How does one compute the sideways velocity of a planet?

[captcass]

A UCSB site says, " However, the Earth is actually moving sideways compared to the center of the Sun at 3 km/second...."

Normally "sideways" refers to the forward orbital direction, which is at 29.79 km/s, so I am assuming this is the velocity towards the center of the Sun due to gravity, that is offset by the acceleration in a stable orbit.  Is that correct? And how does one calculate that sideways velocity for any planet? Thanks.

[Halc]

29 minutes ago, captcass said:

A UCSB site says, " However, the Earth is actually moving sideways compared to the center of the Sun at 3 km/second...."

~30 km/sec actually, as you seem to be aware.  'Sideways' would seem to mean the velocity vector is approximately perpendicular to a vector pointing at the sun from Earth.  This is true of any reasonably circular orbit.  So not sure what this UCSB site is referring to.

Right now the Earth is almost as close to the sun as it's going to get.  It will be closest in a couple weeks.  It's orbital distance is not changing at a rate anywhere near 3 km/sec.  Around April it might be increasing at a max of under 1 km/sec.

Quote

And how does one calculate that sideways velocity for any planet?

If I'm moving thataway, I have no velocity to either side of 'thataway'.  'Sideways velocity' seem to have no meaning out of context. Perhaps if you gave the context of the quote.

Edit: I found the quote, and it seems to have been taken unreviewed from a quora thread.  The '3 km/sec' is blatantly wrong, as is much on quora. They're talking about orbital velocity (off by an order of magnitude) and talking about why the Earth doesn't fall into the sun.

Edited December 29, 2020 by Halc

[swansont]

29 minutes ago, captcass said:

A UCSB site says, " However, the Earth is actually moving sideways compared to the center of the Sun at 3 km/second...."

A link would be appropriate and possibly helpful. Since the tangential speed is 30 km/s, it could just be a typo

[captcass]

Here is a link to the site. It is the top answer. http://scienceline.ucsb.edu/getkey.php?key=770

I should mention that I found statements like, "the sideways velocity is almost entirely due to the forward momentum...", or some such, "sideways" meaning parallel to the Sun's surface, without explaining the other factor(s).  The Earth is always passing tangentially to the Sun, but gravity is accelerating the Earth downward.

In a stable orbit, the acceleration due to gravity manifests as a steady velocity towards the Sun's center, orthogonal to the sideways motion. As the orbit is stable, that downward velocity combines with the tangential velocity and gives us the total free fall velocity along the curve of the orbit. I am trying to figure out how to compute that gravitational velocity component.

 

[Halc]

All orbits are stable unless A) the velocity is small enough that closest approach is less than the radius of the object orbited (collision), or B) the velocity is at least as large as escape veliocity.

All velocity is due to forward momentum, by definition. In other words, the velocity vector of an object cannot have a different direction than its momentum.

The sun's gravity accelerates Earth downward (towards the sun) always, whether the orbit is eccentric or not.

If the Earth is always moving tangentially to the sun, then its orbit is circular (it's not, but close), and there is no 'downward' velocity. The radius remains fixed. There is no other velocity that is the total free fall velocity. There's just the tangential component.

For an eccentric orbit, the force on the orbiting thing is typically not perpendicular to its velocity, which results in a change in radius (distance between objects) and orbital speed. One can compute the speed by integrating F=ma over time, or by computing total mechanical energy at any given point in the orbit, which remains fixed for the entire orbit per energy conservation.

[captcass]

2 minutes ago, Halc said:

The sun's gravity accelerates Earth downward (towards the sun) always,

This is why it has to manifest as just a velocity. Otherwise we'd spiral in. Velocities are accelerated in a gravity well. You are not looking deeply enough at the contributing factors.

I understand what you are saying, but I am not asking that. I am a published cosmologist. Thanks, though.

[MigL]

11 minutes ago, Halc said:

there is no 'downward' velocity. The radius remains fixed

I would agree.
There is an angular speed, the orbital speed of 30 km/s.
But there cannot be a radial velocity, otherwise the orbital radius would change at that rate.

6 minutes ago, captcass said:

I understand what you are saying, but I am not asking that.

Then explain yourself better.
We are confused as to exactly what you are asking.
I'm sure the writings of a published Cosmologist should be much clearer.

[captcass]

14 minutes ago, MigL said:

I'm sure the writings of a published Cosmologist should be much clearer.

You know, one of the reasons I hate asking questions in this particular forum is the snide remarks i get. I come back from time to time because I have had things clarified for me here. But the rude comments really show a lack of professionalism and make it difficult to come here.

I am doing time dilation cosmology. Whereas you see objects "falling thru" space. I see objects evolving along time dilation gradients in an evolving energy field. 

For instance, the reason light bends around massive objects is because it is evolving down the time dilation gradient. It is not because space is bent. It is because the forward direction of evolution is down the dilation gradient.

This is happening to all particle events, regardless of size or complexity.

I am just wondering if anyone here can tell me how to compute the gravitational component without getting into working EFE's? It should be a simple Newtonian calculation, but damned if I can find it.

 

[zapatos]

4 minutes ago, captcass said:

But the rude comments really show a lack of professionalism

No need to be snide about it.

[captcass]

5 minutes ago, zapatos said:

No need to be snide about it.

🤣

[MigL]

9 hours ago, captcass said:

For instance, the reason light bends around massive objects is because it is evolving down the time dilation gradient. It is not because space is bent.

I'm sure a published Cosmologist would know that light 'bends' around massive objects because TIME and space are curved, not just space.

There is a difference between 'accurate' and 'snide'.

[swansont]

11 hours ago, captcass said:

Here is a link to the site. It is the top answer. http://scienceline.ucsb.edu/getkey.php?key=770

I'm not overly impressed by those answers (and others I checked for a different question)

These look to be answers to non-science students, where they avoid using any actual physics.

Quote

I should mention that I found statements like, "the sideways velocity is almost entirely due to the forward momentum...", or some such, "sideways" meaning parallel to the Sun's surface, without explaining the other factor(s).  The Earth is always passing tangentially to the Sun, but gravity is accelerating the Earth downward.

That would be true if the orbit were a perfect circle, but it's not. So there is a radial velocity component. 

 

Quote

In a stable orbit, the acceleration due to gravity manifests as a steady velocity towards the Sun's center, orthogonal to the sideways motion. As the orbit is stable, that downward velocity combines with the tangential velocity and gives us the total free fall velocity along the curve of the orbit. I am trying to figure out how to compute that gravitational velocity component.

In a circular orbit the velocity toward the sun is zero; that will give you the azimuthal component from a = v^2/r

For the overall speed

https://en.wikipedia.org/wiki/Orbital_speed

You can get the radial value from the fact that the orbital energy is constant.

 

 

 

10 hours ago, captcass said:

 I am a published cosmologist. 

 

10 hours ago, captcass said:

You know, one of the reasons I hate asking questions in this particular forum is the snide remarks i get.  

Maybe that's because of people publishing in vanity press who say they are published (Journal of Cosmology, for instance) as if their work has been scrutinized like people who do it for a living.

[captcass]

2 hours ago, MigL said:

because TIME and space are curved, not just space

You are looking at space and time wrong. What we are looking at is the evolving energy field, not a fixed space. Time evolves space, and spatial events, forward in the forward direction of time, which has no depth. So even though we see dimensions, there is no depth, just an evolving energy field. As time evolves the continuum forward, we see the apparent curvature of motion as events evolve down dilation gradients.

 

1 hour ago, swansont said:

as if their work has been scrutinized like people who do it for a living.

My paper was reviewed and published, not paid for. As for the JofC, after I was published, I acted as managing editor/webmaster for nearly a year and saw the inside workings. All papers are reviewed and fees are even waived for authors who cannot afford them. There is a vanity press aspect for the owners of the journal, but not for others they publish. They haven't published anything in about 6 months due to mental "end of life" issues of the Executive Editor. Until that is resolved I suggest people do not submit to them. If you check Vol 27 you will see what the problem is.

 

1 hour ago, swansont said:

That would be true if the orbit were a perfect circle,

The reference was to planetary orbits, not circles.

 

2 hours ago, swansont said:

For the overall speed

https://en.wikipedia.org/wiki/Orbital_speed

Thanks for that, but it doesn't really help.

[swansont]

49 minutes ago, captcass said:

The reference was to planetary orbits, not circles.

Yes. There was a sentence that followed the quoted one that was pertinent to the concept, as each sentence separately referred to orthogonal velocity components.

 

49 minutes ago, captcass said:

Thanks for that, but it doesn't really help.

It should. If you know (or can calculate) orbital speed, breaking it down into the radial and azimuthal components should be straightforward. It should be child’s play for a cosmologist, since cosmologists should know basic physics and math.

And if by “sideways velocity” you mean the azimuthal component (it’s not 100% clear - is “sideways velocity” the standard jargon?) then that’s trivially found from the centripetal acceleration equation.

[captcass]

11 minutes ago, swansont said:

is “sideways velocity” the standard jargon?

Everyone I found referred to the velocity along the orbital path as "sideways" as it parallels the sun's surface, i.e., orthogonal to a line from the planet to the center of the Sun. I've looked at the radial component and it isn't what I am looking for. The 3 km/s in the original post is the range I am looking at confirming, and then solving for the other planets will nail it all down. But I can't do that until I know how they computed the 3 km/s. That is not a typo, either, which is clear from the whole context of the answer. (Also, the earth's average velocity is not 30 km/s, as someone noted above, but 29.79 km/s, often referenced as 29.8, as on this NASA site https://nssdc.gsfc.nasa.gov/planetary/factsheet/)

[Janus]

For those interested in the equations used to work out the radial component for an elliptical orbit:

Orbital velocity: =

V_o = sqrt(u(2/r-1/a)

Where:

u is the gravitational parameter (GM) for the Sun

r is the present radial distance of the orbiting body

a is the semi-major axis of the orbit

"r" for a chosen point of an orbit is found by

r = (a(1-e^2))/(1+e cos(q))

where e is the eccentricity of the orbit

q is the angle from perihelion at the chosen point of the orbit.

The orbital velocity component perpendicular to  the radial line is found by

V_p = 2 sqrt(ua(1+e)/(1-e))/r

V_r, the radial component can derived from V_o and V_p by applying a bit of trig.

 

 

 

[captcass]

18 minutes ago, Janus said:

the equations used to work out the radial component

What do you get for Earth on an equinox?

Edited December 29, 2020 by captcass

[Janus]

51 minutes ago, captcass said:

What do you get for Earth on an equinox?

I'll give you the values to work it out for yourself.

e = 0.0167

a = 1.496e11 meters

u = 1.3275e20 m^3/s^2

q ( at the Vernal equinox) = 102.93 degrees

[captcass]

(Can't seem to get this image to display - just click on it)

for the Earth at perihelion. I don't see how this will help. It is just comparing different speeds at different points in the orbit. I am looking for a gravitational component.

Consider the problem this way, The Sun and Earth are infinitely far away and there are no other planets around the Sun. The Earth is moving at a steady velocity relative to the CMB. As it enters the Sun's gravitational field it appears to accelerate. It continues to accelerate until it makes its closest point of approach and then, because its new momentum equals the force of the gravitational field, it "falls" into orbit. It is still accelerating towards the Sun, but it gets no closer. This means the acceleration is manifesting as part of the overall velocity, the part added as the Earth entered the field.. 

Edited December 29, 2020 by captcass

[swansont]

5 minutes ago, captcass said:

Consider the problem this way, The Sun and Earth are infinitely far away and there are no other planets around the Sun. The Earth is moving at a steady velocity relative to the CMB. As it enters the Sun's gravitational field it appears to accelerate. It continues to accelerate until it makes its closest point of approach and then, because its new momentum equals the force of the gravitational field, it "falls" into orbit. It is still accelerating towards the Sun, but it gets no closer. This means the acceleration is manifesting as part of the overall velocity, the part added as the Earth entered the field.. 

The earth would not fall into orbit under these conditions. You have KE + PE > 0, which means you can’t form a bound system.

“acceleration is manifesting as part of the overall velocity” makes no sense

3 hours ago, captcass said:

(Also, the earth's average velocity is not 30 km/s, as someone noted above, but 29.79 km/s, often referenced as 29.8, as on this NASA site https://nssdc.gsfc.nasa.gov/planetary/factsheet/)

Round it to two significant digits, and what answer do you get?

[captcass]

1 minute ago, swansont said:

The earth would not fall into orbit under these conditions

I realize that. It is a hypothetical to show there is a gravitational velocity component acting orthogonal to the orbit. Borrowing from StackExchange:

"It (the body) does keep accelerating. Its velocity in the direction of the object being orbited keeps increasing.

But this direction keeps changing. The reason the satellite's total speed doesn't increase, at least in the case of a circular orbit, is that while its velocity towards the object increases, its tangential motion moves it forward so that that direction is always perpendicular to the direction of motion. Thus while the satellite is undergoing constant acceleration, that acceleration is always perpendicular to the direction of motion and the speed of the object never changes."

In other words, the acceleration, which is always towards the center of the Sun, manifests as part of the total orbital velocity.

 

[swansont]

IOW, the speed in an elliptical orbit isn’t constant. But the energy is. So you can find the speed at any point in the orbit if you know r, since v^2/2 - GM/r is constant.

From that, you can deduce the radial and azimuthal components, as I stated earlier.

[captcass]

4 minutes ago, swansont said:

you can deduce the radial and azimuthal components, as I stated earlier.

Thank you, I knew that. But I am looking for the other aspect i described.

Consider that I am correct and light appears to bend around a massive object because it is evolving down the gradient, not moving through curved space. At what velocity is the light evolving down the gradient?

[swansont]

2 minutes ago, captcass said:

Thank you, I knew that. But I am looking for the other aspect i described.

What other aspect? There are only two velocity components. Your thread title asks for one of them, which is the azimuthal component.

[captcass]

1 minute ago, swansont said:

There are only two velocity components.

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?