Coriolis effect - time to ground

[SpaceTrekkie]

Coriolis effect -- time to ground

This is not a homework problem, I just have a conceptual question. Does the presence of the Coriolis effect, make a difference in how long it takes for a project tile to hit the ground. Obviously it makes a difference in its final location, but the time?

Thanks!

 

[fatra2]

Hi there,

Never maid the calculation, but from the top of my head, I would say NO. The coriolis effect will bend the trajectory of a free-falling object. Your idea is probably to think, deviation in trajectory = greater distance to hit the ground = more time to get there. The logic behind is quite good, but maybe we are forgetting the coriolis effect itself.

Newton told us that if you apply a force to an object, an acceleration will be developed on it. Therefore, the coriolis effect will increase the speed of the object. My guess would be that it compensates for the increase distance to travel.

But, like I said: never maid the calculation myself. If I have a bit of time, I will look into it. Cheers

 

[vin300]

Does the presence of the Coriolis effect, make a difference in how long it takes for a project tile to hit the ground.
Thanks!

No.It only causes an apparent deflection
An unrotating observer(w.r.t the earth) should measure more time than the observer on the
earth on account of time dilation, but this difference is negligible as the velocity of the surface of the Earth at the equator in inertial space turns out to be approx. 4.65e-5
Even the gravitational time dilation formula is derived from the schwarzschild metric, which is for a non-rotating object

 

[billiards]

At the poles it shouldn't make any difference because the Coriolis force will be exactly zero in the direction of g. (Incidentally this is where Coriolis force will have the greatest "deflecting capacity" for horizontally projected motion -- orthogonal to g). However at the equator the opposite is true -- Coriolis is zero in the horizontal direction but maximized in the vertical direction.

At the equator you get the strange situation where if you fired your projectile to the East the Coriolis force would differ from that if you fired to the West, and so it would make a slight difference to the time.

 

[SpaceTrekkie]

I know that if you fire it at the equator, then it is deflected in its final landing position. But what would cause it to take longer to land? Gravity is still pulling it down at 9.8m/s^2...

or am I totally missing something

 

[A.T.]

I know that if you fire it at the equator, then it is deflected in its final landing position. But what would cause it to take longer to land? Gravity is still pulling it down at 9.8m/s^2...

or am I totally missing something

Consider extreme cases from the inertial frame:

If the planet rotates so fast that the tangential velocity on the equator is almost the orbital velocity, the projectile might never come down, if you fire it east. Because it gets enough tangential velocity to stay in orbit. But firing west reduces it's tangential velocity, so it comes down.

In the rotating frame this difference explained by the vertical Coriolis acceleration.

 

[billiards]

I know that if you fire it at the equator, then it is deflected in its final landing position. But what would cause it to take longer to land? Gravity is still pulling it down at 9.8m/s^2...

or am I totally missing something

Yes I think you are missing something. Your statement that "if you fire it at the equator, then it is deflected in its final landing position" is wrong unless you consider a long range projectile which gets far enough away from the equator. (You're not strictly wrong because the distance traveled will be deflected, but you won't get side-ways deflection which I think is what you meant!)

The coriolis force is a vector that is perpendicular to the Earth's spin axis. That means that at the equator, where the ground is actually parallel to the Earth's spin axis the vector is point straight up or down. Whether it points up or down depends on whether you fire east or west (respectively). The Coriolis force is then added to g so it can make a difference to the landing time.

I'll leave it to you to figure out what is happening at the pole.

 

[Cleonis]

This is not a homework problem, I just have a conceptual question. Does the presence of the Coriolis effect, make a difference in how long it takes for a project tile to hit the ground. Obviously it makes a difference in its final location, but the time?

It depends on whether the projectile is fired straight up or at an angle. If the projectile is fired at an angle then the direction of firing matters.

In spacecraft launches the launch is always in eastward direction. Not only for satellites that are destined for, say, geostationary orbit, but also for missions to other planets. The reason for that, as has been pointed out by other commentators, is that when launching eastward the Earth's rotation works in the spacecraft 's favor.

When a projectile is fired it will impact the Earth after some time, but during the flight the trajectory an orbital trajectory. (That is, if you don't count air resistance, which is actually a significant factor.)
A projectile fired eastward will impact the Earth later than when fired westward.

On my website there is a 3D http://www.cleonis.nl/physics/ejs/ballistics_and_orbits_simulation.php"
You can launch a projectile from any latitude, at any elevation, in any direction. The simulation then computes the trajectory with numerical analysis, and plots it. (The simulation does not take air resistance into account.)

http://www.cleonis.nl

 

[GRDixon]

This is not a homework problem, I just have a conceptual question. Does the presence of the Coriolis effect, make a difference in how long it takes for a project tile to hit the ground. Obviously it makes a difference in its final location, but the time?

Thanks!

Like a magnetic force, the Coriolis force involves a cross product with v, the tile's velocity. As such it causes only radial acceleration; it does not alter the speed. A tile dropped on a non-rotating Earth, and one dropped on our present Earth, will take the same amount of time to reach the ground.

 

[Cleonis]

A tile dropped on a non-rotating Earth, and one dropped on our present Earth, will take the same amount of time to reach the ground.

Let's have a closer look a that.

Let me set up a thought experiment in which a very, very high tower is build. It reaches all the way up to the altitude of geostationary orbit. If you are at the altitude of geostationary orbit, orbiting geostationary, and you release an object to free motion, then it'll never reach the ground.

The point I'm making is that it does matter whether the tile is dropped from some altitude on a rotating planet or from some altitude on a non-rotating planet.

On Earth when you are on the equator you are circumnavigating the Earth's axis with a velocity of about 1600 km/h. For that circumnavigating motion some centripetal force is required, the required acceleration for circumnavigating the Earth in 24 hours is 0.0339 m/s²

That acceleration goes at the expense of gravitational acceleration. In other words: at the equator there is a loss of about 0.3 % of gravitational acceleration. This difference makes objects fall slower at the equator than at the poles.

Note that while this difference is certainly a rotation-of-Earth-effect, it's not Coriolis effect.

 

[vin300]

In spacecraft launches the launch is always in eastward direction. Not only for satellites that are destined for, say, geostationary orbit, but also for missions to other planets. The reason for that, as has been pointed out by other commentators, is that when launching eastward the Earth's rotation works in the spacecraft 's favor.

When a projectile is fired it will impact the Earth after some time, but during the flight the trajectory an orbital trajectory. (That is, if you don't count air resistance, which is actually a significant factor.)
A projectile fired eastward will impact the Earth later than when fired westward.

In the frame of the launcher these velocities don't matter
1) the coriolis force depends on the component of velocity relative to the Earth along the equator and the component along the direction of local rotation
2)the time taken to reach the ground depends on the component of velocity normal to the surface and the latitude,
so one more thing needed to express the dependence of the time on the coriolis force is the latitude

 

[Cleonis]

In spacecraft launches the launch is always in eastward direction. Not only for satellites that are destined for, say, geostationary orbit, but also for missions to other planets. The reason for that, as has been pointed out by other commentators, is that when launching eastward the Earth's rotation works in the spacecraft 's favor.

A projectile fired eastward will impact the Earth later than when fired westward.

In the frame of the launcher these velocities don't matter
[...] the coriolis force depends on the component of velocity relative to the Earth along the equator and the component along the direction of local rotation [...]

Hi vin300, let me address that with the thought experiment that was offered in post #6 by A.T.

Consider extreme cases from the inertial frame:

If the planet rotates so fast that the tangential velocity on the equator is almost the orbital velocity, the projectile might never come down, if you fire it east. Because it gets enough tangential velocity to stay in orbit. But firing west reduces it's tangential velocity, so it comes down.

vin300, I'm really curious what your opionion is about the thought experiment offered by A.T.
What if the Earth would spin 16 times faster? (Low orbit satellites, such as the International Space Station, go around the Earth 16 times per 24 hours.)
If the Earth would spin 16 times faster, what do you expect for satellite launches in eastward versus westward direction?

(For simplicity let's neglect that a faster spinning planet will have a more pronounced equatorial bulge.)

 

[vin300]

Yes, the coriolis force depends on the angular velocity and the velocity of the projectile in the rotating system.Even if the projectile is be fired at greater than the escape velocity this would apply as long as it is to be seen, if it is orbiting there will be the force except when it is geosynchronised

 

[vin300]

Imagine spherical earth, horizontal x, vertical y and z passing through the reader.
1)A projectile is launched from any point on its surface.The velocity of projection is broken into its three components.
2)The component used in kinematics to determine the time to ground may coincide with anyone of the above, or none.
Now, the y-axis is the rotational axis and the component along this does not contribute to the coriolis force.
The constancy of the component in 2) is required for a constant time to the ground.But you can have any combination of component velocities in 1) keeping in mind the constancy of 2).
This clearly indicates, that we cannot have an equation of coriolis force and time to ground.

 

[Cleonis]

This clearly indicates, that we cannot have an equation of coriolis force and time to ground.

It's necessary to distinguish between whether it's practical to use and approach that involves evaluating Coriolis term components, and whether it's possible. (If you use the equation of motion for the co-rotating coordinate systen then Coriolis term components will be involved.)

As to the first, clearly it's impractical. The efficient approach is to evaluate the motion with respect to the inertial coordinate system.

Next: is it possible? Well, any calculation can be performed either for the inertial coordinate system or for the co-rotating coordinate system. The outcome (the duration from launch to impact) is guaranteed to give the same result using either calculation strategy.

So indeed it is possible to use the equation of motion for the co-rotating coordinate system, which involves evaluating Coriolis term components. But as said, it's not practical.

 

[GRDixon]

If you are at the altitude of geostationary orbit, orbiting geostationary, and you release an object to free motion, then it'll never reach the ground.

You're Right!

 

[MotoH]

Just finish playing CoD4?

 

[vin300]

The point is that (for short range projection) the time depends on the initial component normal to the ground, while the coriolis force indirectly depends on the component parallel to the ground, and the instantaneous normal component, so you cannot have a relation