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Re: [Phys-L] apparent weight



-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@www.phys-l.org] On Behalf Of John
Clement
Sent: Thursday, December 11, 2014 2:02 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] apparent weight

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 the 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 bulges.

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?

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_______________________________________________
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