A rocket and the Coriolis acceleration

[brotherbobby]

Homework Statement

Calculate the direction of the coriolis acceleration acting on a rocket moving vertically upwards with a velocity of 2 km/s at a $60^{\circ}\;\text{S}$ latitude.

Relevant Equations

The coriolis acceleration is given by $a_C = -2\Omega_0\times\vec v$, where $\vec\Omega_0$ is the angular velocity of the rotating frame relative to an inertial frame and $\vec v$ is the velocity of the body relative to the rotating frame.

Attempt : I start by copying and pasting the diagram for the problem to the right. The rocket marked with R hurls "up" along the z axis with a speed of $\vec v=v\hat k$ from a place P whose latitude is $\lambda\;\text{(south)}$, or $-\lambda$. The axes are so chosen that $x$ points to the east and $y$ points to the north. The earth rotates in the $y-z$ plane with an angular velocity $\vec\Omega_0 = \Omega_0(\cos\lambda\hat j-\sin\lambda\hat k)$. This can be seen by drawing a line through P parallel to the "east" direction and locating the angle $\lambda$ as the alternate angle equal to the given latitude, and then using vector decomposition. The negative sign for the z-component is consistent with the fact that $\lambda$ is negative, leading to $\sin(-\lambda) = -\sin\lambda$.

The coriolis acceleration of the rocket $\vec a_C = -2m(\Omega_0\times\vec v) =-2m\Omega_0(\cos\lambda\hat j-\sin\lambda\hat k)\times v\hat k = -2m\Omega_0\cos\lambda v \hat j\times\hat k= 2m\Omega_0\cos\lambda v (-\hat i)$. Since $\hat i$ points in the east direction, I find that the coriolis acceleration on the rocket acts to the $\boxed{\text{west}}\quad\color{red}{\Large\times}$.

Answer in the text : The coriolis acceleration of the rocket is towards east.

Request : What is the error I am making?

 

[Hill]

The axes are so chosen that x points to the east and y points to the north. The earth rotates in the y−z plane

The Earth rotates in the north-z plane?

 

[brotherbobby]

The Earth rotates in the north-z plane?

The earth rotates in the Y-Z plane. It's angular velocity vector $\vec\Omega_0$ makes an angle of $\lambda$ with the Y axis and $\pi/2-\lambda$ with the Z axis. It has no component along the X axis, i.e. the east west direction.

 

[Hill]

The earth rotates in the Y-Z plane. It's angular velocity vector $\vec\Omega_0$ makes an angle of $\lambda$ with the Y axis and $\pi/2-\lambda$ with the Z axis. It has no component along the X axis, i.e. the east west direction.

Y-Z plane in your choice of coordinates is North-Z plane. So, you say Yes, the Earth rotates in the North-Z plane?

 

[brotherbobby]

Y-Z plane in your choice of coordinates is North-Z plane. So, you say Yes, the Earth rotates in the North-Z plane?

Yes. The Z plane goes vertically up from the place. Y is along the north direction from the place P.
Rotation of the earth is in the Y-Z plane, same as the North-Up plane.

 

[Hill]

Yes. The Z plane goes vertically up from the place. Y is along the north direction from the place P.
Rotation of the earth is in the Y-Z plane, same as the North-Up plane.

I also think, the "west" is correct.

 

[kuruman]

First a correction. You wrote

Mass $m$ does not belong on the right hand side because you are writing an acceleration not a force.

You put the $z-$axis along a line connecting the center of the Earth to the launching point and define "up" as $\mathbf {\hat z}.$ I would put the $z-$axis along the rotation axis and write $\boldsymbol{\Omega}_0=\Omega_0~\mathbf{\hat z}.$ Since "up" means "away from the center of the Earth" (except for the members of the Flat Earth Society), that would be the radial direction. The position vector of the rocket is simply $\mathbf R=R~\mathbf{\hat r}.$ The Coriolis acceleration is $$\mathbf a_C=-2\Omega_0~\left(\mathbf{\hat z}\times\frac{d\mathbf r}{dt}\right).$$ The axis of the Earth's rotation defines "true North" and it should be easy to figure out the angle between the $z-$axis and the radial "up" direction from the latitude of the launching point. As for the West to East direction, it is along the tangential velocity of the launching point fixed on the Earth; I leave that for you to figure out.

 

[TSny]

Request : What is the error I am making?

Your work looks correct to me (except, as pointed out by @kuruman, $m$ should not be in your expression for the acceleration). I agree with your answer of west.

 

[kuruman]

Your work looks correct to me (except, as pointed out by @kuruman, $m$ should not be in your expression for the acceleration). I agree with your answer of west.

There is a problem with OP's drawing. It is not clear at all which direction is "westward". Usually "west" is opposite to"east" which, according to the drawing, is $-\hat i$ or to the left. The Earth rotates west to east. If one uses the right hand rule rule, launching point P has tangential velocity $\boldsymbol{\Omega}\times \mathbf R_E$ directed into the screen. Thus, the direction west to east is into the screen.

Furthermore, the diagram shows that direction "north" has a component in the direction "east" and "east" has a component in the direction "up". How can that be? Maps are drawn in a plane that is locally tangent to the surface of the Earth and the Cartesian North-East axes have no such components.

 

[TSny]

There is a problem with OP's drawing.

Yes, I agree that the x-axis as drawn is bad. The x-axis (toward the east) should be into the page (perpendicular to the y-z plane). The angle $\lambda$ should be the angle between the equatorial plane of the earth and the z-axis, not the angle between the x and z axes. But the mathematical expressions in the OP look correct to me (except for the factor $m$).

EDIT: Add following figure

 

[kuruman]

Yes, I agree that the x-axis as drawn is bad. The x-axis (toward the east) should be into the page (perpendicular to the y-z plane). The angle $\lambda$ should be the angle between the equatorial plane of the earth and the z-axis, not the angle between the x and z axes. But the mathematical expressions in the OP look correct to me (except for the factor $m$).

EDIT: Add following figure

View attachment 369096

Thanks to your nice drawing it is clear what's what.

To @brotherbobby:
The cross product vector $\mathbf {\Omega}_0\times \mathbf v$ is perpendicular to the plane defined by the axis of rotation and "up". This direction is East. The negative sign in front of the Coriolis term changes the direction to West. If the answer in the text for the Coriolis term says "East", then it is incorrect.

 

[brotherbobby]

Thank you all for your comments. I agree with the corrections to my drawing and with the expression of the coriolis acceleration, namely that the mass of the object must be absent : $\vec a_C = -2(\vec\Omega_0\times\vec v)$.

However, crucial thing is that the answer, that the coriolis force acts on the rocket moving vertically "up" from the point P in question, towards the "west", is correct, far as we think. That opens a can of worms regarding the direction of the coriolis force in both the hemispheres, as I will explain.

What is the direction of the coriolis acceleration on an object that is dropped from a height $h$ at a point P with altitude $\lambda^{\circ}\;\underline{\text{N}}$?

Geometrically, this problem is the same as the one I asked, where the rocket was moving vertically up at a point located south. (I underline the words in both questions to emphasize their similarity).

So let me try and answer this "new" question. The coriolis acceleration $\vec a_C = -2(\vec\Omega_0\times\vec v) = -2\Omega_0(\cos\lambda\hat j+\sin\lambda\hat k)\times (-v\hat k) = 2\Omega_0\cos\lambda v\hat i\rightarrow\boxed{\text{towards east}\;!}$

(I remind you of my co-ordinate conventions - the X axis points east, the Y axis points north, and the Z axis points vertically up from the place in question. These co-ordinates being unfortunately "taken", the angular velocity vector of the earth $\vec\Omega_0$ rests on the YZ-plane, making an angle $\lambda$ with the Y-axis)

The answer should be the same as before, viz. in the western direction. But it isn't!

Request : Please tell me where am going wrong. For I must be.

 

[Hill]

Geometrically, this problem is the same as the one I asked, where the rocket was moving vertically up at a point located south.

No, it is not. Toward east is correct in this new case.

 

[brotherbobby]

No, it is not. Toward east is correct in this new case.

Isn't going "up" in the south hemisphere, far as the coriolis force is concerned, the same as going "down" in the northern? (up and down being relative to the place(s) in question, relative to land)

 

[Hill]

Isn't going "up" in the south hemisphere, far as the coriolis force is concerned, the same as going "down" in the northern? (up and down being relative to the place(s) in question, relative to land)

No, it is not the same.
If it were, there wouldn't be any coriolis force in the equatorial plane, regardless of moving "up" or "down", right?

 

[brotherbobby]

No, it is not the same.
If it were, there wouldn't be any coriolis force in the equatorial plane, regardless of moving "up" or "down", right?

Yes good question. Sorry I was under the impression that, precisely for my thinking up there, namely that "going up in the northern" is the same as "going down in the southern" and vice versa, that I believed that the coriolis force would disappear in places on the equator. I was also "helped" in this thinking as a result of symmetry - if there's such a force at the equator, which does it point? East or west? Why choose one direction over the other, when it's located right in between both hemispheres?

Of course the mathematics doesn't doesn't side with me : $\cos\lambda = \cos 0 = 1$, thus it points due east for a body freely falling down at the equator, and towards the west for a body projected vertically up. I hope I am correct regarding the directions. Thank you for the correction.

But the point raises another of those ugly can of worms I mentioned. Look at this query which has been appearing across undergraduate examinations in my country (India).

Q. For a particle projected vertically up, show that the deflection would be westwards. If the particle reaches a vertical height of $\boldsymbol h$ at a point on the earth with latitude $\lambda\;\text{N}$, show that the amount of deflection upon reaching the ground is $\boldsymbol{\boxed{d = \dfrac{4}{3}\Omega_0 \cos\lambda \sqrt{\dfrac{8h^3}{g}}}}$, where $\Omega_0$ is the angular velocity of the earth and $g$ the (effective) acceleration due to gravity.

Curiously enough, despite my grave doubts, I can solve the problem, finding first the deflection for the upwards motion and multiplying it by 2.

But how do we account for it? Let's stick to the same place. Doesn't the deflection due to the coriolis force while going up cancel the defection due to the force while going down? If it does, wouldn't the body drop at the same point?

 

[Hill]

Yes good question. Sorry I was under the impression that, precisely for my thinking up there, namely that "going up in the northern" is the same as "going down in the southern" and vice versa, that I believed that the coriolis force would disappear in places on the equator. I was also "helped" in this thinking as a result of symmetry - if there's such a force at the equator, which does it point? East or west? Why choose one direction over the other, when it's located right in between both hemispheres?

Of course the mathematics doesn't doesn't side with me : $\cos\lambda = \cos 0 = 1$, thus it points due east for a body freely falling down at the equator, and towards the west for a body projected vertically up. I hope I am correct regarding the directions. Thank you for the correction.

But the point raises another of those ugly can of worms I mentioned. Look at this query which has been appearing across undergraduate examinations in my country (India).

Q. For a particle projected vertically up, show that the deflection would be westwards. If the particle reaches a vertical height of $\boldsymbol h$ at a point on the earth with latitude $\lambda\;\text{N}$, show that the amount of deflection upon reaching the ground is $\boldsymbol{\boxed{d = \dfrac{4}{3}\Omega_0 \cos\lambda \sqrt{\dfrac{8h^3}{g}}}}$, where $\Omega_0$ is the angular velocity of the earth and $g$ the (effective) acceleration due to gravity.

Curiously enough, despite my grave doubts, I can solve the problem, finding first the deflection for the upwards motion and multiplying it by 2.

But how do we account for it? Let's stick to the same place. Doesn't the deflection due to the coriolis force while going up cancel the defection due to the force while going down? If it does, wouldn't the body drop at the same point?

In these two cases, i.e. "up and down" vs. "down", the tangential velocities at the point when the particle starts going down are different.

 

[brotherbobby]

In these two cases, i.e. "up and down" vs. "down", the tangential velocities at the point when the particle starts going down are different.

Due to the coriolis force, the particle acquires a velocity due east (along X) on its way up. For a place in the northern hemisphere.
Likewise, doesn't it acquire an equal and opposite velocity (along -X towards) west on its way down, that should result in no net horizontal displacement?

 

[Hill]

doesn't it acquire an equal and opposite velocity (along -X towards) west on its way down

I don't see why it would. There is nothing to change the tangential velocity, in the inertial frame.
In the situation you describe in the post #12, I assume that the particle starts falling being at rest in the rotating frame.

 

[brotherbobby]

I don't see why it would. There is nothing to change the tangential velocity, in the inertial frame.
In the situation you describe in the post #12, I assume that the particle starts falling being at rest in the rotating frame.

I should reply to you later when time and the advantage of having done some mathematics first permits. For the case is involved. The particle has acquired a speed (and deflection) due east projected up, after having reached its highest point (P in the northern hemisphere). On its way down, it will acquire a (negative) velocity $v_z$ due to gravity. But how about the coriolis force? It will have a velocity of the type $\vec v = v_x\hat i-v_z\hat k$, $v_x$ as a result of the coriolis acceleration on its way up. What is the coriolis force on this object on its way down? While $\hat k\times\hat j = -\hat i$, which would act in the west direction, $-\hat i\times\hat j = -\hat k$, which is along gravity! This would mean that the time of descent would be lower, and is relevant here. Because the question is whether the deflection due west would be enough to more than compensate for the deflection due east. The time of fall being less, it seems that it wouldn't be, making the particle suffer a net eastwards displacement from the point P where it's launched, in contradiction to the question.

 

[kuruman]

But how do we account for it? Let's stick to the same place. Doesn't the deflection due to the coriolis force while going up cancel the defection due to the force while going down? If it does, wouldn't the body drop at the same point?

The diagram on the right shows the Earth rotating with angular speed $\Omega_0$. We are looking down the axis of rotation in an inertial frame. A tower of height $h$ is built at the Equator. Point A at the base of the tower is moving west to east with tangential speed $v_A=\Omega_0R_E$. Point B at the tip of the tower moves west to east with tangential speed $v_B=\Omega_0(R_E+h).$ Note that every point above A along the tower moves faster relative to point A.

Now imagine firing a rocket at point A aimed radially out towards point B with initial velocity $\mathbf V_{\!R}.$ In our inertial frame, the rocket has initial radial velocity $\mathbf V_{\! R}$ and initial west to east velocity $\mathbf v_{\!A}.$ As it rises next to the tower, it falls farther and farther behind the points along the tower in the west to east direction because its speed in that direction stays constant at $v_{\!A}$ whilst the speed of the points on the tower increases. Now suppose the rocket reaches maximum height and falls back down. Its west to east velocity is still $\mathbf v_A$ so it will fall even farther behind, not cancel the deviation on its way down and land west of point A.

Mathematically, the Coriolis acceleration $\mathbf a_c=-2\mathbf{\Omega}_0\times \mathbf V$ changes sign when the rocket changes direction and moves down after reaching maximum height. This does not mean that its west to east velocity relative to the tower changes direction. It becomes smaller and reaches zero when the rocket lands but the rocket moves away from the tower throughout the trip. An observer, at rest in the non-inertial frame of the tower, will see the rocket move sideways away from the tower on its way up and will deduce that there is a sideways (Coriolis) acceleration. After the rocket reaches maximum height, this observer will see that the sideways speed of the rocket away from the tower will start diminishing and will deduce that the sideways (Coriolis) acceleration has changed direction.

Of course, if a rock is released from point B and allowed to fall freely, it will have higher west to east initial velocity than point A and will therefore land east of point A.

 

[A.T.]

Doesn't the deflection due to the coriolis force while going up cancel the defection due to the force while going down?

Why would you think that? Froce is not the same as deflection.