From: gpkatch@alcor.concordia.ca (Gary Katch)
Newsgroups: rec.org.mensa
Subject: Re: The Providence of Astology
Date: 9 Oct 92 16:04:06 GMT
  1 "sichase@csa1.lbl.gov
    sci.skeptic
    Some Real Physics (Was: Re: Of Egg Yolks and Astrology)
    30 Sep 92 19:49:00 +0000"
    <30SEP199211494188@csa1.lbl.gov>
  2 "arromdee@jyusenkyou.cs.jhu.edu
    30 Sep 92 20:02:35 +0000"
    <1992Sep30.20
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  "edp@math.zk3.dec.com (Eric Postpischil)"
  "06 Oct 92 18:04:19 GMT"
  <1992Oct6.180419.3423@nntpd.lkg.dec.com>
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Eric Postpischil brings up a common protest about astrology:
> ...
>But where was the doctor!?  The gravitational influence of the doctor is
>greater than that of Mars or most other astronomic bodies.

There was a discussion of this in sci.skeptic recently.  Relevant
sections follow and show that things are never really so simple.
[See References header for references]

In an article [1], SCOTT I CHASE explains:

  Come on. This is just not true.  Have any of you gugs who believe this
  actually tried to work out the numbers?   Geez.

  Sun vs. Doctor:

  Doctor:  100 Kg at 2 feet from the baby.  Sun:     2E30 Kg at 9.3E7
  miles from the baby.

  The gravitiational force, G, of the Sun is down by a factor of ratio
  of distances squared = ((9.3E7*5280)/2)^2 = 6E22.  That's a big loss
  due to distance.  On the other hand, G for the sun is larger due to
  it's HUGE mass by a factor of the ratio of the masses = (2E30/100) =
  2E28.  So the bottom line is that the Sun has a pull more than 300,000
  times the pull of the doctor.

  Moon vs. Doctor:

  Moon:         7E22 Kg at 384,000 Km from the baby.

  I'll spare the calculation, which any of you can do.  The bottom line
  is that the moon pulls more than 10,000 more strongly than the doctor.
  You all should have realized this - after all, your doctor's mass
  can't affect the tides, even at close distances, but the moon's mass
  can.

But Ken Arromdee protests [2]:

  Do you know what tides are?

  The tidal forces are indeed greater from the doctor than the sun.
  Tides are caused by the difference in gravitational forces between
  the front and the back.  Although the sun's gravity is strong, all of
  a person is essentially the same distance from it, which is not so
  for nearby things.

And Nichael Cramer supports this [3]:

  The problem is that this is mistated.  Properly, it should be that
  "the gravitation _tidal_ force of the doctor..."  (actually the tidal
  gradient, to be fussy).

  Since this goes as r^-3, when you do the numbers you see that the
  doctor wins big.

  To anticipate what will obvious be your next approach, why should we
  believe that the tidal force to be any less significant than the
  gravitational force.

SCOTT I CHASE considers all this and continues [4]:

  Most of us who got physics degrees from MIT and are now graduate
  students at Berkeley know quite a bit about tides.

  That's not what the original poster was talking about, as you can see
  from the question which you yourself reposted.  He was making a claim
  about the net gravitational pull of the Sun, vs. the Doctor.  That's
  the question to which I responded.

  But you're still doing the same thing that I complained about in the
  first place - what good are your words when a little calculation is
  at hand? Can't you see that your mumbings about "all of a person is
  essentially the same distance from it" are useless to a skeptic?
  Physics is built upon mathematics, not philosophy.

  Let's consider the tidal stress:

  Of the Sun on the Baby:

  Length of the baby = 1 foot.  Fractional change in the Sun's pull over
  the length of the baby =
	(9.3E7*5280+1)/(9.3E7*5280) - 1 = 2E-12

  Of the Moon on the Baby:

  Fractional change = (384,000,000*3+1)/(384,000,000*3) - 1 = 9E-10

  Of the Doctor on the Baby:

  I assumed in my previous post that the doctor was 2 feet from the
  baby, so the fractional change in the force is:
	(2+1)/2 - 1  = .5

  Now we scale by the relative average pull to get the relative tidal
  stress:

  Sun/Doctor  = 2E-12/0.5 * 300000 = 1E-6 Moon/Doctor = 9E-10/0.5 *
  10000  = 2E-5

  So the Doctor indeed produces more tidal stress than either the Sun
  or the Moon.

  Incidentally, you can divide these two ratios to get:

  Moon/Sun = 20.

  This is roughly correct, and demonstrates why the Moon dominates the
  Earth's tides and why the Earth-Moon-Sun alignment produces small
  corrections which lead to neap tides, etc.

SCOTT I CHASE continues [7]:

  Yesterday, I posted calculations comparing the Moon to a Doctor, and
  found that the tidal force of the Moon is larger than that of the
  Doctor.  But now you want to talk about mountains, which are millions
  of times more massive than the doctor, versus planets which are tens
  to thousands as far away as the Moon.  So, without doing the calculation,
  since I don't have hard numbers, I think that the Mountains will win
  this contest.

In an article [8] by Carl J Lydick also considers tides:

  But since the patient/Earth system is in free fall wrt the Sun, while
  the doctor and the patient are not in free-fall wrt each other, we
  should be comparing the gravitational effect of the doctor with the
  tidal effect of the sun.  I think you'll find that in that comparison
  the doctor loses by only a factor of 26 or thereabouts, since the
  tidal force goes down as the cube of th e distance between objects,
  and is proportional to the size of the object experiencing the tides
  (in this case the patient/Earth system).  With the sun barely winning
  out over the doctor, the planets don't stand a chance.

  And tides are so much weaker that even large objects (e.g., the sun)
  at large distances exert tidal forces that are comparable to the
  gravitational forces exerted by nearby skeptics, astrologers, etc.

In an article [9] by Carl J Lydick it says:

  ... let's not forget that the force to be considered between
  the doctor and the baby is the gravitational force, not the tidal
  force.  They're not in freefall around each other (unless, of course,
  we're assuming a space habitat).  Furthermore, the net tidal force
  exerted on the baby will be a function of the distance of the baby
  from the center of the earth (that *IS* approximately the thing that's
  orbiting the sun), not a function of the length of the baby.

You can't get away with much in sci.skeptic...

-- gk