# Conservation of momentum close to the Earth vs. far from the Earth

1. Apr 14, 2013

### Lelephant

There is something I don't fully comprehend. When I throw a ball into the air while I'm sitting in a train, I won't accelerate past the ball because the ball carries the momentum of the train just as well as I do. Furthermore, the ball, the train and I carry the momentum of the Earth, which is orbiting the sun.

When we launch a spacecraft, assuming that we launch one normal to the ground, that craft does not carry the momentum of the Earth by the time it's reached orbit even if it is still subjected to the pull of gravity. By the time the spacecraft is pulled back down from orbit by the acceleration of Earth's gravity, it obviously lands in a completely different place on the surface of the Earth because it hasn't carried the momentum of Earth's rotation into space with it.

Why not? If the ball conserves the train's momentum, shouldn't the space shuttle conserve the Earth's momentum?

Last edited: Apr 14, 2013
2. Apr 14, 2013

### Bandersnatch

Why do you think it doesn't?

3. Apr 14, 2013

### Lelephant

Because if it did, it seems as though the space craft would be following the rotation of the Earth. As in, as the space craft got further from the Earth, it would be rotating around the Earth at the same velocity as the Earth was rotating along its axis previously.

4. Apr 14, 2013

### Bandersnatch

But that would mean it gained momentum.

Consider the concept of angular velocity ω

$$ω=\frac{V}{r}$$

V is the linear velocity,
r is the distance from the centre of the rotation

For the craft to remain directly above the same point on the Earth's surface, it's angular velocity must remain constant at all times, and equal

$$ω=\frac{2\pi}{T}$$

Where T is the period of rotation equal to 24 hours in our case.

So, if

$$\frac{2\pi}{T} = \frac{V}{r}$$

and everything on the left hand side is constant, then for any increase in r, V must increase by the same factor.

Since the momentum(velocity) stays the same, the angular velocity goes down the farther away from the centre of the Earth you go.

This is, of course, assuming the vertical takeoff, which is not how the rockets are launched, since you need to increase their tangential velocity to keep them in orbit.

5. Apr 14, 2013

### Staff: Mentor

They are launched in the direction of earth's rotation for exactly that reason in order to keep the linear momentum they had when sitting on the earth.
Oh, I see what you mean. You're talking about maintaining the angular speed. The rotation rate. Objects do not keep the angular rotation rate, it is just that for objects near earth and at low speed, you don't notice. So:
If you throw a ball straight up from earth, it will not land at the exact spot it was thrown from.

This effect is called the Coriolis effect. http://en.wikipedia.org/wiki/Coriolis_effect

6. Apr 14, 2013

### Lelephant

Why isn't the angular velocity 2*pi*r/T? It seems like the speed at which you travel around the perimeter of the Earth ought to be the circumference (distance traveled in one Earth rotation) divided by 24 hours (the period of one of Earth's rotations) given that velocity = distance/time.

So basically, the spacecraft does maintain angular velocity, but as it leaves the atmosphere and moves away from Earth, its angular velocity becomes negligible?

7. Apr 14, 2013

### Bandersnatch

You should read up on the difference between angular velocity and velocity.

In short, while velocity is in the form of distance_change/time_change, and so has units of m/s, angular velocity(as the name suggests) is angle_change/time_change, and has units of radians(or degrees) per second.

8. Apr 14, 2013

### Lelephant

And it's 2pi because it's in terms of radians. Alright, I think I understand. Thanks!