# Is g affected by other sources of gravity?

## Main Question or Discussion Point

Earth shouldn't alone make the value of g=9.81m/sec^2. There is the moon, all the planets and our sun (ignoring the outstanding multitude of outer space for my simplicity) and all of these bodies must have some influence on everything that's on Earth. So if we want to find the real acceleration due to gravity on Earth, wouldn't we have to find the net acceleration after taking into account all these sources of gravity that are trying to pull me toward their respective selves?

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It's a similar question to the one : Would you include the change in g if you were to calculate the height at which you'd throw a ball?

berkeman
Mentor
Earth shouldn't alone make the value of g=9.81m/sec^2. There is the moon, all the planets and our sun (ignoring the outstanding multitude of outer space for my simplicity) and all of these bodies must have some influence on everything that's on Earth. So if we want to find the real acceleration due to gravity on Earth, wouldn't we have to find the net acceleration after taking into account all these sources of gravity that are trying to pull me toward their respective selves?
Sure. Are you familiar with the equation you can use to calculate the effects of each of those other masses on the value of g that you feel on the surface of the Earth? You can calculate the effect of the moon and the sun for starters...

mfb
Mentor

Earth: ~9.81 m/s^2
Sun (average): 0.0059 m/s^2
Moon (average): 0.000033 m/s^2
Jupiter at closest approach: 3.7*10-7 m/s^2 = 0.00000037m/s^2
Venus at closest approach: 2.3*10-7 m/s^2

But: Those forces are very uniform throughout earth. You cannot measure an absolute acceleration - only a difference between two points (here: earth and you). This changes numbers a lot:
Earth: ~9.81 m/s^2
Moon (average): 6*10-7 m/s^2
Sun (average): 3*10-7 m/s^2
Venus at closest approach: 4*10-11 m/s^2
Jupiter at closest approach: 4*10-12 m/s^2
And just for fun:
Proxima Centauri (closest star): ~10-23 m/s^2
1000 kg in a distance of 2m: ~2*10-8 m/s^2

Best relative gravimeters according to Wikipedia article: ~10-11 m/s^2 (at least within some hours of measurement: source, pdf)

sophiecentaur
Gold Member

Earth: ~9.81 m/s^2
Sun (average): 0.0059 m/s^2
Moon (average): 0.000033 m/s^2
Jupiter at closest approach: 3.7*10-7 m/s^2 = 0.00000037m/s^2
Venus at closest approach: 2.3*10-7 m/s^2

But: Those forces are very uniform throughout earth. You cannot measure an absolute acceleration - only a difference between two points (here: earth and you). This changes numbers a lot:
Earth: ~9.81 m/s^2
Moon (average): 6*10-7 m/s^2
Sun (average): 3*10-7 m/s^2
Venus at closest approach: 4*10-11 m/s^2
Jupiter at closest approach: 4*10-12 m/s^2
And just for fun:
Proxima Centauri (closest star): ~10-23 m/s^2
1000 kg in a distance of 2m: ~2*10-8 m/s^2

Best relative gravimeters according to Wikipedia article: ~10-11 m/s^2 (at least within some hours of measurement: source, pdf)
Those figures can be used to knock on the head the ideas of Astrology, based on the effect of planetary positions when you are born. The gravitational field of the midwife, right next to your Mum is greater so 'serious' Astrology would need to take that into consideration too!
Not that I'm suggesting a midwife could be 1000kg!!

But the measured value is different all over the World.
i think its the average

sophiecentaur
Gold Member
Of course. And it is measurably affected by the Moon and Sun - hence the tides.

rcgldr
Homework Helper
There is a standard value for g by definition, 9.80665 m / s^2 or 32.1740 ft / s^2. Wiki article:

http://en.wikipedia.org/wiki/Standard_gravity
But the measured value is different all over the World.
Hmm, part of my post never made it, maybe I lost an edit update. Anyway, what is missing from my post is already mentioned in the other posts in this thread, that the sun and moon have enought effect to affect the least significant digits of the standard value. I didn't know how accurately gravity can be currently measured.

Sure. Are you familiar with the equation you can use to calculate the effects of each of those other masses on the value of g that you feel on the surface of the Earth? You can calculate the effect of the moon and the sun for starters...
Yes. I know of two methods that can be used to calculate g.
1) Using Newton's law of gravity and his 2nd law of motion. (I think, g calculated using this approach would be the value that it only Earth alone is responsible for producing).
2) Using a pendulum whose time period is known. (This one should automatically take into account all the other forces present )

I did perform those experiments in school, but I wasn't enough interested back then to check how well the two techniques agree with each other.

Thanks for your answers everyone! You guys are awesome! (and that was a very informative post mfb. Thanks!)

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D H
Staff Emeritus
Anyway, what is missing from my post is already mentioned in the other posts in this thread, that the sun and moon have enought effect to affect the least significant digits of the standard value.
No, they don't. The least significant digit in the defined value of g0 is a muliplier of 10-5 m/s2. The largest of the third body effects come from the moon, and even that is very small. The moon decreases g by about 10-6 m/s2 when the moon is directly overhead or underfoot, increases g by about half that amount when the moon is directly on the horizon. This is an order of magnitude smaller than the least significant digit in g0.

I didn't know how accurately gravity can be currently measured.
Some very sensitive gravimeters can tell when the roof of the building they are housed in has been cleared of snow.

jbriggs444
Homework Helper
2019 Award
Yes. I know of two methods that can be used to calculate g.
1) Using Newton's law of gravity and his 2nd law of motion. (I think, g calculated using this approach would be the value that it only Earth alone is responsible for producing).
Hmm. Second law is often expressed as f = ma and you're trying to derive the g in f = mg. A little algebra and that gives... g = a

So the experiment is to drop a lead weight and see how long it takes to fall a measured distance? Or to see how fast it is falling after a measured time?

No, such a measurement is not dependent entirely on the gravity from the mass of the Earth. It would also depend on the motion of the Earth and on the gravity from the Sun, Moon and planets. The contribution from the rotation of the Earth is typically far more significant than the contribution from the tidal force of the Moon.

Pythagorean
Gold Member
Tidal forces wouldn't affect a point particle, it's the differential force across the earth that causes tides. Otherwise the sun would be our tidal driver.

sophiecentaur
Gold Member
Tidal forces wouldn't affect a point particle, it's the differential force across the earth that causes tides. Otherwise the sun would be our tidal driver.
If the Sun were not a contributor to tides, there would be no spring and neap tides.

No, such a measurement is not dependent entirely on the gravity from the mass of the Earth. It would also depend on the motion of the Earth and on the gravity from the Sun, Moon and planets. The contribution from the rotation of the Earth is typically far more significant than the contribution from the tidal force of the Moon.
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.

The planets would have very little effect, but the Sun and Moon do, hence tides and King tides

To the original question, the experiment would have been done countless times over years so you would assume that the value would be an average. To calculate the real force, you can mathematically factor in all the planets to give you a true value for that present moment.

D H
Staff Emeritus
Tidal forces wouldn't affect a point particle, it's the differential force across the earth that causes tides. Otherwise the sun would be our tidal driver.
If the Sun were not a contributor to tides, there would be no spring and neap tides.
sophiecentaur, you misunderstood Pythagorean's point. The Sun does contribute to the tides, but it is not the dominant effect (the driver). The Moon is the driver. The gravitational acceleration at a point on the Earth's surface toward the Sun is over 170 times the gravitational acceleration toward the Moon. This would not only make the Sun the driver, it would make the Sun essentially the only body of concern.

This isn't the right metric, however. The tidal force caused by some body is the difference between the gravitational accelerations toward that body at the surface and center of the Earth. This makes the contributions from the Sun a bit less than half of those from the Moon.

Pythagorean
Gold Member
If the Sun were not a contributor to tides, there would be no spring and neap tides.
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.

mfb
Mentor
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
Gold Member
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
Gold Member
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
Mentor
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.

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.

D H
Staff Emeritus