Calculating Coriolis Force for Projectile on Rotating Earth

[quasar987]

I have this assignement (due tomorrow by the way) where I must compare the range of a projectile on the non-rotating Earth with the range of the projectile on the rotating earth. The question reads:

A projectile is lauched towards east from a point located north of the equator at the geographical latitude \lambda. Its initial speed v_0 makes and angle \alpha with respect to the ground. If the range of the projectile is R for \omega = 0, calculate the change in range cause by the rotation of the earth. Neglect terms of order \omega^2.

I asked the teacher and he said that the range is rather small, so we can consider the Earth locally flat.

For a rotating system at the surface of the earth, the equation of motion in the rotating coordinate system reads (Symon pp. 278 equ. 7.36)

m\frac{d^{*2}\vec{r}}{dt^2}= \vec{F} - m\vec{\omega}\times(\vec{\omega}\times\vec{r})-2m\vec{\omega}\times\frac{d^{*}\vec{r}}{dt}-m\frac{d\vec{\omega}}{dt}\times\vec{r}

We neglect terms containing \omega^2. This means we can neglect the centripetal force in the above equation. And the rotation of the Earth is constant so the last term vanishes too.

I chose a coordinate system located at the surface of the Earth and whose 3 axis points towards nord, east and vertical, each equiped with unit vectors respectively \hat{n}, \hat{e} and \hat{v}

I find that the angular speed is

\vec{\omega} = \omega cos\lambda \ \hat{n} + \omega sin \lambda \ \hat{v}

My dilemna lies in that by calculating the coriolis force, I get

-2\omega \left[ \left(cos \lambda \frac{d^*v}{dt}-sin\lambda \frac{d^*n}{dt} \right) \hat{e}-cos\lambda \frac{d^*e}{dt}\ \hat{v}+sin\lambda \frac{d^*e}{dt}\ \hat{n}\right]

So solving the equation of motion for either e(t), v(t) or n(t) would means solving 3 second order D.E. simultaneously! I doubt this is what is asked of me in this introductory course to special relativity (!)… On the other hand, my friend is convinced, though he cannot explain why, that the term d^*\vec{r}/dt in the coriolis force is to be taken as the speed function for a regular projectile, that is

\frac{d^*\vec{r}}{dt} = (v_0 cos \alpha)\hat{e} + (v_0 sin \alpha - gt)\hat{v}

So either way, it makes no sense to me. Please tell me what you think.

 

[vincentchan]

checking this type of question is extremely time consuming and tired... however, i did it
first... assuming you are on the northern hemisphere and \hat{v} pointing southward...(your original post define \hat{n} and \hat{e} very well but didn't show the direction of \hat{v}, in order to elimilate confusion.. I defined the \hat{v} for you in my way, sure you can assume \hat{v} point northward and change my sign)
secondl, people measure latitute from the equater... that's mean at the equater, \lambda=0 degree and at north pole, \lambda = 90 degree... you did it the other way... so I corrected you
\vec{\omega} = \omega sin\lambda \ \hat{n} - \omega cos \lambda \ \hat{v}
the coriolis force is just
-2m\ver{w}\times\ver{v}
the little star in your equation means ROTATIONAL FRAME...so
\ver{v}=v_{0} \hat{e}
the rest should be simple

 

[vincentchan]

oh, no sorry,
\ver{v} = v_{0} ( sin\omega \hat{n} + cos \omega \hat{e})-gt\hat{n}
forgot it is throwing up...

edit
is alpha instead of omega

 

[quasar987]

You're right, I should have included a graphic. Well here's one. The axis are defined by how a human located at the origin would say where vertical, north and east are.

 

[quasar987]

And here's a justification of omega.

 

[vincentchan]

well, then your coordinate is fine.. so do w and v... but the cross product seems funny...

 

[vincentchan]

replace
e by x,
n by y
v by z
and do it again... i am so sure you did something funny in you cross product

 

[quasar987]

Yea, so ok.. you're basically agreeing with my friend as to what d^*\vec{r}/dt is.

But how can you know what it is since finding d^*\vec{r}/dt comes from "integrating" d^{*2}\vec{r}/dt^2 once?!

 

[quasar987]

In other words, your v comes from the D.E.

\frac{d^{*2}\vec{r}}{dt^2} = -g\hat{v}

which "contradicts" the big equ with the coriolis force.

 

[vincentchan]

after you do the cross product, you will get something like
d^{*2}\vec{r}/dt^2 = ?? \hat{e} + ?? \hat{n} + ?? \hat{v}
those ?? is all combination of constant and gt...since w and v contain gt and some sine and cosine only
because you are in rotational frame... you can imagine you are now on a FLAT PLANE.. solve for v direction and get the time t the ball stays in air... you have the force and intitial velocity for the e and d direction... findint the distance shouldn't be difficult for you

 

[quasar987]

I'll do the cross product before your eyes.

Let's put d^*\vec{r}/dt = \vec{v} for simplicity.

\vec{v} = v_e(t)\hat{e} + v_n(t)\hat{n} + v_v(t)\hat{v}

\vec{\omega} = \omega cos\lambda \ \hat{n} + \omega sin \lambda \ \hat{v}

\Rightarrow \vec{\omega}\times \vec{v} = \omega cos\lamnda v_e (\hat{n}\times \hat{e}) + \omega cos\lambda v_v(\hat{n}\times \hat{v}) + \omega sin\lamnda v_e (\hat{v}\times \hat{e}) + \omega sin\lamnda v_n (\hat{v}\times \hat{n})
=\omega cos\lamnda v_e (-\hat{v}) + \omega cos\lambda v_v(\hat{e}) + \omega sin\lamnda v_e (\hat{n}) + \omega sin\lamnda v_n (-\hat{e})

= what i have written in post #1

 

[quasar987]

after you do the cross product, you will get something like
d^{*2}\vec{r}/dt^2 = ?? \hat{e} + ?? \hat{n} + ?? \hat{v}
those ?? is all combination of constant and gt...since w and v contain gt and some sine and cosine only

Yes, IF the speed is in fact given by \ver{v} = v_{0} ( sin\alpha \hat{n} + cos \alpha \hat{e})-gt\hat{n}, like you said. But what about my arguments of post 8 & 9 ? They are the reason I can't believe this is the expression of the speed in the rotating coord. sys.

 

[vincentchan]

i don't understand what you mean..
\frac{d^{*2}\vec{r}}{dt^2} = -g\hat{v}
the above equation is simply wrong because you ignore the corislis force

 

[quasar987]

Yes, that is my point, you will see. Let's solve the equation:

First, it is equivalent to the 3 scalar d.e.

\frac{d^*v_e}{dt} = 0
\frac{d^*v_n}{dt} = 0
\frac{d^*v_v}{dt} = -g

Let's solve each of these... we know what the answers are already, they are

v_e(t) = v_e_0 = v_0cos(\alpha)
v_n(t) = v_n_0 = 0
v_v(t) = v_v_0 t - gt = v_0sin(\alpha) - gt

Or, if we but it back to vectorial form,

\ver{v} = v_{0} ( sin\alpha \hat{v} + cos \alpha \hat{e})-gt\hat{v}

Which is the speed you assumed. That is to say, it is obtained by solving a wrong differential equation of motion. Like you said, the right one has the coriolis force term in it.


(Sorry for the hundreds of latex errors in the original message; I try to answer as fast as possible so you don't have to wait since you'Re kind enough to asist me in this mess ^^)

 

[vincentchan]

good point, because the v here is not the REAL v... the v here is assumed the coriolis force doesn't exist... and I understood what is your original question now
since we are doing approximation here only... you can replace the REAL v by your v... the error should be minimal since the real v and our v is very close...
I don't think there is an exact solution for your problem... and you are not asking for the exact solution here anyway...

 

[vincentchan]

easily see, the answer (if you do it by my method) is of the first order of omega... any higher order correction should be of the order of omega square... and the question is asking you to neglect them...

 

[quasar987]

AAAAaaahhhh ! Yes that's it! Well, thanks a lot... vincentsama .