Is g affected by other sources of gravity?


by ViolentCorpse
Tags: affected, gravity, sources
mfb
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#19
Nov30-12, 10:12 AM
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Quote Quote by Vaxx View Post
Velocity doesn't affect acceleration to the center of the earth. A bullet fired out of a gun and a mass dropped at the same time will hit the ground at the same time.
On a level where you care about tides, this is not true any more. In a rotating frame, you get coriolis and centrifugal forces. A bullet fired with the rotation of earth will travel longer than a bullet fired in the opposite direction.
sophiecentaur
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#20
Nov30-12, 01:12 PM
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Quote Quote by Pythagorean View Post
I dont mean it doesn't contribute, I'm just saying the sun pulls much harder on point masses on Earth than the moon, but doesn't generate as much tidal force. It's mass makes up for its distance (compared to the moon) for total force.

The moons tidal strength comes from the ratio of the distance of the moon from nearest edge of Earth to the distance at the farthest edge of Earth. That ratio is practically 1 for the Sun.
The overall effect is much the same on the Sun side as for the opposite side (i.e.both tiders of the day) and it is very comparable with the effect of the Moon so I can't see how you can assert that it "doesn't generate as much tidal force" - as if it is negligible. Of course the Moon's contribution is more but not by that much.
You'd need to explain that bit about point masses.
Pythagorean
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#21
Nov30-12, 01:25 PM
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I didn't say it was negligible... you're putting words in my mouth for why? The point is really simple: the moon only has 1e-6 (that's micro!) the force the sun does on a point particle on Earth right? Yet, it has 2-3 times the tidal force.

I was responding to a post where somebody had set up a thought experiment using a point particle (a lead weight is effectively a point particle for astronomical bodies) and had gone on to talk about the tidal forces on the point particle, but there are no tidal forces on a point particle, as one might recognize if they understood how tidal forces are generated. And a particular fact that highlights that caveat is how much more massively the sun pulls on point particles (a million times more, using DH's numbers) yet still loses to the moon in tidal forces.
mfb
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#22
Nov30-12, 02:40 PM
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Quote Quote by Pythagorean View Post
I didn't say it was negligible... you're putting words in my mouth for why? The point is really simple: the moon only has 1e-6 (that's micro!) the force the sun does on a point particle on Earth right? Yet, it has 2-3 times the tidal force.
The ratio is about 1/200 for the gravitational force (check post 4).
Gravitational potential differs by a factor of ~100 000.
Pythagorean
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#23
Nov30-12, 03:19 PM
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ah, you're right, my bad:

http://www.wolframalpha.com/input/?i...+to+sun%5E2%29
Choragos
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#24
Dec4-12, 10:09 AM
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You are correct: all astronomical bodies do exert a force and thus affect 'g', as has been discussed. I work with gravity methods every day, and we have to account for the position of the sun and moon over the course of the day or our measurements will be thrown off quite a bit. The easiest way is to assume the change is linear and periodically go back to a base station with the gravimeter, compute the 'drift' and subtract it out. More sophisticated ways use tide tables to model the effect.

Today's relative gravimeters that we use in exploration geophysics are good to the microGal range. A Gal (after Galileo) is equal to one centimetre/second^2. The average gravitational acceleration on Earth is therefore about 981 Gal, so we're measuring about one part in a billion. For reference, a microGal is about the gravitational force a worm feels from the apple it's crawling over. If you stand over these instruments while they are measuring you can throw them off. However, I digress.

The astronomical bodies affect gravity measurements of g in two ways. 1. the direct attraction from the body itself, and 2. from the affect those bodies have on the Earth. Tides affect the measurement because there's a bunch of mass (due to water) that moves (remember that everything that has mass exerts a gravitational pull). Also, even on solid rock there are tides, and depending on where you are the ground level relative to the centre of mass of Earth can change by as much as a metre. If I remember correctly, gravitational acceleration decreases by about 300 microGal/m as you move away from the surface of the Earth (when you're close to the surface).

And finally, just as commentary, g changes as you move over the surface of the Earth. One, as you are at different elevations (in Denver, g is about 9.79 m/s^2), and two, as you move over denser areas. For example, if you measure g over a dense orebody it will be higher than if you move a kilometre away and measure it off of the body.

What's the practical application of this? Well I just did an experiment where I took a garden gnome (http://www.gnomeexperiment.com/) and weighed it in Golden, CO (about 5800 feet elevation) and then drove to the top of Mt. Evans (~14300 feet elevation). The 350 gram gnome weighed something like 0.15 grams less at the top of Mt. Evans. That doesn't really have anything to do with your question, but I thought it might be interesting.
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#25
Dec4-12, 10:48 AM
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Quote Quote by Choragos View Post
If I remember correctly, gravitational acceleration decreases by about 300 microGal/m as you move away from the surface of the Earth (when you're close to the surface).
That 300 microGal/m is the free air correction. Since the person is still standing on solid ground, you also have to incorporate the Bouguer correction into this 2nd order tidal effect. For surface rock of typical density, this reduces the effect to about 110 microGal/m.

There's also a third order effect caused by the Earth tides. That the ground is undulating means that an object at rest on the surface is accelerating due to these undulations. With a frequency of about twice per day and an amplitude of about 0.5 meters, this comes out to about 1 microGal.

Aside #1: These tidal forces also affect satellites orbiting the Earth. From the perspective of an Earth-centered frame of reference, bodies such as the Moon and Sun exert 1/R3 tidal forces on the satellite, much as those bodies affect objects on the surface of the Earth. (When viewed from the perspective of an inertial frame, these tidal gravitational forces of course become good old 1/R[sup]2[sup] gravitational forces, but nobody in their right mind models the behavior of a satellite orbiting the Earth from the perspective of an inertial frame.)

Aside #2: People who do precision satellite tracking don't call these perturbing gravitational forces from the Moon, the Sun, and other planets "tidal forces." They instead call them third body forces. The term "tidal force" is reserved for those second order effects you talked about. The solid body tides in the Earth caused by the Moon, Sun, etc. have a subtle effect on the orbits of satellites in low Earth orbit. The ocean tides also have an effect on satellite orbits, but these are about an order of magnitude smaller than the perturbations from the solid body tides.


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