> Actually, rough calculations (guessing at the size of the moon as I
> don't have figures handy) show that the moon is only producing about
> 10000 times as much gravitational force as the person sitting next to
> you, so Dawkins should be right for any planet other than Earth.
Grrr. On rec.arts.sf.science people show their math. "Astrological"
calculations based on Newton's law below.
F=G*M*m/R^2. G and m (your mass) are constant, so I'll look at M/R^2.
M moon[1]: 7e22 kg. R (semimajor axis): 384,000 km == 3.8e8 m.
M/R^2= 4.8e5.
M person: 100 kg. R: 1 meter.
M/R^2= 100.
Ouch. But you know, people do get closer than one meter.
R= 1cm. M/R^2= 1e6.
So your sex partner, or mother, seem to be twice as significant as the moon.
Never mind riding a crowded train. Actually, since their mass and yours are
spread out over several centimeters (along a line going through two mashed
torsos) we'd actually have differential effects in the bodies, as the force
fades rapidly.
So Dawkins wasn't far off. And any other planet is even less significant.
Jupiter M is 1.9e27 kg, R=7.8e11m. R/M^2= 3122.
Venus M= 4.8e24 kg, R=1e11 m, R/M^2= 480
and everything else will be smaller than the biggest and closest planets.
One cubic kilometer of rock masses 3 trillion kilogram, so the number for a
mountain three kilometers away would be 1e5, and my horoscope should be
affected by whether I was born (how about live?) in Chicago or Pasadena.
And if we throw G (6e-11) back into the equation, we can see just how big
these forces actually are.
[1]
http://www.oikos.warwick.ac.uk/~plumbago/Planetary_data/planets.html
-xx- GCU Librarian Bound in Pale Leather X-)
"Bastards or not, the U.S.A. would at least give my children passports."
-- Freeman Dyson, _Disturbing the Universe_