Question about gravity wells

[Peron]

As we see in this video at 1:00

the moons gravity well is very small, hardly extending towards the Earth. So then how does the moon pull on the Earth?

[swansont]

As we see in this video at 1:00

the moons gravity well is very small, hardly extending towards the Earth. So then how does the moon pull on the Earth?

Didn't watch the video, but if the implication is that the well has a finite range it's wrong. The moon's pull is a force and the gravity well is energy. Further, if it's representing the gravitational potential, it ignores the mass of outside bodies.

 

The potential varies as 1/r and force as 1/r^2. They have an infinite range.

[md65536]

As we see in this video at 1:00

the moons gravity well is very small, hardly extending towards the Earth. So then how does the moon pull on the Earth?

Every mass pulls on every mass (gravity has an infinite range).

The moon's gravity well is not "small" but "shallow", especially at the range of the Earth.

The slope of the well corresponds to escape velocity (I think), ie. how much effort is required to escape from the mass.

 

If you could accelerate the Earth, it would be much easier to make it fly off from the moon than from the sun.

 

Or another way to think of it: The Earth's velocity around the sun keeps it in orbit, but it would be fast enough to escape the moon. (We don't escape the moon however, because the earth/moon barycenter also goes around the sun, while the Earth's velocity around that barycenter is relatively much slower).

[MigL]

We can't escape the moon because its gravity well is located inside the gravity well of the earth, which is located inside the gravity well of the sun,which is located inside the gravity well of the galaxy, etc. In effect if the earth were to escape the sun's gravity well, it would take the moon with it. Just semantics, I know, but sometimes you need to be anal about word meanings on these forums.

Edited August 21, 2011 by MigL

[swansont]

The slope of the well corresponds to escape velocity (I think), ie. how much effort is required to escape from the mass.

If we're talking about the potential, the slope represents the acceleration. [math]F= -\nabla{U}[/math] If you divide through by the mass of the object, you are left with the acceleration and the potential

[md65536]

If we're talking about the potential, the slope represents the acceleration. [math]F= -\nabla{U}[/math] If you divide through by the mass of the object, you are left with the acceleration and the potential

That makes more sense.

If you try to accelerate away from the mass with that force, it will balance gravity and you won't go anywhere.

 

If you accelerate away with a higher force, you'll move away but it will take infinite time and energy to escape it, until you reach the escape velocity (which will be smaller the farther you are away from the mass, corresponding with decreased gravitational acceleration which can be seen in the video as the slope of gravity wells getting shallower the farther you are from the mass). At escape velocity, your momentum will carry you further from the mass and its diminishing gravitational force will not ever be able to slow you to a stop, and you can escape without any additional energy spent.

 

 

Just out of curiosity, would escape velocity instead correspond to the height of the well at any point? Which would also correspond to the integral of the slope of the well wrt. time, along the infinite line of an escape path of an object moving directly away from the mass?

[swansont]

That makes more sense.

If you try to accelerate away from the mass with that force, it will balance gravity and you won't go anywhere.

 

If you accelerate away with a higher force, you'll move away but it will take infinite time and energy to escape it, until you reach the escape velocity (which will be smaller the farther you are away from the mass, corresponding with decreased gravitational acceleration which can be seen in the video as the slope of gravity wells getting shallower the farther you are from the mass). At escape velocity, your momentum will carry you further from the mass and its diminishing gravitational force will not ever be able to slow you to a stop, and you can escape without any additional energy spent.

 

 

Just out of curiosity, would escape velocity instead correspond to the height of the well at any point? Which would also correspond to the integral of the slope of the well wrt. time, along the infinite line of an escape path of an object moving directly away from the mass?

 

It's related. The potential*mass gives you the potential energy, which is equal in magnitude to the kinetic energy you'd need to have. So the potential is equal to v^2/2, where v is the escape velocity. Since the potential energy is negative, going up the well gives you a smaller value.