Earth/Moon Gravity

Nick Wedd writes:
You may assume knowledge of Newtonian physics, and of the general
structure of the solar system, but not of any of its dimensions. Using
only your own body as a measuring instrument, find which is denser, the
sun or the moon.

Very cute question! Possible spoiler...

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Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

First I observe either an annular solar eclipse or the start or end of
a total one. Either way, I deduce that the Sun and the Moon have almost
the same apparent size, so R and D are very nearly equal.

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is. Therefore it's the Moon that's driving the tides and therefore its
tidal force is significantly larger. But the tidal forces are in the
ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is
significantly less than 1, or in other words, the Moon is denser.

I don't know if this was the intended solution, though. It seems to me
that if I have to observe the behavior of the tides, then I'm not only
using my eyes as an instrument, which is allowed, but also the ocean
and the land, which isn't. For that matter, observing the eclipse
would be a lot safer if I was allowed to use a solar filter, but that
might be excluded as well -- and I don't have so much dedication to
rec.puzzles that I want to risk my eyesight for it. (A pinhole
projection system, of course, would be Right Out, even if the pinholes
are naturally formed by tree leaves.)
--
Mark Brader "I can say nothing at this point."
Toronto "Well, you were wrong."
-- Monty Python's Flying Circus

My text in this article is in the public domain.

In message , Mark Brader
writes
Nick Wedd writes:
You may assume knowledge of Newtonian physics, and of the general
structure of the solar system, but not of any of its dimensions. Using
only your own body as a measuring instrument, find which is denser, the
sun or the moon.

Very cute question! Possible spoiler...

40
39
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Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

First I observe either an annular solar eclipse or the start or end of
a total one. Either way, I deduce that the Sun and the Moon have almost
the same apparent size, so R and D are very nearly equal.

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is. Therefore it's the Moon that's driving the tides and therefore its
tidal force is significantly larger. But the tidal forces are in the
ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is
significantly less than 1, or in other words, the Moon is denser.

I don't know if this was the intended solution, though. It seems to me
that if I have to observe the behavior of the tides, then I'm not only
using my eyes as an instrument, which is allowed, but also the ocean
and the land, which isn't. For that matter, observing the eclipse
would be a lot safer if I was allowed to use a solar filter, but that
might be excluded as well -- and I don't have so much dedication to
rec.puzzles that I want to risk my eyesight for it. (A pinhole
projection system, of course, would be Right Out, even if the pinholes
are naturally formed by tree leaves.)

Very good answer.

I wouldn't have bothered waiting for an eclipse, I would just have held
my hand out at arm's length in front of the setting sun, and observed
that the last joint of my little finger subtends about the same angle as
the sun does. Then wait for a full moon, and do the same, showing that
it's pretty much the same apparent size. Not as accurate as your
method, but good enough for our purposes, as it happens.

Then go and stand in the sea for a month or two, noting how far up your
body each tide comes. Don't try this in the Bay of Fundy.

Nick
--
Nick Wedd

"Mark Brader" wrote in message
...
Nick Wedd writes:
You may assume knowledge of Newtonian physics, and of the general
structure of the solar system, but not of any of its dimensions. Using
only your own body as a measuring instrument, find which is denser, the
sun or the moon.

Very cute question! Possible spoiler...

40
39
38
37
36
35
34
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2

Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

First I observe either an annular solar eclipse or the start or end of
a total one. Either way, I deduce that the Sun and the Moon have almost
the same apparent size, so R and D are very nearly equal.

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is. Therefore it's the Moon that's driving the tides and therefore its
tidal force is significantly larger. But the tidal forces are in the
ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is
significantly less than 1, or in other words, the Moon is denser.

I don't know if this was the intended solution, though. It seems to me
that if I have to observe the behavior of the tides, then I'm not only
using my eyes as an instrument, which is allowed, but also the ocean
and the land, which isn't. For that matter, observing the eclipse
would be a lot safer if I was allowed to use a solar filter, but that
might be excluded as well -- and I don't have so much dedication to
rec.puzzles that I want to risk my eyesight for it. (A pinhole
projection system, of course, would be Right Out, even if the pinholes
are naturally formed by tree leaves.)

Or you just post a link to this page.

http://www.math.sunysb.edu/~tony/tides/sun-moon.html

Saves all that tedious typing donchaknow
--
Dave Baker
Puma Race Engines
www.pumaracing.co.uk
Camp USA engineer minces about for high performance specialist (4,4,7)

Nick Wedd and I (Mark Brader) write:

You may assume knowledge of Newtonian physics, and of the general
structure of the solar system, but not of any of its dimensions. Using
only your own body as a measuring instrument, find which is denser, the
sun or the moon.

Very cute question! Possible spoiler...

40
39
38
37
36
35
34
33
32
31
30
29
28
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...
First I observe either an annular solar eclipse or the start or end of
a total one. ....
I now go to a coastal location and observe the behavior of the ocean
over a few days. ...

I don't know if this was the intended solution, though. It seems to me
that if I have to observe the behavior of the tides, then I'm not only
using my eyes as an instrument, which is allowed, but also the ocean
and the land, which isn't. For that matter, observing the eclipse
would be a lot safer if I was allowed to use a solar filter...

Very good answer.

Thanks.

I wouldn't have bothered waiting for an eclipse, I would just have held
my hand out at arm's length in front of the setting sun, and observed
that the last joint of my little finger subtends about the same angle as
the sun does. Then wait for a full moon...

Ah, good point. I thought about using a tree or building or something,
but not my finger. Anyway, I already *have* observed some solar eclipses
(admittedly with suitable protection), so I'm covered on that part.

Then go and stand in the sea for a month or two, noting how far up your
body each tide comes.

If you like, but this still does not answer my objection: I'm using the
ocean and the land as an instrument, not just my body.

Don't try this in the Bay of Fundy.

Hear, hear!
--
Mark Brader, Toronto "...one man's feature is another man's bug."
--Chris Torek

My text in this article is in the public domain.

In article ,
Nick Wedd wrote:

In message , Mark Brader
writes
Nick Wedd writes:
You may assume knowledge of Newtonian physics, and of the general
structure of the solar system, but not of any of its dimensions. Using
only your own body as a measuring instrument, find which is denser, the
sun or the moon.

Very cute question! Possible spoiler...

.............................

Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

First I observe either an annular solar eclipse or the start or end of
a total one. Either way, I deduce that the Sun and the Moon have almost
the same apparent size, so R and D are very nearly equal.

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is. Therefore it's the Moon that's driving the tides and therefore its
tidal force is significantly larger. But the tidal forces are in the
ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is
significantly less than 1, or in other words, the Moon is denser.

I don't know if this was the intended solution, though. It seems to me
that if I have to observe the behavior of the tides, then I'm not only
using my eyes as an instrument, which is allowed, but also the ocean
and the land, which isn't. For that matter, observing the eclipse
would be a lot safer if I was allowed to use a solar filter, but that
might be excluded as well -- and I don't have so much dedication to
rec.puzzles that I want to risk my eyesight for it. (A pinhole
projection system, of course, would be Right Out, even if the pinholes
are naturally formed by tree leaves.)

Very good answer.

I wouldn't have bothered waiting for an eclipse, I would just have held
my hand out at arm's length in front of the setting sun, and observed
that the last joint of my little finger subtends about the same angle as
the sun does. Then wait for a full moon, and do the same, showing that
it's pretty much the same apparent size. Not as accurate as your
method, but good enough for our purposes, as it happens.

Then go and stand in the sea for a month or two, noting how far up your
body each tide comes. Don't try this in the Bay of Fundy.

Nick

One could use the same method do find out that the density of M31 is
much smaller than the density of the Sun, or the Moon: M31 has an
apparent diameter which is (very approximately) the same as the
apparent diameter of the Sun or the Moon. Now, M31 causes no noticeable
tides at all -- therefore the mean density of M31 must be much smaller
than the density of the Sun, or the Moon.

The experiment can be repeated for any object having approximately the
same apparent diameter as the Sun or the Moon, since the tidal force can
be considered to depend only on the density and the apparent diameter
of the object, no matter how far away it is.

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://stjarnhimlen.se/

In sci.astro message , Tue, 5 Dec
2006 23:29:19, Mark Brader wrote:
Nick Wedd writes:
You may assume knowledge of Newtonian physics, and of the general
structure of the solar system, but not of any of its dimensions. Using
only your own body as a measuring instrument, find which is denser, the
sun or the moon.

Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

Tidal field is proportional to the density of the attractor and to the
cube of its angular diameter, only.

First I observe either an annular solar eclipse or the start or end of
a total one. Either way, I deduce that the Sun and the Moon have almost
the same apparent size, so R and D are very nearly equal.

Eclipse not needed. The angular size of the Sun can be estimated by
pointing the head at the Sun, making a pinhole aperture with the fingers
in front of the face, and observing the projected image on a toe. It
will probably fit the size, in at least one direction, of at least one
toe. Then, later, lie with the head away from the Moon and compare its
apparent size with the standard toe.

Of course, the Sun can be observed quite comfortably by the eye at
sunset, or at other times when clouds co-operate.

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is.

As stated, not true in *all* locations. Coastal geometry may mean that
high tides are doubled, e.g. to the North and West of the Isle of Wight.

Therefore it's the Moon that's driving the tides and therefore its
tidal force is significantly larger. But the tidal forces are in the
ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is
significantly less than 1, or in other words, the Moon is denser.

And by using the relative amplitude (measured, of course, in feet) of
Spring & Neap Tides, one gets a figure for the density ratio.

--
(c) John Stockton, Surrey, UK. Turnpike v6.05 MIME.
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

In article id, says...
In sci.astro message , Tue, 5 Dec
2006 23:29:19, Mark Brader wrote:
Nick Wedd writes:
....

First I observe either an annular solar eclipse or the start or end of
a total one. Either way, I deduce that the Sun and the Moon have almost
the same apparent size, so R and D are very nearly equal.

Eclipse not needed. The angular size of the Sun can be estimated by
pointing the head at the Sun, making a pinhole aperture with the fingers
in front of the face, and observing the projected image on a toe. It
will probably fit the size, in at least one direction, of at least one
toe. Then, later, lie with the head away from the Moon and compare its
apparent size with the standard toe.

Of course, the Sun can be observed quite comfortably by the eye at
sunset, or at other times when clouds co-operate.

only in certain parts of the world. If the air is clean and relatively free of dust, dirt, smoke etc, you wi;ll still
ruin your eyes by staring at the setting sun for more than a second or two.

Mike

Mark Brader:
Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

John Stockton:
Tidal field is proportional to the density of the attractor and to the
cube of its angular diameter, only.

I said tidal force was proportional to M/R^3, which is true. Angular
diameter is D/R and density is M/D^3, so yes, tidal force can also be
described as proportional to (M/D^3)*((D/R)^3) -- in effect, my posting
was deriving that. So what's the "only" supposed to mean?

(In the preceding paragraph I'm deliberately being loose as to whether
M is a mass or a ratio of masses, and likewise the other variables --
this doesn't affect how they relate to each other.)

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is.

As stated, not true in *all* locations. Coastal geometry may mean that
high tides are doubled, e.g. to the North and West of the Isle of Wight.

True, but the cycles will still relate to the Moon's position and not
the Sun's.
--
Mark Brader | "Unless developers are careful, good software
Toronto | attracts so many improvements that it eventually
| rolls over and sinks..." --Ben & Peter Laurie

My text in this article is in the public domain.

In sci.astro message , Thu, 7 Dec
2006 06:28:34, Mark Brader wrote:
Mark Brader:
Say that the Sun is R times as far away as the Moon, D times the diameter
of the Moon, and M times the mass. Then it must have M/D^3 times the
density.

John Stockton:
Tidal field is proportional to the density of the attractor and to the
cube of its angular diameter, only.

I said tidal force was proportional to M/R^3, which is true. Angular
diameter is D/R and density is M/D^3, so yes, tidal force can also be
described as proportional to (M/D^3)*((D/R)^3) -- in effect, my posting
was deriving that. So what's the "only" supposed to mean?

That the tidal field depends on no other variable. The sentence states
more compactly what you proved.


(In the preceding paragraph I'm deliberately being loose as to whether
M is a mass or a ratio of masses, and likewise the other variables --
this doesn't affect how they relate to each other.)

I now go to a coastal location and observe the behavior of the ocean
over a few days. I find that alternate high tides occur when the Moon
is in a similar position between rising and setting, not when the Sun
is.

As stated, not true in *all* locations. Coastal geometry may mean that
high tides are doubled, e.g. to the North and West of the Isle of Wight.

True, but the cycles will still relate to the Moon's position and not
the Sun's.

Yes, but instead of alternate high tides occurring at an interval of
12.5 hours or so, they (alternately) have intervals of (roughly, for
Bournemouth, from ancient memory) about 2 and 10.5 hours. The simplest
fix is to go to a small mid-ocean island.

--
(c) John Stockton, Surrey, UK. /
Web URL:http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, & links.
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