Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-L] apparent weight

You are absolutely right. It is obvious once one thinks about it, which I
did not pursue originally. DUH
John M. Clement
Houston, TX


The accleration of the object closer to the Sun is greater
because the
gravitational force is greater. Of course this is ignoring the
rotation of the Earth which will also have to be factored in for a
more exact calculation. That this is correct is easily seen by the
tides. At noon the water is rising because the pull of the
Sun/mass
is greater on the water which is closer to the sun than the Earth
underneath. Again this is a simple model which should appeal to a
beginning physics student and ignores complications and
second order
effects. The tidal forces are certainly complicated by
geography and the Coriolis pseudo force.


Okay, let's consider the case of a non-rotating earth except
for the one rotation per year corresponding to the
orbit--that is, let's consider an earth that always has the
same side facing the sun. Further, let's treat it as being
in a circular orbit about the sun. I claim that, relative to
the sun, an object resting on a scale on the earth at that
point closest to the sun, has an acceleration directed toward
the sun that is smaller in magnitude than the magnitude of
the toward-the-sun acceleration of the center of the earth.
Here is my reasoning. In the non-rotating reference frame
centered on the sun, the earth is rotational motion about the
sun with an angular velocity of magnitude omega = (2 pi
radians)/(1 year). The classical acceleration, relative to
the sun, of any particle of the earth is r*omega^2 where r is
the distance of the particle from the center of the sun. The
bigger r is the bigger the acceleration. The center of the
earth is farther from the sun so it's r is
bigger than that of the object resting on a scale on the
earth at that point closest to the sun. Hence, the center of
the earth has a greater acceleration relative to the sun than
the object resting on a scale on the earth at that point
closest to the sun. This means that the acceleration of the
object resting on a scale on the earth at that point closest
to the sun.

How does that jive with the tides? Consider a sample of the
earth's core at the center of the earth having the same mass
as the object resting on a scale on the earth at that point
closest to the sun. The gravitational force of the sun on
the object resting on a scale on the earth at that point
closest to the sun has a greater gravitational force being
exerted on it by the sun as you pointed out, and yet it has a
smaller toward-the-sun acceleration. The net force on the
object resting on a scale on the earth at that point closest
to the sun is the vector sum of the gravitational force of
the sun on the object, the gravitational force of the earth
on the object, and the normal force of the scale on the
object. If the earth were not in orbit about a star, the
normal force of the scale on the object and the gravitational
force of the earth on the object would both have the same
magnitude. Given that the toward-the-sun gravitational force
of the sun on the object is too great for th e prevailing
acceleration, the sum of the other two forces must be in the
away-from-the-sun direction. The gravitational force of the
earth on the object is in the away-from-the-sun direction and
the normal force of the scale on the object is in the
toward-the-sun direction. In order for these two forces to
add up to a force that is directed away from the sun, the
normal force exerted on the object by the scale must be of
smaller magnitude than the gravitational force exerted on the
object by the earth. By Newton's third law, the normal force
exerted by the scale on the object has the same magnitude as
the normal force of the object on the scale. This means that
the scale reading is lower than it would be if the earth were
not in orbit about a star. This means that the object has an
apparent weight that is less than the magnitude of the
gravitational force of the earth on the object. This lower
apparent weight is consistent with the solar contribution to
one of the tidal bulge s.

If the effect were the reverse, then when the sun and moon
were at 90
degrees the tides would be greatest, but it is when they
are lined up
with the Earth that you get the highest tides or what is
often termed a "spring tide".

I am curious as to the reasoning you used to deduce that
the effect is
the reverse of what I claim.

John M. Clement
Houston, TX


Defining the acceleration of the person as the time rate
of change
of the velocity of the person relative to a reference
frame in which
the sun is at rest at the origin and the earth is orbiting the
origin once per year, you have it exactly backwards. The
person's
acceleration is greater than that of the center of the earth at
midnight when the person is farther from the sun, and
smaller than
that of the center of the earth at noon when the person
is closer to the sun.


The lighter scale reading will happen for both the Sun
above and the
Sun below, just as the Moon produces 2 tides. This
simple analogy
argument might be helpful with students. The simple physics
explanation is that since you are closer to the Sun when above
than the center of the Earth, you have a greater acceleration
toward the sun. When the sun is below the Earth is
accelerating
more
towards the
sun than you are because it is closer to the Sun. In both
cases this
results in a lower scale reading. I will leave the
complicated explanations to others, as they will confuse most
students.
Unfortunately most simple accounts of the tides ignore the
true effect
and just talk about the pull. It is a gradient effect.

Of course a simple bathroom scale would be inadequate for
seeing this
effect, and a gym scale which is a balance would show no effect.

John M. clement
Houston, TX



If you stand on a sensitive scale, will you be slightly
lighter at
noon (with the Sun above you and pulling opposite to Earth's
gravity), and slightly heavier at midnight (with the Sun
below you
and pulling in the same direction as Earth's gravity)?
Ignore any lunar effects.

This came up the other day, and nobody could really agree.
One side said no because the Earth is in freefall around the
Sun, and it is the Earth that pulls us to it. The other side
said yes due
to tidal effects as seen in the oceans.

I searched online, and found the same conflicting arguments!

Can anyone help with this?

_______________________________________________
Forum for Physics Educators
Phys-l@www.phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l

_______________________________________________
Forum for Physics Educators
Phys-l@www.phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@www.phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l

_______________________________________________
Forum for Physics Educators
Phys-l@www.phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@www.phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l