Velocity of the Moon (magnitude and direction)

[corykowa]

Homework Statement

1. Homework Statement

The velocity of the Moon relative to the center of the Earth can be approximated by varrowbold(t) = v [−sin (ωt) xhatbold + cos (ωt) yhatbold], where v = 945 m/s and ω = 2.46 multiplied by 10−6 radians/s. (The time required for the Moon to complete one orbit is 29.5 days.) To approximate the instantaneous acceleration of the Moon at t = 0, calculate the magnitude and direction of the average acceleration during the following two time intervals.

(a) between t = 0 and t = 0.200 days
______ m/s
______ degrees (counterclockwise from the +x axis)

(b) between t = 0 and t = 0.0020 days
_____ m/s
_____ degrees (counterclockwise from the +x axis)

The Attempt at a Solution

When I try to solve for Vx and Vy I get the same result for each one, yielding zero, and if I try another method, i get very minute numbers. None of which appear to be correct. I'm at a loss for how to do this problem

 

[tiny-tim]

Welcome to PF!

Hi corykowa! Welcome to PF!

(on this forum, just use the bold (B) tag for vectors: v = v[−sin (ωt)i + cos (ωt)j] )

… calculate the magnitude and direction of the average acceleration during the following two time intervals.

What formula did you use for average acceleration?

Show us what you tried, and then we'll know how to help!

 

[corykowa]

Hi thanks for the help. I was thinking since the moon is moving around us, that it would be a centripetal acceleration problem? But maybe that's where I was going wrong. So I was using that formula in relation to vectors...

a = -lim (v/2*theta)2vsin(theta)ey as lim of theta approaches zero = (-v^2/R)ey

 

[tiny-tim]

Hi corykowa!

(just got up :zzz: …)

Hi thanks for the help. I was thinking since the moon is moving around us, that it would be a centripetal acceleration problem? But maybe that's where I was going wrong. So I was using that formula in relation to vectors...

a = -lim (v/2*theta)2vsin(theta)ey as lim of theta approaches zero = (-v^2/R)ey

(have a theta: θ and try using the X² tag just above the Reply box )

No, you're misreading the question.

It's only asking you for the average acceleration.

(a), over 0.2 days, will give you a figure close to the instantaneous acceleration, and (b), over 0.002 days, will give you a figure even closer.

The formula you need is average acceleration = (v(t₂) - v(t₁))/(t₂ - t₁)

 

[corykowa]

Oh thank you much! I just tried that equation and my response "differs from the correct answer by orders of magnitude". I'm not too sure what that means... I feel like i must be off by 10^x in my answer but not too sure how to go about figuring that out?? I got 945 when t=2 and 1.89 when t=.002. Also, I still don't understand how to solve for the direction (in relation to the +x axis). I figure it has something to do with arctan?? Thank you!

 

[tiny-tim]

Hi corykowa!

(just got up :zzz: …)

Oh thank you much! I just tried that equation and my response "differs from the correct answer by orders of magnitude". I'm not too sure what that means... I feel like i must be off by 10^x in my answer but not too sure how to go about figuring that out?? I got 945 when t=2 and 1.89 when t=.002. Also, I still don't understand how to solve for the direction (in relation to the +x axis). I figure it has something to do with arctan?? Thank you!

(try using the X² tag just above the Reply box )

Did you remember to convert the days into seconds?

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

(and y/x is the tangent of the angle from the x-azis)

 

[corykowa]

I had completely spaced converting days into seconds. However, what I ended up doing was:

differentiate v(t) = v( -sin(wt) + cos(wt)) which is
-v(w) (cos(wt) + sin(wt)) you have the acceleration
v(w) = 2.32 x 10^-3 meter/sec^2
At t = 0 days, the acceleration is - 2.32 x 10^-3 meters/sec^2

This appears to be what my homework was asking me as webassign accepted this as correct.

I'm still in the process of figuring out the direction part though. I'm trying: \theta = artan Vy/Vx

Thank you for your help!