How does one compute the sideways velocity of a planet?

[swansont]

17 minutes ago, captcass said:

How about this: The Earth is always falling towards the Sun. For each km it moves forward in its orbit, how many km does it fall?

Define “fall” 

It’s an imprecise term, which can be interpreted differently depending on your coordinate system. 

[captcass]

2 minutes ago, swansont said:

Define “fall” 

I used that as that is how people commonly see objects, as "falling" due to gravity. I don't see them as falling, but as evolving through the continuum. The Earth is always accelerating towards the center of the Sun. Hence it is in free "fall" in space. 

One of the problems we have is that we still refer to events  in classical terms, when it is a quantum world. Things do not fall "thru" anywhere. They evolve within the continuum. We currently have to use external forces to shift masses within it. We need to be focusing instead on the continuum and how to evolve masses through it. This is the only way our "visitors" can maneuver the way they do in the air and water. I think this could possibly be a gravity drive based on virtual dilation gradients.

[MigL]

I'm still very confused.
There are two velocities experienced by an orbiting body, in spherical polar coordinates ( for simplicity ).
There is the angular component, the change in the angles of SP coordinates, which is also known as the orbital speed.
This speed varies somewhat, to satisfy Kepler's orbital equation, but is approx. 30 km/s ( rounded off ! ).

The radial component of SP coordinates also changes, going from positive to negative, since the orbit is slightly elliptical
But it is very slight, and there is no net ( around the whole orbit ) change of radius, or else the orbit would be either decaying, or escaping.

Can you please tell us which 'change of distance', per unit time, refers to the velocity you want to know about ????

[captcass]

11 minutes ago, MigL said:

which 'change of distance', per unit time

km/s. In the answer originally quoted from UCSB, it is stated the Earth moves "sideways" towards the center of the Sun at a rate of 3 km/s.

If the Earth suddenly began to continue straight along its current tangent line, at what rate would the orbit drop away below it?

[swansont]

1 hour ago, captcass said:

I used that as that is how people commonly see objects, as "falling" due to gravity. I don't see them as falling, but as evolving through the continuum. The Earth is always accelerating towards the center of the Sun. Hence it is in free "fall" in space. 

But you’re a published cosmologist. Surely you can be more precise in your language, rather than relying on how non-scientists describe things.

Does fall mean the change in y as x changes, in a Cartesian system, or does fall mean a change in r, in a spherical coordinate system.

 

1 hour ago, captcass said:

One of the problems we have is that we still refer to events  in classical terms, when it is a quantum world.

Absolutely none of what’s been discussed is related to QM.

 

[captcass]

1 minute ago, swansont said:

Absolutely none of what’s been discussed is related to QM.

Actually, it does. You just aren't understanding what you are seeing. The continuum isn't hidden from view on some tiny scale. We are looking at it all the time.

 

 

Soooooo....after re-reading the original UCSB answer, I, too, think it was an error on the part of the person answering the question. Sooooo. Thanks to all anyway, but it looks like it is my bad.

I was excited because 3.2 km/s would have made sense to what I am working on in one respect. My calculations based on the curve of the orbit gives me only  5.97042055×10⁻⁶ km/s, which seems way too small.

 

So. Sorry and thanks.

[Phi for All]

14 hours ago, captcass said:

Soooooo....after re-reading the original UCSB answer, I, too, think it was an error on the part of the person answering the question. Sooooo. Thanks to all anyway, but it looks like it is my bad.

I was excited because 3.2 km/s would have made sense to what I am working on in one respect. My calculations based on the curve of the orbit gives me only  5.97042055×10⁻⁶ km/s, which seems way too small.

 

So. Sorry and thanks.

!

Moderator Note

Your apology for criticizing the "snide remarks" of the members trying to help you see this is accepted.