If a 200 pound person stands on a scale on the equator does the scale
show the same weight as if the person would stand on one of the poles?
Stefan, Switzerland

Stefan,

Nice question.

Gravity varies around the world, so that a person's weight varies as well.  

At the poles, the force of gravity is about one-half of one percent
greater than at the equator, so that a person weighing in at 200
pounds at the equator would add about a pound, and would weigh in at
about 201 pounds at the poles.

Of course (and as noted below, in the comments), it's unlikely that
such a change would be noted, since a person's weight varies
throughout the day anyway.  But for a fixed weight of 200 pounds, the
variation would be noticeable if the scales were carefully calibrated.

Here's a site discussing the variation in gravity at the equator and poles:


http://www.seed.slb.com/qa2/FAQView.cfm?ID=991
How much does gravity vary around the Earth and why?


Hope that's what you needed...let me know if there's anything else I
can do for you.


pafalafa-ga


search strategy -- Google search on: [ gravity variation pole equator ]

A person's weight varies anyway over the course of a day so it would
be impossible to prove empirically.

Moreover, clothing is a factor and a person at the pole might be well
wrapped up whereas he/she might opt for nudity at the equator.

Depending on the type of scale used, your are comparing masses, and
the mass does not vary, so an scale like the above ones will show the
same weight, I guess:
http://healthjournal.upmc.com/0105/Images/PhotoWeightLossSurg.jpg
http://www.ancasterjewellers.com/diamonds/carat%20weight/scale.jpg

"Depending on the type of scale used, your are comparing masses, and
the mass does not vary, so an scale like the above ones will show the
same weight, I guess:
http://healthjournal.upmc.com/0105/Images/PhotoWeightLossSurg.jpg
http://www.ancasterjewellers.com/diamonds/carat%20weight/scale.jpg"

No, not really. If this were true then scales would indicate the same
weight even if they were, for example, situated on the surface of the
Moon. In fact, were this true, bathroom & hospital scales would show
the same weight even under conditions of free-fall, i.e.,
"weightlessness".  Indeed, on the equator or on a pole, the mass
(inertia) of an object is the same either place. But the average
radius of the equator (from Earth's mass center) is greater than at
the poles. Newton's law of gravitation is

F = GMm/r^2

   r is the radius from the mass center. And if that's the only
variable to change between the two locations, then F will be less on
the equator than on a pole.

   And yet, by all accounts the inertial and gravitational masses of
an object are the same amount. They are the same thing. So if a
hospital scale measures my mass, how does it show two different
weights for different places? The answer is that such devices don't
measure mass. They measure the value of F in the equation above, and F
is determined not just by r, but also by the product of two
gravitating masses: Mm. If M is Earth, and m is either me or my
daughter, then indeed the number shown on the scale will be different
(given a certain location) from that of my pre-teen child, who is
smaller and has less body mass. The product Mm for me is greater than
the product Mm for my kid. This is even though there is a law of
physics that, under otherwise identical conditions, two different
masses fall due to gravity with equal acceleration.

   So how do you measure the actual mass of a body? How do astronauts
falling freely in a space station determine their body masses? (They
need to know this for purposes of collecting medical data.) They
measure not their weight, but their inertia. Newton's basic force law
is this:

F = ma / Force equals the product of mass & acceleration

   This says that if I push on an object with a given mass and it
changes speed with a certain acceleration, then multiplying the mass
and the acceleration tells the force. But this can be rearranged to be

F/a = m

   So if I give an astronaut a shove with a known force and measure
how much its speed changes over a given time, then the acceleration
can be derived. Doing the division F/a then tells the astronaut's body
mass.

Don't forget to take into account the centrifigal force of a rotating
planet. If you're wondering about the reading on the scale (and not
just the differing gravitational pull), this has some effect, however
small it may be.


Given a equatorial circumference of 24902 miles, a person standing on
the equator is traveling around the earth's axis at roughly 1038 miles
per hour. The guy standing on the pole isn't moving any more than
someone standing on the centerpoint of a merry-go-round.

Somebody else will have to do the math. My high school physics and
calculus are haven't gotten much use these past twenty years.

Cheers.

qed100, you have completely missed livioflores' point!

Did you even look at the pictures linked? They show two types of
balance scales, which work by comparing two mass against each other.
These scales will indeed give the same reading on the Moon as they
would on the Earth, because the 200-pound person (who now weighs 33
pounds) will still be, for example, 10 times heavier than the 20-pound
reference weight (which now weighs 3.3 pounds).

They won't work in free-fall, however, contrary to your opinion; the
gravitational pull has to be strong enough to overcome frictional
forces in the pivot and allow the arms to swing.

Hi manuka,

   Yes, I did look at the illustrations in the links. My intuition
told me that there must be a continuous change in the behavior of the
scales with the strength of gravity. After thinking about it I see
that you & livioflores are correct that the balance scales, upon
reaching equilibrium, will tend to show the same value both on Earth's
surface and the Moon's. But, my intuition was correct; the scales will
still behave differently, and it'll still be possible to measure the
differences in weight for the two different surface gravities. Here's
why:

   The two masses on a balance scale oppose each other with their
respective torques. The masses on the ends of the moment arms stay
equal. The lengths of the arms stay equal. The accelerations due to
gravity change. Thus, the torques on the masses change. And although
the ratio of the two masses' torques remains the same in both
situations for all horizontal projections of the moment arms, the
actual torques scale with gravity, and the time required for the scale
to reach equilibrium will also change. It is this length of time which
will tend to approach zero as the value of g approaches zero, and by
measuring the time, the mass in question can be determined as a
multiple of the reference mass.

   Thanks. I appreciate seeing the problem more clearly now.

"Don't forget to take into account the centrifigal force of a rotating
planet. If you're wondering about the reading on the scale (and not
just the differing gravitational pull), this has some effect, however
small it may be."

Great point, It's not insignificant.

Equatorial radius is roughly 6378135 meters and the velocity is about 468.5 m/s.

Angular acceleration is vē/r which offers an opposing acceleration of
0.0344 m/sē. That's about a third of a percent of the calculated
acceleration due to gravity at the equator/sealevel of 9.822, and
agrees closely to the measured value of g at the equator of 9.789
m/sē.

Though upon reference to the link the answer, the value of g used in
the response was from a measurement and not a calculation and thus was
included in the result. Oh well, fun anyway.