Re: [Phys-L] apparent weight/earth moon sun

[Bill Nettles <bnettles@uu.edu>]

Okay, so sometimes I get impatient.  I wrote my own VPython program, integrating the Newtonian gravitational accelerations of earth from sun  and of moon from sun and earth. I wanted to see the path of the moon around the sun.

I scaled to variation of the earth moon separation to be 20x (adjustable in the program) larger in the view screen than the orbital radius and used verlet velocity integration with an adjustable time step (1 hour is moderately slow, 3 hours is a nice speed).  I included solar radial lines to aid in counting conjunctions and was able to adjust initial conditions to have a near correct synodic period.  Now I'm wondering whether the earth acceleration due to the moon would affect things dramatically.... hmmm.... about 1% of the moon accel due to earth.  I'll check the effect later.

Anyway, The moon orbit is nice with curvature toward the sun, reaching near cusps at the new moon positions. Not at all sinusoidal viewed w.r.t the sun.
This probably isn't the cleanest code, the I like the pictures.  Requires VPython from vpython.org, but doesn't have to have the latest and greatest version.

from visual import *
## set up visual objects
scene.width=700
scene.height=700

sun=sphere(color=color.yellow,radius=.2)
earth=sphere(color=color.white,radius=.11)
moon=sphere(color=color.red,radius=.11)
sun.pos=(0,0,0)

##set physical constants in SI units
G=6.67e-11
earth.mass=5.98e24
moon.mass=7.342e22
sunmass=1.98e30
edist=149.17e9
mdist=240000*1609.

## the .r properties are the positions in SI units
## the .pos properties are the scaled positions and are used for plotting
##  scale is the scaling factor for the earth-moon separation versus the
##  orbital radius around the Sun
scale=20

## the 1e9 factor was introduced early in development to handle viewing scale problems
##  I simply didn't want to clear it up

earth.r=vector(edist/1e9,0,0)

moon.r=vector(edist/1e9+mdist/1e9,0,0)
deltar=moon.r-earth.r

## use earth.pos=(0,0,0) to view moon motion around earth. Also,
##  make sun.visible=false

earth.pos=earth.r/scale
#earth.pos=(0,0,0)
#sun.visible=false

## set initial speeds at full moon position

moon.pos=earth.pos+deltar
espeed=edist*2*pi/3.156e7
mspeed=mdist*2*pi/(27.32*24*3600)
earth.p=vector(0,espeed*earth.mass,0)
## adjust moon speed to get synodic period close to 29.5 days...it's not exact
moon.p=vector(0,(espeed+0.99*mspeed)*moon.mass,0)

## time interval in seconds...3 hours runs well, 1 hour is slow but more precise
dt=3600.*3
factor=1.00e9
mdist=mdist/factor
##leave a trail
earth.trail=curve(color=earth.color,width=.5)
moon.trail=curve(color=moon.color,width=.5)
efac=-G*earth.mass*sunmass/factor/factor
mfac=-G*moon.mass/factor/factor
earth.force=efac*earth.r/mag(earth.r)**3
delta=mag(earth.r-moon.r)
moon.force=mfac*(sunmass*moon.r/mag(moon.r)**3-earth.mass*(earth.r-moon.r)/delta**3)
earthf0=earth.force
moonf0=moon.force
##draw radials to the sun; aids in counting conjunctions
linemoonsun=curve(pos=[moon.pos,sun.pos],color=moon.color)
lineearthsun=curve(pos=[earth.pos,sun.pos],color=earth.color)
while true:
    rate(200)
## use the velocity verlet integration method
    earth.r=earth.r + (earth.p+0.5*earthf0*dt)*dt/factor/earth.mass
    earth.force=efac*earth.r/mag(earth.r)**3
    earth.p=earth.p+0.5*(earth.force+earthf0)*dt
 
    moon.r=moon.r + (moon.p+0.5*moonf0*dt)*dt/factor/moon.mass
    delta=mag(earth.r-moon.r)
    moon.force=mfac*(sunmass*moon.r/mag(moon.r)**3-earth.mass*(earth.r-moon.r)/delta**3)
    moon.p=moon.p+0.5*(moon.force+moonf0)*dt

    earthf0=earth.force
    moonf0=moon.force

    deltar=moon.r-earth.r
##  change earth.pos depending on which view you want, sun centered or earth dentered

    earth.pos=earth.r/scale
#    earth.pos=(0,0,0)

    moon.pos=earth.pos+deltar
    earth.trail.append(pos=earth.pos)
    moon.trail.append(pos=moon.pos)

    linemoonsun.pos=[moon.pos,sun.pos]
    lineearthsun.pos=[earth.pos,sun.pos]

________________________________________
From: Phys-l [phys-l-bounces@www.phys-l.org] on behalf of Bill Nettles [bnettles@uu.edu]
Sent: Sunday, December 14, 2014 7:30 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] apparent weight/earth moon sun

You should check out the University of Nebraska/Lincoln astronomy simulation site.  Lot's of interesting exercises and good demos for explaining a variety of things from eclipses to sidereal vs solar time.

http://astro.unl.edu


http://astro.unl.edu/classaction/animations/lunarcycles/moonphases.html

This demo gives a CONCEPTUAL idea of the moon's motion along with Earth. It is NOT to scale, so you should draw any conclusions about the real path of the moon. JD's plot is much better.  You should do the exercise yourself, or even better, use Visual Python to build your own demo.
________________________________________
From: Phys-l [phys-l-bounces@www.phys-l.org] on behalf of Anthony Lapinski [Anthony_Lapinski@pds.org]
Sent: Sunday, December 14, 2014 9:33 AM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] apparent weight

Thanks for the detailed reply and notes. Very helpful.

One more thing. I heard/read once somewhere that
because the Sun's force on the Moon is about 2x the
Earth's force on the Moon, the Moon orbits the Sun more
than it orbits the Earth. In other words, the Moon's true
orbital path is always concave around the Sun.

How would the Moon's orbit actually look? Similar
to the epicycle model to show planetary motion? Would the
Moon's orbital path ever cross itself? Trying to visualize this...
Is there an onlne simulation anywhere?


Phys-L@Phys-L.org writes:
> On 12/13/2014 09:10 PM, Anthony Lapinski wrote:
> >  I might not ask this question to my HS students!
> > Seems a bit advanced for them.
>
> If left strictly to their own devices, they're not likely
> to figure it out anytime soon.  It took Isaac Newton a
> good long while to figure it out ... using calculus.
>
> OTOH we don't have to leave them to their own devices.
> We can leave a trail of breadcrumbs for them to follow.
>
> The basic idea can be understood using basic principles
> of physics and a little bit of algebra.  (It would be
> a whole lot easier using calculus, but we can get by
> without that.)   Basically all you need is the equivalence
> principle plus some simple scaling ideas.  I wrote up
> the details at
>  https://www.av8n.com/physics/tides.htm#sec-low-tech
>
> The math is not rocket science.  You need to convince
> yourself that the difference between 1/60^2 and 1/61^2
> is very nearly 2/60^3.  That is to say, the tidal stress
> scales like inverse distance /cubed/.
>
> > We do calculate the forces of the Sun and Moon on the Earth. The
> > Sun's force is 175x more, but the Moon's tidal force is 3x more.
>
> Closer to 2.15x more, I would think.
>  https://www.av8n.com/physics/periods.m
>
> More importantly:  The force scales like distance squared,
> but the tidal stress scales like distance cubed, so we would
> expect them to be different.
>
> > Kids wonder about this.
>
> As well they should.  It's a good question, and deserves
> a good answer.  The answer is that in view of the equivalence
> principle, the only thing that could possibly matter is the
> /difference/ between the moon's field at the center of the
> earth and the moon's field at the surface of the earth.
> The field scales inversely like r squared, but the difference
> scales inversely like r cubed.
>
> > The Sun's force is 2x more, yet the Moon "appears" to orbit the
> > Earth. It actually orbits the Sun more! Kids wonder about this, too.
>
> That's basically the same question.  The same answer applies,
> mutatis mutandis.
>
> That is, we invoke the equivalence principle again.  If the
> earth and the moon were freely falling toward the sun at
> the same rate, there would be no observable consequences.
> Therefore the only thing that could possibly matter is the
> /difference/ between the sun's field at the two locations.
> This scales like distance cubed.  The pair/sun distance is
> 389 times the intra-pair distance, so we can pretty much
> ignore the sun when analyzing the dynamics of the pair.
>
> There is "some" tidal stress that messes with the orbital
> dynamics, but it's really small.
>
> ======================
>
> In this context, the best way to handle the centrifugal
> force is to not mention it at all.
>
> If somebody insists on bringing it up, the answer is:
> Remember, the centrifugal field exists in the rotating
> reference frame and not otherwise.  It does not matter
> at all whether the earth is rotating;  the only thing
> that matters is whether the reference frame is rotating.
> Repeat as many times as necessary:  The centrifugal field
> exists in the rotating reference frame and not otherwise.
>  https://www.av8n.com/physics/rotating-frame.htm
>
> For the tide question, choose an appropriate /non rotating/
> reference frame and turn the crank.
>
> ======================
>
> This is what I consider "conceptual physics".  You take two
> really basic concepts -- the equivalence principle plus scaling
> laws -- and combine them using a bit of reasoning.   It really
> offends me when I see books and courses that use the term
> "conceptual physics" as a euphemism for /bonehead physics/.
> The concepts are useless if you don't know how to combine
> them.  If folks are going to teach the concepts without the
> reasoning, they might as well not bother.  The students will
> (rightly) judge the disconnected concepts to be useless, and
> will forget them as quickly as possible.
>
> ======================
>
> BTW the tide question, if handled correctly, helps motivate
> kids to sign up for calculus.  Newton invented calculus for
> a reason, so he could do physics.
>
> The idea that the tidal stress scales like 1/r cubed was
> not just some lucky guess.  You can predict that with
> zero effort if you know a little bit of calculus.  I
> plotted it out so I could show people, but I knew in
> advance what the plot was going to look like.
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