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

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Discussion Overview

The discussion revolves around the conservation of momentum in the context of objects moving near Earth compared to those in orbit. Participants explore the implications of momentum transfer when a ball is thrown from a moving train versus a spacecraft launched into orbit, examining how these scenarios relate to Earth's rotation and gravitational effects.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that when a ball is thrown from a moving train, it retains the momentum of the train, suggesting a similar expectation for a spacecraft launched into orbit.
  • Another participant questions the assumption that the spacecraft does not carry Earth's momentum once in orbit, prompting a discussion about the relationship between the spacecraft's motion and Earth's rotation.
  • It is proposed that for the spacecraft to remain above a fixed point on Earth, its angular velocity must remain constant, leading to a discussion about the relationship between linear velocity and angular velocity.
  • Some participants introduce the Coriolis effect as a factor that influences the trajectory of objects thrown from a moving reference frame, such as Earth.
  • There is a clarification about the definitions of angular velocity and linear velocity, with emphasis on their different units and implications for motion.
  • A participant raises a question about the formula for angular velocity, suggesting that it should consider the distance traveled around the Earth's circumference.
  • Another participant acknowledges the distinction between angular velocity and linear velocity, indicating a growing understanding of the concepts discussed.

Areas of Agreement / Disagreement

Participants express differing views on whether a spacecraft retains Earth's momentum in orbit, leading to an unresolved discussion with multiple competing perspectives on the relationship between angular and linear velocities.

Contextual Notes

Participants discuss various assumptions regarding the motion of objects in relation to Earth's rotation and gravitational influence, with some mathematical steps remaining unresolved. The discussion also highlights the complexity of defining momentum in different reference frames.

Lelephant
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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?
 
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Lelephant said:
that craft does not carry the momentum of the Earth by the time it's reached orbit
Why do you think it doesn't?
 
Because if it did, it seems as though the spacecraft would be following the rotation of the Earth. As in, as the spacecraft 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.
 
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.
 
Lelephant said:
Because if it did, it seems as though the spacecraft would be following the rotation of the Earth.
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.
As in, as the spacecraft 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.
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:
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.
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
 
Bandersnatch said:
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.

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.

Bandersnatch said:
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.

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

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

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