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.

Question

(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.

Answer 1

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