How Does Earth's Helical Trajectory Affect Its Calculated Speed Through Space?

[TrickyDicky]

Usually the mean translation speed of the Earth around the sun is calculated (approximating its elliptic orbit with a circular orbit) to be around 29.7 Km/s.
This is fine if we consider the sun is still, but we know this is not the case (see for ilustration ), the sun is also moving thru space and the Earth traces a helical path with very clear speed cycles according to the Earth's distance to the sun.
I was wondering if anybody knows about a more realistic calculation of the Earth's speed in its helical trajectory, I would think it should be less than the 29.7 Km/s calculated with the circular orbit approximation taking into account that at certain points the sun surpasses the earth, and subsequently the Earth has to accelerate to catch up with the sun and surpass it.
Heuristically it would seem like the Earth's accelerates and deccelerates a maximum around +/- 15 Km/s that adds up to the total of around 30 Km/s usually calculated.
This would fit well with the pattern of the CMB dipole from COBE (see George Smoot astrophysics page where it says: "the COBE DMR observations clearly show the change in velocity at the 30 kilometers per second due to the motion of the Earth around the Sun. One can see a clear sinsusodial pattern in the amplitude and direction of the dipole with a one year period in the four years of COBE DMR data. Differencing maps taken six months apart produces the familar dipole pattern with the amplitude and direction of the Earth's motion."

Any thoughts? corrections?

 

[HallsofIvy]

You cannot talk about "speed" without saying what it is to be relative to. The first figure you give, 29.7 km/s, is correct for speed relative to the sun.

You want "earth's speed in its helical trajectory"- okay, relative to what? Yes, the sun is moving, relative to nearby stars- but with different speeds depending upon which you take it "relative to".

 

[TrickyDicky]

You cannot talk about "speed" without saying what it is to be relative to. The first figure you give, 29.7 km/s, is correct for speed relative to the sun.

You want "earth's speed in its helical trajectory"- okay, relative to what? Yes, the sun is moving, relative to nearby stars- but with different speeds depending upon which you take it "relative to".

Sorry, I thought it was evident from the set up, relative to the "moving" sun, not to the "still" sun. So if we picture the sun as a "moving target", the Earth has an accelerating period in which surpasses the sun and crosses to the other side where the opposite happens, it lags until their speed is equal and then it crosses again to the other side increasing its speed until it surpasses again. The point of the cycle where their speeds are the same for a moment is what I'd call the "zero speed" just for reference so that at the point of maximum speed advantage wrt the sun it goes near 15 Km/s. wrt the sun.
Actually the 30 Km/s mean speed is not realistic, it is the mean speed it would need an object to travel a trajectory whose length is that of one circunference with radius the mean distance earth-sun. But since in fact the sun is moving and the earh's path is helical its speed is more realistically modeled as a cyclic variation from 0 to 15 Km/s and from 15Km/s to 0 wrt the moving sun.
Is this correct?

 

[harrylin]

Sorry, I thought it was evident from the set up, relative to the "moving" sun, not to the "still" sun. [..]

I'm afraid that you misunderstood the question; in order to define a trajectory and a velocity one has to choose a reference system. Logically you supposedly mean "relative to our Galaxy", since you are referring to that video.

 

[TrickyDicky]

I'm afraid that you misunderstood the question; in order to define a trajectory and a velocity one has to choose a reference system. Logically you supposedly mean "relative to our Galaxy", since you are referring to that video.

Well, yes, taking the galaxy motion in the usual idealized sense of the comoving frame (CMB frame, thus my reference to the COBE dipole picture).

 

[Janus]

Sorry, I thought it was evident from the set up, relative to the "moving" sun, not to the "still" sun. So if we picture the sun as a "moving target", the Earth has an accelerating period in which surpasses the sun and crosses to the other side where the opposite happens, it lags until their speed is equal and then it crosses again to the other side increasing its speed until it surpasses again. The point of the cycle where their speeds are the same for a moment is what I'd call the "zero speed" just for reference so that at the point of maximum speed advantage wrt the sun it goes near 15 Km/s. wrt the sun.
Actually the 30 Km/s mean speed is not realistic, it is the mean speed it would need an object to travel a trajectory whose length is that of one circunference with radius the mean distance earth-sun. But since in fact the sun is moving and the earh's path is helical its speed is more realistically modeled as a cyclic variation from 0 to 15 Km/s and from 15Km/s to 0 wrt the moving sun.
Is this correct?

I'm sorry, but this just seems like a complete jumbling up of reference frames.

The 30 km/sec is the Earth's speed relative to the Sun.
The Earth's orbit is tilted some 60° to the galactic equator.
The Sun orbits the Galaxy's center at some 220 km/sec.

So, relative to the galactic center, the Earth travels in a helical path that looks something like a slightly flattened corkscrew. Relative to the galactic center, the Earth's velocity oscillates around 220 km/sec, (between ~206 and ~236 km/sec).

Now, while at first, this seems to indicate that the Earth's speed with respect to the Sun varies from roughly + or - 15 km/sec, as measured from the galactic core, this is misleading. You have to do vector addition to get the speed difference and when you do, you get 30 km/sec.

So you can say the the Earth travels at 30 km/sec with respect to the Sun, or travels between ~206 and ~236 km with respect to the galactic center, but you cannot say the the Earth moves at 15 km/sec with respect to the Sun.

 

[TrickyDicky]

Thanks Janus, I see what you mean.
I guess what I was saying would only be valid if one makes the unrealistic assumption that the sun is representing exactly the CMB frame.