Is g affected by other sources of gravity?

ViolentCorpse

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?

rudolfstr

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

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

To add some numbers:

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

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!!

rcgldr

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

sophiecentaur

But the measured value is different all over the World.

Nstraw

i think its the average

sophiecentaur

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

rcgldr

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.

ViolentCorpse

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!)

D H

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.


Some very sensitive gravimeters can tell when the roof of the building they are housed in has been cleared of snow.

jbriggs444

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

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

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

Vaxx

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

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.