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.

 

[brotherbobby]

View attachment 369126The 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.

Brilliantly explained. Thank you and sorry for the delay in responding.

 

[brotherbobby]

I have solved the problem I mentioned in post #1 at the top, which was about the direction of coriolis deflection of a rocket fired "up" from a point in the southern hemisphere. It would be deflected west. Addtionally, I raised two similar cases of the deflection for a body dropped from a height and a projectile fired, both from the northern hemisphere. The answer to the first is towards east and that for the second is towards west. Thank you all for your help.

Instead of starting a new thread, I thought I'd write a new problem here, much the same as the problems above.

Problem : A circular platform rotates with a uniform angular velocity $\omega$ counter-clockwise about an axis through its center and normal to its plane, as shown. A man of mass $m$ walks radially inward with a uniform speed $v_0$ with respect to the platform. Calculate the magnitude of the coriolis force and its direction relative to the direction along which the man works.

Solution : (The magnitude is trivial, but it's the direction where I find myself disagreeing with the text. )

The angular velocity of the platform $\vec\omega= \omega\hat k$ and the velocity of the man $v=-v_0\hat i$, considering the direction "due east" being along the $+x$ axis.

The coriolis force $\small{\vec F_C = -2m(\vec\omega\hat k\times -v_0\hat i) = -2m\omega v_0(\hat k\times -\hat i)=\boxed{2m\omega v_0\hat j}}$, implying that, according to me, a coriolis force of $\boxed{2m\omega v_0}\;{\color{green}\checkmark}$ acts on the man in a direction relative to his $\boxed{\text{right}}\;\color{red}{\large\times}$.

Text answer : The force acts on the man towards the left.

I hope I am correct and the book mistaken, but a confirmation would be relieving.

Doubt : (Arguing conceptually, I tend to agree with the text.)

The man has a uniform velocity in the rotating frame of the platform, towards its center. In a small time that he moves towards the center, he has also moved to his right, relative to an inertial frame, for the platform is rotating towards "north" and therefore his "right". But the rotating frame sees no such deflection. This can be explained by an observer in the rotating frame by assuming that there is a fictitious force (the coriolis force) acting to the left (or "due south") that compensates for such a deflection.

Where am I mistaken in my argument above, which seems to contradict my mathematics?

 

[TSny]

The coriolis force $\small{\vec F_C = -2m(\vec\omega\hat k\times -v_0\hat i) = -2m\omega v_0(\hat k\times -\hat i)=\boxed{2m\omega v_0\hat j}}$, implying that, according to me, a coriolis force of $\boxed{2m\omega v_0}\;{\color{green}\checkmark}$ acts on the man in a direction relative to his $\boxed{\text{right}}\;\color{red}{\large\times}$.

Yes, the Coriolis force is toward the right.

In the video below, you can see that when the rotating frame rotates counterclockwise relative to the inertial frame of the room, the ball deflects "toward the right" relative to the person tossing the ball. So, the direction of the Coriolis force is initially toward the right, as you calculated.

Text answer : The force acts on the man towards the left.

When the man walks with constant velocity toward the center of the platform, there is no accelertion of the man relative to the platform. So, the sum of real and fictitious forces in the rotating frame must be zero. Therefore, there must be a real force acting on the man toward his left in order to cancel the fictitious Coriolis force toward his right. The real force might come from the interaction of his feet with the platform. Could the text be referring to this real force rather than the Coriolis force?

There would also be a real, radially inward force on the man that cancels the fictitious centrifugal force.

 

[kuruman]

The real force might come from the interaction of his feet with the platform. Could the text be referring to this real force rather than the Coriolis force?

Perhaps, but I don't think so. The real force in this case is static friction which adjusts itself to provide the observed acceleration $a_{rot.}=0$ in the rotating frame. That's because in the rotating frame the man walks along a straight line at constant velocity relative to the platform in a direction radially in. So now we have to find the direction of friction that is necessary to provide zero acceleration in the rotating frame and deduce that the fictitious force(s) must be equal and opposite to that.

Suppose that the man is just standing on the platform. The radial component of friction is needed to cancel the fictitious centrifugal acceleration and is radially in. This centripetal component of friction is always there whether the man is just standing or walking; it is proportional to the distance of the man from the center of the platform and goes to zero at the center. We will set that aside and consider what happens in the azimuthal direction.

In the inertial frame, the man has azimuthal (tangential) velocity $~\mathbf v_{\phi}= \boldsymbol{\omega}\times \mathbf r.$ This component is also proportional to his distance from the center and goes to zero at the center.

Imagine the man moving towards the center at constant speed. In the inertial frame, if he is to stay on a rotating diameter (red dotted line in the figure), he has to rely on static friction to reduce his tangential velocity. This means that the azimuthal component of friction must be directed to his left. In the rotating frame, the man will say that there is a Coriolis force acting on him to his right which is cancelled by the force of static friction to provide the zero observed acceleration. In other words, the man senses the static friction and interprets it as the reaction to the action of the Coriolis force. This direction for the Coriolis force is supported by the math as already established.

I suspect that the author of the answer has confused the real force with the fictitious force. It is more expedient to consider the real force(s) in the inertial frame first and then consider how they complement the fictitious forces to provide the observed acceleration in the non-inertial frame.

For the interested reader, the math behind the problem of figuring out the forces that must act on the man so that he can move at constant speed along a fixed diameter on the rotating platform from one end to the other and emerge at a point opposite to the entry point is treated in this article with the additional constraint of an upper limit to the magnitude of the acceleration.