0) =0)
r2= ry ( because cos 90 = 0 and sin 90=1)
note: you have to convert 7.38 days in to 7.38x24x60x60 seconds
and then convert the product of w and t into degrees to calculate the sin and cosine
thus we have ;
average velocity as ;
Vavg = ry-rx divided by t2-t1
the magnitude of this is given as ;
squareroot of {(r^2) + (-r^2)} divided by 7.38x24x60x60
= 852
and the direction is tan(inverse) of -1= 135
To find the average velocity of the Moon between t = 0 and t = 3.12 days, we need to calculate the displacement and divide it by the time interval.
(a)
Step 1: Calculate the displacement
The displacement is given by the difference in position at the two times:
Δr = r [cos(ωt2) + sin(ωt2)] - r [cos(ωt1) + sin(ωt1)]
Step 2: Calculate the time interval
The time interval is given by t2 - t1:
Δt = t2 - t1
Step 3: Calculate the average velocity
The average velocity is the displacement divided by the time interval:
v_avg = Δr / Δt
Now let's plug in the values:
r = 3.84 x 10^8 m
ω = 2.46 x 10^-6 radians/s
t1 = 0
t2 = 3.12 days = 3.12 x 24 x 60 x 60 seconds
Δr = r [cos(ωt2) + sin(ωt2)] - r [cos(ωt1) + sin(ωt1)]
Δr = 3.84 x 10^8 [cos(2.46 x 10^-6 x (3.12 x 24 x 60 x 60)) + sin(2.46 x 10^-6 x (3.12 x 24 x 60 x 60))] - 3.84 x 10^8 [cos(2.46 x 10^-6 x 0) + sin(2.46 x 10^-6 x 0)]
Δt = t2 - t1 = 3.12 x 24 x 60 x 60 seconds
v_avg = Δr / Δt
Calculate the above expressions to find the magnitude and direction of the Moon's average velocity between t = 0 and t = 3.12 days.
(b)
To determine if the instantaneous speed of the Moon is greater than, less than, or the same as the average speed found in part (a), we need to consider the nature of the given position equation.
The given position equation is of the form r(t) = r [cos(ωt) + sin(ωt)]. This equation represents simple harmonic motion, as it is a superposition of a cosine and sine wave.
In simple harmonic motion, the instantaneous speed of an object at any given time is equal to the magnitude of its velocity. The magnitude of the average velocity calculated in part (a) represents the average speed, which is the total distance traveled divided by the time taken.
Since the position equation represents simple harmonic motion, the instantaneous speed at any given time will vary. It will be maximum at the extremities of motion (when the sine and cosine terms reach their maximum values) and minimum at the middle point (when the sine and cosine terms reach their minimum values).
Therefore, the instantaneous speed of the Moon will be greater than, less than, or equal to the average speed depending on the specific time considered.