Compensating for Earth's Rotation With v = ωr

[nmnna]

Homework Statement

At what speed and in what direction must the plane fly over the equator at height $h$ so that the Sun is always in the same place in space for it?

Relevant Equations

$$v=\omega r$$

We know that $v = \omega r$ where $r = R_{\text{E}} + h$. To compensate for the motion, the plane must fly along the equator at the same speed as the Earth but in the opposite direction, i.e. from east to west, so
$$\vec{v} = -\vec{ v}_{\text{E}}$$
$$v_{\text{E}} = \omega_{\text{E}} R_{\text{E}}$$
By using $\omega = \frac{2\pi}{T}$, we can find that $\omega_{\text{E}} = \frac{\pi}{12}$
Please give me some hints for the solution, I'm struggling to find the right steps from here so I'd really appreciate your help.
The answer in my textbook is $v = \frac{R_{\text{E}} + h}{2h}$

 

[PeroK]

The answer in my textbook is $v = \frac{R_{\text{E}} + h}{2h}$

The dimensions of that answer are wrong. Also, the rotation of the Earth must be a factor, surely?

 

[nmnna]

the rotation of the Earth must be a factor, surely?

I think so

 

[PeroK]

I think so

So, the textbook answer can't be right. What answer do you get?

 

[nmnna]

So, the textbook answer can't be right. What answer do you get?

$v = (R_{\text{E}} + h) \cdot \frac{\pi}{12}$

 

[PeroK]

$v = (R_{\text{E}} + h) \cdot \frac{\pi}{12}$

The dimensions are equally wrong. You can't have a speed equal to a distance.

 

[nmnna]

The dimensions are equally wrong. You can't have a speed equal to a distance.

I think my assumptions eariler were wrong, $\vec{v}$ does not have the same magnitude as $\vec{v}_{\text{E}}$. The angular velocity of the plane must be equal to the angular velocity of the Earth, then the Sun will be the center of the circle relative to which the Earth with the plane is moving. So $\omega = \omega_{\text{E}} = \frac{2\pi}{24} \frac{\text{rad}}{\text{h}}$, then $v = \omega_{\text{E}} (R_{\text{E}} + h)$
If we take $(R_{\text{E}} + h)$ in kilometers then the dimensions are correct
Is this right?

 

[hmmm27]

I think you should start by defining the scope of the system, and the reference frame of the dimensions. Not to sound too soapboxy, but defining the problem goes a long way towards finding the answer.

 

[PeroK]

I think my assumptions eariler were wrong, $\vec{v}$ does not have the same magnitude as $\vec{v}_{\text{E}}$. The angular velocity of the plane must be equal to the angular velocity of the Earth, then the Sun will be the center of the circle relative to which the Earth with the plane is moving. So $\omega = \omega_{\text{E}} = \frac{2\pi}{24} \frac{\text{rad}}{\text{h}}$, then $v = \omega_{\text{E}} (R_{\text{E}} + h)$
If we take $(R_{\text{E}} + h)$ in kilometers then the dimensions are correct
Is this right?

It depends what format the answer should take. I would have said that $$v = \frac{2\pi(R_E + h)}{T}$$ where $T = 1$ day. Then you can plug in $R_E, h$ and $T$ as appropriate. But, yes, $w_E = \frac{2\pi}{T}$, so that would work as well.

 

[haruspex]

$v = (R_{\text{E}} + h) \cdot \frac{\pi}{12}$

When you plug a number into a formula and that number has dimension, you should include the units. E.g. if you walk at speed v for 30 seconds then the distance traveled is 30v seconds, not 30v.