the geometric center of the Earth and the center of mass

pixel01

Hi all,

We are now going to do a kind of experiment in which we use a plumb bob to identify the vertical direction.
But I think, the pumb string will point to the center of mass, while we need the geometric center . I mean the COM of the Earth is diffent from the geometric center and it depends on the position of the Moon.

Am I right? I if we still use the plumb bob, what is the errorness at max?

tiny-tim

mmm … I never thought of that before!

Yes, if the moon is on the horizon, then the centre of mass will be shifted sideways.

From http://en.wikipedia.org/wiki/Mass_of_the_Moon: Moon mass = 0.0123 Earth mass = 1/80 approx.

Earth radius = 4,000 miles approx, moon orbit radius = 240,000 miles approx.

So displacement of centre of mass from centre of earth = 1/80 x 240,000 = 3000 miles.

But the force diagram will be different, since it depends on the square of the distance.

So the force from the moon will be 0.0123 squared, = 0.00015, times the force from the earth, and the tangent of the angle of deviation from the direction of the centre of the earth will be 1/6600 approx.

pixel01

Thank tiny tim.

About the accuracy of plumb bob, at first I think about the in homogenous density of the earth, and then about the moon. You are right that the moon has very little effect on the plumb. But how about the density? Can we just forget it?
Anyway, I am not a student.

tiny-tim

Yes, the earth is inhomogenous (and maybe the moon is too - I don't know), but it is spherically symmetrical (well, more or less), so that doesn't matter - we can regard the entire mass as being concentrated at the centre!

Um - you originally said "We are now …", so I assumed you were in a class!

arildno

It will point at neither, due to the rotation of the Earth.

pixel01

Yes, and it depends on what latatude you are in. Thanks

tiny-tim

hmm … good point … let's see …

At the equator, which is the simple case, there's no difference: in equilibrium, the bob will go with uniform speed round a circle in the plane of the equator and whose centre is the centre of the earth (and of the equator).

So the acceleration is towards the centre of that circle, which is the same direction as the force from gravity: vertical. So, using a force triangle, the third (and only remaining) force, the tension in the string, must also be vertical!

However, at latitude ±a, where 0 < a < 90º, the acceleration is towards the centre of the circle of latitude, at an angle a to the vertical, and it's r.cosa.w^2, where r = radius = 4000 miles = 20,000,000 feet approx, w = angular velocity = 2π/24 hours^-1 = 1/14,000 sec^-1 approx, while g = 32 feet per sec per sec, so one side of the force triangle is approx 20,000,000 x cosa / 200,000,000 x32 = cosa / 320 times the other.

So the tangent of the angle of the third side from the vertical, which is the angle at which the string must hang in equlilibrium, is cosa x sina / 320 approx, or a maximum (at 45º latitude) of 1/640.

So the string hangs, towards the nearer pole, at a maximum angle (at 45º latitude) of about 1/40 degree to the vertical.

oops! correction! I think it's the further pole!

D H

The Earth is not an inertial reference frame. It is rotating and accelerating. The effective force on the plumb bob from the Moon is the difference in the inertial acceleration (i.e., the acceleration as observed in an inertial frame) of the Earth toward the Moon and the inertial acceleration of the plumb bob toward the Moon. This is more-or-less an inverse cubic relationship, not an inverse square relationship.

The Moon is very inhomogeneous. You might find this article on lunar mascons of interest: http://science.nasa.gov/headlines/y2...v_loworbit.htm:

The mascons' gravitational anomaly is so great—half a percent—that it actually would be measurable to astronauts on the lunar surface. "If you were standing at the edge of one of the maria, a plumb bob would hang about a third of a degree off vertical, pointing toward the mascon," Konopliv says.

tiny-tim

hmm … no, I don't get that.

I know the tidal force from the Moon (or Sun) follows a roughly inverse cubic relationship (http://nostalgia.wikipedia.org/wiki/Tidal_force), but surely that would be the same even if the earth wasn't rotating?

And isn't the tidal bulge related to the force each bit of water receives from the water around it (which is also free to move) - thus making the gradient vital - while the bob gets force only from the string above it (which is fixed)?
ooh … my brain hurts!

Cool link! Thanks, DH!

D H

A plumb bob just above the surface of the Earth is accelerating toward the Moon, but so is the Earth itself. The effective acceleration toward the Moon is the difference between the acceleration of the plumb bob and the Earth toward the Moon:

[tex]
\begin{aligned}
a_{\text{eff}} &=
GM_{\text{moon}}
\left(
\frac {R_m\hat m + r_e\hat b}{|| R_m\hat m + r_e\hat b||^3} -
\frac {\hat m}{{R_m}^2}
\right) \\
&\approx
\frac{GM_{\text{moon}}}{{R_m}^2}\frac{r_e}{R_m}
(\hat b - 3(\hat m\cdot\hat b)\hat m)
\end{aligned}
[/tex]

where

[tex]
\begin{aligned}
R_m &\quad\text{is the Earth-moon distance} \\
\hat m &\quad\text{is the unit vector from the center of the Earth to the moon} \\
r_e &\quad\text{is the radius of the Earth} \\
\hat b &\quad\text{is the unit vector from the plumb bob to the center of the Earth}
\end{aligned}
[/tex]

Ignoring the acceleration of the Earth leads to an acceleration toward the Moon of about 38 microns/square second and a maximum plumb bob deviation when the Moon is at the horizon. The effective acceleration is about 1 micron/square second (rather than 38) and the maximum deviation of the plumb bob due to the Moon occurs when the Moon is a bit over 45 degrees above the horizon. The deviation is tiny: about 21 milliarcseconds when the Moon is at perigee.