# Can rotational force of the earth have any effect on flying projectiles?

• danjroman
In summary: Coriolis force (which points to the right of the velocity vector) is m*v*w^2 . If you're at the equator, the centrifugal and Coriolis forces are equal and opposite. So, for a particle at the equator, F = -m*v*w^2 .
danjroman
Well we know guys that the Earth spins to the east. So i was thinking, if you launch projectiles (lets say golf balls) of equal speed and angle to the east and launch some to the west, will the ones thrown to the east land farther from the launching point compared to the ones thrown to the west. The idea here is that since the Earth is spinning to the east, it will give an extra force exerted on the ball going to the east. Contrastingly, the balls launched to west will have to deal with the opposite force of Earth acting to the east which will cause it to land closer to the origin. Opinions?

do u know why the shuttle launch site is situated in Florida? Well, it is closer to the equator!

uhmmm, what does that mean. So do u think the golf ball thing might have some results?

Indeed it does.
Your effect will be most strongly present the closer you are to equator.
That's why Florida is used rather than, say, Alaska.

Oh great! I am planning to do this experiment here in singapore, just 1 degree north of the equator! yay!

I don't know if it gives a force as such (neglect air resistance), but rather that the time the ball is in the air, the Earth has rotated, so it lands in a different spot than it would have if the Earth were not rotating.

In the Earth's frame, assuming we're in the northern hemisphere, we will have two forces in addition to gravity... the centrifugal force, and the Coriolis force. The centrifugal force is directed away from the axis of rotation, and it's effect is that the direction of g is not exactly towards the centre, but slightly south of it. The coriolis force is directed to the right of the velocity vector. What effect you get depends of the direction you're throwing the ball.

Isn't the centrifugal and coriolis forces fictitious?

The Coriolis force is real, but the effect is negligible for any small-scale system (such as water going down a sink). It starts to make itself known in large storms or ocean currents.

@ Danjroman: fictitious or not, if you're on the rotating earth, you cannot neglect them. They're fictitious in the sense that from a frame outside the earth, it'll appear as though you're correcting for unreal forces. On the earth, however, the forces will appear to be real. As Danger said, the Coriolis force is the reason why hurricanes rotate.

They may be introduce small corrections, but I think those are the the very effects the OP is asking for (ie, the effects of the rotation of the earth).

If you want to calculate the actual effect, try this: take the w-axis to be the z-axis. calculate Fcor = -2m(w x v) and Fcent = -m wx(wxr) using suitable axes. Then use F = ma and calculate the motion. You might want to refer some book on undergrad classical mech, which has a chapter on fictitious forces.

thanks a lot rahuldandekar but I am just a grade 11 student in high school. Do u think u can give me an easy step by step way in calculating for this with what the level of physics i should have. thanks

danjroman said:
Isn't the centrifugal and coriolis forces fictitious?
Danger said:
The Coriolis force is real,
The Coriolis force is a ficticious force. Ficticious forces are forces that are added to an accelerating reference frame in order to make Newton's first and second laws work correctly. Because of the Earth's rotation you can consider a person on the Earth to be in a rotating reference frame where the two ficticious forces are the centrifugal and Coriolis forces.

When working in a rotating reference frame you cannot ignore the ficticious forces any more than you can ignore gravity. In the rotating reference frame they can accelerate projectiles, do work, cause stress and strain, and basically everything else that you can think of a force doing. The only way that they do not behave as a real force is that they violate Newton's 3rd law, i.e. there is no equal and opposite reaction force to the Coriolis force.

DaleSpam said:
The Coriolis force is a ficticious force. Ficticious forces are forces that are added to an accelerating reference frame in order to make Newton's first and second laws work correctly.

Oops! I misunderstood what the term meant (I have no formal education). I thought he was suggesting that there is no such thing. Sorry.

Often, they are called auxiliary forces instead (at least in Norwegian)

OK. To make the discussion easier (and as was mentioned earlier), I'll assume you're on the equator.

Then the centrifugal force (which points away from the rotational axis) is m*r*w^2 , where R is the radius of the earth, and w is the angular acceleration of the earth. At the equator, this points directly vertically at the surface, so all it does is make a tiny correction to gravity. we'll call that corrected value g (it'll actually be the gravitational g minus r*w^2).

The coriolis force, at the equator, is zero... so it simplifies the discussion somewhat. (The coriolis force depends as the cosine of the angle from the north pole. It is maximum at the poles, zero at the equator, and in opposite directions on the northern and southern hemispheres).

So, all we have is that gravity is slightly reduced. What will be the effect? Look at the formulae for projectile motion, both the maximum height and range are proportional to 1/g... so if g is reduced, both will be increased. R*w^2 is 3.39 x 10^-2, which is quite a bit smaller than g. The correction is therefore very small, about 0.3 % of g.

I don't think the rotation of the Earth has any effect like the one you asked about, that is, if you throw a ball in the direction opposite to the Earth's rotation, the distance traveled won't be increased. Why? For the same reason that a ball throw vertically from a moving car lands in the car again. Think about it.

There will also be some effect from the curvature of the earth, etc, but it'll be very small. All these effects will be negligible against air resistance.

rahuldandekar said:
The Coriolis force, at the equator, is zero... so it simplifies the discussion somewhat. (The Coriolis force depends as the cosine of the angle from the north pole.

...

I don't think the rotation of the Earth has any effect like the one you asked about, that is, if you throw a ball in the direction opposite to the Earth's rotation, the distance traveled won't be increased.

Ok, I was kind of wrong there. The rotating Earth isn't exactly like a moving car, because, um, well, it rotates. There are some terms in the Coriolis force which depend on cos of the angle from the pole, and some which depend on sine. The sine terms do matter that the equator. (The cos terms are zero at the equator, and opposite on both hemispheres. )

The sine terms give effects directed towards the west, and also vertically upwards. So, a body will be deflected towards the west, and there will be some further correction to gravity.

The west deflection depends on the vertical velocity, and the vertical Coriolis correction term depends on the eastward velocity.

Dollars to donuts the Abrams M1A1 corrects for a rotating Earth in calculating a firing solution.

danjroman said:
Well we know guys that the Earth spins to the east. So i was thinking, if you launch projectiles (lets say golf balls) of equal speed and angle to the east and launch some to the west, will the ones thrown to the east land farther from the launching point compared to the ones thrown to the west. The idea here is that since the Earth is spinning to the east, it will give an extra force exerted on the ball going to the east. Contrastingly, the balls launched to west will have to deal with the opposite force of Earth acting to the east which will cause it to land closer to the origin. Opinions?

Sorry to bump an old thread , but I think the simple answer to this one is yes. Throw something East it flies for longer simply because going east it has a higher orbit speed around the centre of gravity of Earth (you get some free from already present rotation of earth).

So it will fly higher for the same energy input even though relative to Earth's surface the initial launch speed looks the same.

Talking about coriolis and centrifugal corrections just adds unnecessary complexity. The way to analyse it IMHO is to look from a point out in space. Not from one bound to the surface of the Earth.

..edit :.. I just realized the guy answered his own question in his original post. lol

Last edited:
rahuldandekar said:
The coriolis force, at the equator, is zero...

The horizontal components of the coriolis force for objects that only move horizontally is zero. This is important for emptying bathtubs and for the weather (the reason you do not get hurricanes on the equator)

There is vertical movement, and a vertical components of the coriolis force.
In general the equation is $$F_c = -2m \omega \times v$$

where \omega is the angular momentum vector of the earth, pointing in the direction of the north pole.
For movement on the equator

If you move north, the coriolis force is 0. $$\omega$$ is parallel with $$v$$
if you move east, the coriolis force is directed upwards
if you move south, the coriolis force is 0
if you move west, the coriolis force is downwards
if you move up, the coriolis force is directed west.
if you move down, the coriolis force is directed east.

danjroman said:
Well we know guys that the Earth spins to the east. So i was thinking, if you launch projectiles (lets say golf balls) of equal speed and angle to the east and launch some to the west, will the ones thrown to the east land farther from the launching point compared to the ones thrown to the west. The idea here is that since the Earth is spinning to the east, it will give an extra force exerted on the ball going to the east. Contrastingly, the balls launched to west will have to deal with the opposite force of Earth acting to the east which will cause it to land closer to the origin. Opinions?
Absolutely. Recall the big guns on WW-II battleships. To hit a target which is miles away they had to account for the coriolis force.

Pete

danjroman said:
Well we know guys that the Earth spins to the east. So i was thinking, if you launch projectiles (lets say golf balls) of equal speed and angle to the east and launch some to the west, will the ones thrown to the east land farther from the launching point compared to the ones thrown to the west. The idea here is that since the Earth is spinning to the east, it will give an extra force exerted on the ball going to the east. Contrastingly, the balls launched to west will have to deal with the opposite force of Earth acting to the east which will cause it to land closer to the origin. Opinions?
There would be no difference in the distance the balls traveled on Earth.
Imagine you were on a moving conveyor belt.
Throw a ball in front of you in the direction of travell.
Throw another with the same force in the opposite direction.
Both balls will travell the same distance on the conveyor.
They will only appear to travell different distances to an observer not on the conveyor.
Or in the case of the golf balls from a reference point in space.

When I was in the military I worked in the fire direction control part of the artillery. We would get a location in grid coordinates and calculate how the gun should be aimed to hit that target. There are many variables that you put in the calculations and one of them was the movement of the Earth as the shell flies through the air to the target. I recall that this was a very small number. If you are looking to learn what effects projectiles flying through the air, you might want to look at artillery firing tables. This would give you a good idea of all of the variables involved.

When I was in the military I worked in the fire direction control part of the artillery. We would get a location in grid coordinates and calculate how the gun should be aimed to hit that target. There are many variables that you put in the calculations and one of them was the movement of the Earth as the shell flies through the air to the target. I recall that this was a very small number. If you are looking to learn what effects projectiles flying through the air, you might want to look at artillery firing tables. This would give you a good idea of all of the variables involved.

vibjwb said:
When I was in the military I worked in the fire direction control part of the artillery. We would get a location in grid coordinates and calculate how the gun should be aimed to hit that target. There are many variables that you put in the calculations and one of them was the movement of the Earth as the shell flies through the air to the target. I recall that this was a very small number. If you are looking to learn what effects projectiles flying through the air, you might want to look at artillery firing tables. This would give you a good idea of all of the variables involved.
Bet there are no variables when fireing a shell in direct east or west direction at the equator.

In guns, the Coreolis force will cause a projectile to curve slightly to the right when fired in the northern hemisphere. At the equator, the surface of the Earth is moving about 1000 miles per hour, or 44,700 cm per sec. The effect will cause the projectile to move several cm to the right. So does right-hand rifling, and for this reason, long range guns have to correct for both of these effects.

Bob S said:
In guns, the Coreolis force will cause a projectile to curve slightly to the right when fired in the northern hemisphere. At the equator, the surface of the Earth is moving about 1000 miles per hour, or 44,700 cm per sec. The effect will cause the projectile to move several cm to the right. So does right-hand rifling, and for this reason, long range guns have to correct for both of these effects.
So in the southern hemisphere it must move to the left.
There must be a point between north and south (somewhere near the equator) where rifling and the Coreolis effect does not happen and no correction is needed.

Buckleymanor said:
So in the southern hemisphere it must move to the left.
There must be a point between north and south (somewhere near the equator) where rifling and the Coreolis effect does not happen and no correction is needed.
If you consider only horizontal motion, the Coriolis force is zero at the equator. But if you shoot something up vertically at the equator, it will be diverted westwards.
http://en.wikipedia.org/wiki/Coriolis_effect#Rotating_sphere
I guess ship artillery flies pretty high, so they have to account for this even on the equator.

A.T. said:
If you consider only horizontal motion, the Coriolis force is zero at the equator. But if you shoot something up vertically at the equator, it will be diverted westwards.
http://en.wikipedia.org/wiki/Coriolis_effect#Rotating_sphere
I guess ship artillery flies pretty high, so they have to account for this even on the equator.
Because the Earth moves to the east as the shot is moveing up vertically.
So it will land in the west a distance proportunal to the time of flight of the object plus the distance the Earth has rotated east during the the time of flight.
You would only have to account for this if you were doing some pretty fancy shooting.

Last edited:
Buckleymanor said:
Bet there are no variables when fireing a shell in direct east or west direction at the equator.
The Coriolis effect most certainly does come into play when firing a shell due east or due west at the equator. The Coriolis acceleration is $-2{\boldsymbol{\Omega}}\times{\boldsymbol v}$. For a shell fired west at the equator, the Coriolis acceleration is directed downward. For shell fired east at the equator, it is directed upward.

Some examples, all calculated per
• No air friction
• Spherical gravity
• 200 step leap frog integration
• μ = 398,600.44183 km3/second2
• Re = 6378.137 km
• Ωe = 2pi/sidereal day (*not* 2pi/24 hrs) = 7.29211585×10-5 s-1

1. A handgun with a muzzle velocity of 465.1 meters per second is fired horizontally 1 meter off the ground due west at the equator. The Coriolis acceleration is 0.0678 m/s2 downward. That's small compared to gravity, but not negligible. I chose 465.1 meters/second for this first example because that is the rotational velocity of someone at the Equator. From the perspective of an inertial observer instantaneously above the gun, the bullet leaves the gun at pretty close to a dead stop. The inertial observer sees the bullet drop straight to the ground, elapsed time = 0.4518 seconds. The Earth rotates by 210.1 meters while the bullet falls. An Earthbound observer sees the westbound bullet as having traveled 210.1 meters to the west.

2. Same gun, fired due east. The inertial observer sees the bullet leave the chamber at 930.2 meters/second. This puts the bullet in a highly elliptic orbit. This orbit intersects the Earth 0.4550 seconds later and 423.2 meters downrange (from the perspective of the inertial observer). The Earth however has rotated by 211.6 meters while this bullet falls.
An Earthbound observer sees the eastbound bullet as having traveled 211.6 meters to the east, or 1.5 meters further than the westbound bullet.

3. A west-facing tank fires a shell 45 degrees from horizontal. The shell leaves the gun 3 meters off the ground with a muzzle velocity of 1800 meters/second. The inertial observer sees the shell as having an upward velocity component of 1272.8 meters/second and a horizontal component of 807.7 meters/second (westward). The inertial observer sees the shell's orbit intersecting the Earth 212.0 km to the west. Meanwhile, the Earth has rotated by 124.3 km during the shell's flight. An Earthbound observer sees this shell as having traveled 336.3 km to the west.

4. The tank does the same thing as case 3, but this time facing to the east. The inertial observer sees the shell as having an upward velocity component of 1272.8 meters/second and a horizontal component of 1737.9 meters/second (eastward). The inertial observer sees the shell's orbit intersecting the Earth 474.2 km to the east. Meanwhile, the Earth has rotated by 129.3 km during the shell's flight. An Earthbound observer sees this shell as having traveled 345.0 km to the west, or 8.7 km further than the westbound shell.Final remark: The European Space Agency launch site is the Guiana Space Centre, 5o9'25" North latitude. The preferred launch direction is due east for the simple reason that doing so maximizes the extra boost the rockets get for free thanks to the Earth's rotation.

Last edited:
Buckleymanor said:
Because the Earth moves to the east as the shot is moveing up vertically.
Yes, that is the explanation for a inertial frame, where there is no Coriolis force.
Buckleymanor said:
So it will land in the west a distance proportional to the time of flight of the object plus the distance the Earth has rotated east during the the time of flight.
Not the time of flight of the object is crucial, but how high it flies. It has the same tangential speed as the surface, but with greater height a lower angular speed, so it stays behind the Earth's rotation.

A.T. said:
Yes, that is the explanation for a inertial frame, where there is no Coriolis force.

Not the time of flight of the object is crucial, but how high it flies. It has the same tangential speed as the surface, but with greater height a lower angular speed, so it stays behind the Earth's rotation.
When a projectile or bullet is fired up vertically the height to which it flies would be dependant on the time of it's flight.The longer it took the higher it's flight.
So the height is crucial but the time of flight should tell you how high it flew.

Buckleymanor said:
There would be no difference in the distance the balls traveled on Earth.
Buckleymanor, this is incorrect. Read posts #16, #18, and #29.

D H said:
Buckleymanor, this is incorrect. Read posts #16, #18, and #29.
Yes I agree, your explanation #29 explains in some detail.
What I was not taking into consideration was the highly eliptical orbit the bullet is put in caused by the rotation of the Earth when the bullet is fired eastwards.
Obviosly this effects the time of flight and should be taken into consideration(which you have).
KB

Final remark: The European Space Agency launch site is the Guiana Space Centre, 5o9'25" North latitude. The preferred launch direction is due east for the simple reason that doing so maximizes the extra boost the rockets get for free thanks to the Earth's rotation.
Presume the time of launch is important.
At certain times the launch direction might be pointing east but would be pointing in the wrong direction due to the Earths orbit around the Sun.
So any extra boost would be minimized.
Thanks.KB.

The Sun is pretty much irrelevant when it comes to the energy needed to launch a vehicle into low Earth orbit. The Sun does have a small, perturbative effect on vehicles in low Earth orbit. This effect is small but cumulative. The cumulative effect during the 10 to 15 minutes needed to go from the launch platform to orbit insertion is incredibly small, so small that it is not taken into account. The uncertainties in atmospheric drag and thruster performance vastly overwhelm this tiny effect.

For a vehicle going on to other planets, the direction in which it is pointing when it performs its Earth departure burn is very critical. An analogous situation to launching east versus west applies.

## 1. How does the rotational force of the earth affect the trajectory of flying projectiles?

The rotational force of the earth, also known as the Coriolis force, can have a significant effect on the trajectory of flying projectiles. This force causes objects to appear to deflect to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. This means that projectiles will experience a slight change in direction as they travel through the air, which can impact their final destination.

## 2. Can the Coriolis force cause a projectile to change direction mid-flight?

No, the Coriolis force does not cause a projectile to change direction mid-flight. It only causes a slight deviation in the trajectory of the projectile, which is typically too small to notice in short-distance flights. However, over longer distances, the effect of the Coriolis force can become more significant and may need to be taken into account for accurate targeting.

## 3. Does the rotational force of the earth affect all flying objects, regardless of their size?

Yes, the Coriolis force affects all flying objects, regardless of their size. However, the effect is more noticeable in larger objects with longer flight times, such as airplanes or missiles. Smaller objects, like a thrown baseball, may also experience the Coriolis force, but the effect is typically too small to be significant.

## 4. Can the rotational force of the earth be used to increase the range of a projectile?

No, the Coriolis force cannot be used to increase the range of a projectile. While it may slightly alter the trajectory of the projectile, it does not provide any additional force or propulsion. The range of a projectile is primarily determined by its initial velocity, angle of launch, and air resistance.

## 5. How can the Coriolis force be accounted for in projectile calculations?

The Coriolis force can be accounted for in projectile calculations by using the appropriate equations and considering the direction and magnitude of the force. This may require additional calculations or adjustments to the initial velocity and angle of launch. In some cases, computer simulations or specialized software may be used to accurately account for the Coriolis force in projectile calculations.

• Classical Physics
Replies
23
Views
343
• Optics
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
2K
• Mechanics
Replies
10
Views
4K
• Astronomy and Astrophysics
Replies
10
Views
2K
• Mechanics
Replies
5
Views
1K
• Mechanics
Replies
14
Views
2K
• Mechanics
Replies
37
Views
5K
• Mechanics
Replies
35
Views
16K
• Mechanics
Replies
34
Views
5K