# Planes affected by earth's rotational spin

1. Sep 23, 2009

I was wondering, when planes are flying at an altittude of 10km above sea level,, does the earth's rotational spin affect the planes velocity? e.g. fly against rotational spin would it increase the velocity of the plane?

2. Sep 23, 2009

### tony873004

It doesn't even have to be 10 km high. Earth rotation affects a deep pass in a football game, by a few millimeters if I remember correctly.

Earth's rotation causes a coriolis force, causing airmasses to move in circular motions. Any plane flying through a moving airmass will be affected. This is why the speed of planes (and boats too) are measured in knots rather than mph or km/hr. Knots tells you how fast you are moving relative to the air, (or in the case of a boat, the water), rather than ground speed.

3. Sep 24, 2009

ohh thank you i understand now :)

that's why space ship launches are situated closer to the equator in which they can utilize the earth's rotational spin more efficiently

4. Sep 24, 2009

### FredGarvin

The unit of knots is simply a measure of speed. It has nothing to do with relative ground speed or airspeed. It is equivalent to 1 nautical mile per hour.

5. Sep 24, 2009

### tony873004

Spaceships launched at the equator get an extra 1000 mph boost that they would not have if they were launched from the poles. Additionally, launching from the equator allows you to launch into an orbit of any inclination, from equatorial to polar. Launching from the poles only allows you to enter polar orbits (without burning a lot of fuel correcting).

Thanks for the correction.

6. Sep 24, 2009

### DaveC426913

Can you please clarify this second statement? Other than the extra 1000mph, I don't see how it is any easier to go from an equatorial launch to a polar orbit than it is to go from a polar launch to an equatorial orbit. Either way, you burn the same fuel to alter your orbit.

I can see if you're suggesting that the 1000mph would make it easier, sure, but your second statement said "additionally", suggesting it is cumulative with the first. I think that you're double-dipping.

7. Sep 24, 2009

### willem2

you can fly a polar orbit from the equator by starting of north or southwards. your extra 1000mph will be wasted. If you set of from the pole the only way to get into an equitorial orbit, is to fly to the equator and kill all your north/south velocity and start of accelerating east or westwards

8. Sep 24, 2009

### D H

Staff Emeritus
They only get an extra 1000 mph boost at the equator if they are launching into a prograde equatorial orbit. For any other inclination, only the projection of the velocity due to earth rotation onto the desired orbital plane is of use. The component of that initial velocity due to Earth rotation that is normal to the desired orbital plan is worse than useless. It has to be negated by the rocket. For an inclination of 45 degrees or more, it is better to launch from somewhere other than the equator. For a polar orbit, the best place to launch is somewhere far in the north.

Aside: Most polar orbiters are not in a true polar orbit. They are in an orbit with an inclination greater than 90 so they can be sun synchronous. Launching from the equator into a 98 degree orbit is downright silly.

Last edited: Sep 24, 2009
9. Sep 24, 2009

### DaveC426913

10. Sep 24, 2009

### nucleus

The word Knots goes back to the sailing days. They used to have a rope with knots tied in it a certain distance apart. This rope was wound on a drum and the other end contained a wooden block that would drag in the water. By counting the number of knots in a certain time they could get a measure of speed.

More here:
http://en.wikipedia.org/wiki/Chip_log

11. Sep 25, 2009

### tony873004

I agree with all this. But the OP asked why we have launches situated from close to the equator. Pretty much it's because we are attempting to launch into near-equatorial, prograde orbits.

Such a method would measure the speed relative to the water, rather than relative to Earth's surface. This is probably how it was once explained to me, which is why I assumed it was speed relative to the air or water mass. But as Fred pointed out, (and Wiki agrees with him :) ), it's simply a measure of speed.

12. Sep 25, 2009

could you say the use of the earth's rotational spin to increase the velocity of a space craft is the "slingshot affect", or does that only apply as centripetal force? e.g using the planets orbit for momentum like Apollo 11 (i think it was Apollo 11)

13. Sep 25, 2009

### D H

Staff Emeritus
It sounds like your talking about the gravity assist or gravity slingshot maneuvers used to help unmanned probes go to other planets with minimal fuel expenditures. The Apollo missions did not use a gravity assist; they went direct from the Earth to the Moon. This approach was used for the first time with Mariner 10, which NASA sent to Mercury by way of Venus. Recently, the Cassini probe used Venus twice, then the Earth, and finally Jupiter to help get the probe to Saturn. This reduced the delta-v requirements from 15.7 km/s for a simple Hohmann transfer to 2.0 km/s.

14. Sep 25, 2009

### Cleonis

The fact that the Earth rotates has no effect on the velocity of planes relative to the Earth.

However, there are some rotation-of-Earth effects that do affect objects with a velocity relative to the Earth. This can be illustrated more clearly with the example of an airship (also known as 'Zeppelin').

An airship maintains it altitude by regulating its buoyancy: to sustain its altitude it must be neutrally buoyant in the air layer where it is floating.

Now imagine an airship positioned above some location that is on the Equator. The airship is co-rotating with the Earth; it's circumnavigating the Earth's axis in a full day. The amount of centripetal force that is required for that circumnavigating motion is about 3% of the Earth's gravitational force. (So you're not remotely in danger of being flung form the Earth, but on the Equator you do weigh a bit less than on the poles.)

Now imagine that airship cruising in westward direction along the equator, at a velocity of 25 meters per second. (That is about 55 miles per hour)

Going westward, against the rotation of the Earth, the airship is still circumnavigating the Earth's axis, but slower than the Earth itself. Now less centripetal force is required, so the airship is a bit heavier, so it needs more lift to be neutrally buoyant.

Finally, imagine that the airship makes a U-turn, to going 25 meters per second to the East. Now it's circumnavigating the Earth's axis faster than the Earth itself, and the airship needs to adjust its lift in order to remain neutrally buoyant.

In geophysics this rotation-of-Earth effect is called the Eötvös effect, after the geophysicist Eötvös. He first noticed the effect in gravimetric readings that had been taken onboard a sailing vessel. Al over the world many gravimetric readings had been taken. Readings during motion westward were consistently somewhat higher than when moving eastward.

Cleonis

15. Sep 25, 2009

### D H

Staff Emeritus
Sure it does, in two ways. The first is so obvious that no one has mentioned it yet: The Earth's atmosphere rotates with the planet, more-or-less. The second way in which the Earth's rotation effects the velocity of a plane relative to the Earth is that expressing velocity with respect to the surface of the Earth means that one is working in a rotating reference frame. One of the terms in the equations of motion for a rotating reference frame is the Coriolis acceleration, $-2\boldsymbol{\Omega}\times\mathbf v$.

This contradicts your initial statement. The Eötvös effect is simply the vertical component of the Coriolis acceleration.

16. Sep 25, 2009

### Cleonis

Let me put it this way: let's say you have a small airplane, that needs to reach a velocity of 100 km/h to get safely airborne. To accelerate the airplane to 100 km/h is the job of the propulsion.

Question: the amount of work that the propulsion must do to bring the airplane up to speed, is it the same on a non-rotating and a rotating planet? Yes, it's the same: it's the change of velocity that matters. That is: changing velocity from 1600 km/h to 1700 km/h requires the same amount of work als changing velocity from 0 km/h to 100 km/h.

Indeed the Eötvös effect can be seen as an aspect of the Rotation-of-Earth effect that is taken into acccount in meteorology and oceanography. Remarkably, this connection had remained unrecognized very long. As far as I know the first one to point it out explicitly was the meteorologist Anders Persson, around 1998.

Cleonis

Last edited: Sep 25, 2009
17. Sep 25, 2009

### D H

Staff Emeritus
So what? The centrifugal and coriolis forces don't do work. That does not mean that these apparent forces do not affect the behavior of a plane. It just means that the effect does not exhibit itself in the form of a change in energy.

18. Sep 25, 2009

### Cleonis

The original question informed about airplane velocity, and given the indications that the questioner had no physics background I took that as a question about airplane speed. (Still, technically the word 'velocity' includes direction; the usual definition of 'velocity' is as vector quantity.)

In the wider sense, taking velocity as vector quantity: the rotation-of-Earth effect does have influence on direction of motion.

In the case of the Eötvös effect: if the airship is trimmed for neutral buoyance while flying eastward, and the airshap makes a U-turn, it will have to be retrimmed. If not then the Eötvös effect will make the airship lose altitude, quite a dangerous situation.

When an airship loses altitude gravity is doing work. This illustrates that the Eötvös effect does not exclude change of energy.

Cleonis