Earth's Angular Velocity and Various Object Launching

[erok81]

First...hopefully this is in the right section and that gem of a title makes sense. :rofl:

We were having a discussion at work on the topic of launching satellites and ships launching missiles.

First off the satellites.
The comment was made that it is better to launch a space bound object nearest to the equator as the angular velocity is greatest there and therefore easier to obtain an escape velocity. This makes to me and seems correct. If it isn't correct, why not? I understand why it would be correct.

Second...ships and missile launches.
A comment was made that ships, while launching missiles, have to account for the spin of the Earth when launching at other ships/continents. This didn't make sense to me. Since everything is in the same rotating reference frame (earth) this didn't factor into calculating targets. Is it correct that ships have to take this into account when targeting?

 

[tiny-tim]

hi erok81!

The comment was made that it is better to launch a space bound object nearest to the equator as the angular velocity is greatest there and therefore easier to obtain an escape velocity. This makes to me and seems correct.

yes

A comment was made that ships, while launching missiles, have to account for the spin of the Earth when launching at other ships/continents. This didn't make sense to me. Since everything is in the same rotating reference frame (earth) this didn't factor into calculating targets. Is it correct that ships have to take this into account when targeting?

in a uniformly rotating frame of reference, two fictitious forces are needed to restore the validity of Newton's laws: centrifugal force and Coriolis force …

i think Coriolis force would make a very slight difference

(centrifugal force is already incorporated into the local value of g, but Coriolis force can't be, since it depends on velocity )

 

[beelzebot]

Go stand on the north or south pole. You will be rotating about your own center of mass and have zero translational speed relative to the center of the planet.

Now take a stroll down to the equator. Once you get there, you'll be traveling at some ~500m/s relative to the center of the Earth because you'll be in uniform circular motion around it.

What happened in between? How did you reach such a speed? The answer is that the friction between the Earth and your feet sped you up.

Now consider what would happen if you tried to walk (slide) down to the equator across an ocean of frictionless ice. No force could act on you, so by the time you get down there you will still be at rest with respect to the center of the Earth - but the surface of the Earth at the equator is moving beneath you! Therefore to someone moving along with the surface of the Earth it looks like you are moving because a force acted on you - the Coriolis force.

Something similar to this is experienced by objects moving through air or water (although there is still a little friction), which is why artillerymen sometimes have to account for it.

 

[daumphys]

Second...ships and missile launches.
A comment was made that ships, while launching missiles, have to account for the spin of the Earth when launching at other ships/continents. This didn't make sense to me. Since everything is in the same rotating reference frame (earth) this didn't factor into calculating targets. Is it correct that ships have to take this into account when targeting?

This is Coriolis force. This force is related to the direction of wind and ocean current in the Earth. At the Northern hemisphere of the Earth, wind and ocean current moving direction is clock wise. And Southern hemisphere is counter clock wise. When we throw a stone from the equator to the North pole, the stone is bended to the right direction.

 

[AlephZero]

This has nothing specifically to do with ships. It affects aiming of any heavy artillery with a range of more than a few miles.

Supppose you are at latitude theta on the Earth's surface. Call the Earth's radius R.

The Earth is rotating and you are traveling round a circle of radius R cos theta, once every 24 hours.

Suppose you are about half way between the equator and the north pole, and you aim a gun due north and fire it. When the shell leaves the gun, it has a sideways velocity that will carry it a distance of 2 pi R cos theta in 24 hours, the same as the gun.

But as it travels north, theta increases, cos theta decreases, and the sideways velocity of an object fixed to the Earth decreases. Therefore relative to the earth, the shell is not moving due north, but travels on a curved track.

The real situation is more complicated, because the a long range shell does not travel at ground level, it also has a vertical motion relative to the earth, and that means the effective value of R is also changing as well as theta. This affects the shell even if you fire east or west, and the latitude theta is constant. If you fire at the same angle above the horizon, the ranges aiming due east and due west will be different, because of the rotation of the Earth "underneath" the shell.