# Gravity shift due to sun & moon passing overhead

1. Dec 10, 2012

### corndodger

I would like to know a function or at least an approach toward developing a function that describes the change of gravity on the earth's surface due to the sun, moon and if possible, the tides.

I am quite familiar with the fundamentals but my math isn't up to this.

the corndodger

2. Dec 10, 2012

### Staff: Mentor

Newton's law of gravitation: $F=\frac{GMm}{r^2}$
As acceleration of a small mass on earth (small compared to the mass of earth) and if we ignore the direction, this can be simplified to $a=\frac{GM}{r^2}$ where G is the gravitational constant, M is some big mass and r is the distance to the center of this mass.
Moon and sun are not just attracting your small mass, they are attracting earth as well. If earth and the small mass are accelerated in the same way, you do not see this (on earth), so the difference between the acceleration of earth and the acceleration of your mass is relevant. If the object is just over your head:
$\Delta a=\frac{GM}{r^2} - a=\frac{GM}{(r+r_e)^2}$ where re is the radius of earth (it is possible to simplify the formula a bit, if necessary). You can use the known parameters of the solar system to estimate those numbers for moon and sun. I did the same a few weeks ago here, where the second list is relevant.

3. Dec 10, 2012

### corndodger

Thanks for your reply. I am building a precision pendulum complete with data logging to try to measure these microscopic gravitational changes. This was done once before in the '80s using an old astronomical pendulum from the 1920s. He collected a month or two of data, then fft'd and it showed 6 (I believe) different frequencies, sun, moon, tides (he was on the west coast) and some identifiable earth resonances.

I have the electronics controller built and operational and am working on the pendulum. I understand the Newtonian formulae you've provided, but I'm still stuck calculating the difference between having the sun directly overhead and being on the other side of the world. Plus the difference the moon makes as it crosses overhead, say, between you and the sun.

This mathematical exercise is just to try to get an idea of the order of magnitude between the two signals. I really wonder how deep in the noise these signals are buried.

4. Dec 10, 2012

### Staff: Mentor

Sun/Moon directly below gives (nearly*) the same influence, just in the opposite direction.
Moon/Sun in horizontal direction attract earth and the test mass in the same way, so this cancels.
(Direct) tidal effects are ~1 part in 10 millions, this will be hard to achieve with a pendulum I think.

*they are not always in the same distance anyway, and that gives more deviation from the averaged values.

5. Dec 12, 2012

### corndodger

Well made pendulums and mechanical clockworks give far better resolution of time than you might think. Consider a day of 84600 seconds, pendulum clocks can consistently give time within a second a day...that's a resolution of 1/84600 = 0.00001182. so we're talking about millionths. It takes only in the millionths of an inch change in pendulum length to make a second or two a day change in timekeeping.

So humor my steam punkness, I want to see for myself what a pendulum can actually do. For all I know they may be sensitive to solar flares, dark matter or detect gravity waves given the right performance analysis.

jim

6. Dec 12, 2012

### Staff: Mentor

You need a precision of 10 milliseconds per day to see the direct attraction by moon and sun. And there is an additional problem: Sun has a period of one day, so your clock has to be able to display the time with a precision of a few milliseconds or better.
The tides on earth should be easier to measure, but they are more complicated to evaluate.