### What is the gravitational force felt on Earth from the other planets in our solar system?

How much gravitational force is felt on earth from the other planets in the solar system? The sun exerts the strongest g-force, holding us in it's orbit, followed by the moon which affects the tides on earth, but how much force do we feel from Jupiter, Saturn, Venus, etc?

I know I can calculate with the mass of the planet and the distance from it, i was just hoping these are well known figures i could find on the internet, without having to calculate all of them myself. It is a simple calculation though, I will if I have to, just trying to save myself some time. though i probably could have done it by now myself :)

@MarcusQuinnRodriguezTenes : Please post your results if you decide to do the calculations yourself. I think I might be a little lazy... :p

@MarcusQuinnRodriguezTenes Remember that all planets form a corotational system together with the sun, so the distances between two planets - or a planet and a point of observation on Earth - *is not constant*. Henceforth, the values you calculate with and get for the gravity change with time, **but** you can fairly easily create a program to compute the exact values at a given time, as the "exact" positions of the planets with respect to time can be found on various freely available databases :)

Because of the inverse square law for Newtonian gravity we have the acceleration due the gravity $g_b$ at the surface of the Earth due to a body of mass $m_b$ at a distance $d_b \gg r_e$ (where $r_e\approx 6371 \mbox{km}$ denotes the radius of the Earth, note all distances will need be in $\mbox{km}$ in what follows) is:

$$

g_b=g\times \frac{m_b}{m_e}\times \left(\frac{r_e}{d_b}\right)^2

$$

where $g$ is the usual accelleration due to gravity (from the Earth at the Earth's surface $\approx 10 \mbox{m/s}^2$, and $m_e\approx 6.0 \times 10^{24} \mbox{kg}$. We get the maximum acceleration due to a body when that body is at its closest to the Earth, which is what we do from now on (except for the Sun and Moon where the mean distance is used).Now for the Moon $r_b\approx 0.384 \times 10^6 \mbox{km}$, and $m_b\approx 7.3 \times 10^{22} \mbox{kg}$, so the accelleration at the Earth's surface due to the Moon $g_b\approx 3.3 \times 10^{-5} \mbox{m/s}^2$

Then putting this relation and Solar-System data into a spread sheet we get:

Thanks for this. Looking at column D, does this infer that when Mars is closed (every two years?), the gravitational effect on Earth is twice that of the moon?

No, look at the exponents the Moon has a "g" of $\approx 6\times 10^{-3} \mbox{ m/s}^2 $ and Mars has a "g" of $\approx 7\times 10^{-9} \mbox{ m/s}^2 $, that is about six orders of magnitude lower.

You might want to add, that you actually can't "feel" the gravity of the sun, as earth is on a stable orbit around the sund the zentrifugal force ~= gravitational force (on the surface of the earth).

That clears it up. Ty

Any ideas what those forces would feel like in layman's terms? How little force is 1.88E-7 ?

@joseph.hainline in layman's terms, a g force of 1.88e-7 couldn't be felt. Not close. A 200 lbs man under that low a g force would several times lighter than a feather, you could lift a truck, in that g-force, with your pinky. You might be able to lift a 747. Now, heavy objects still have inertia, so you couldn't, for example, throw a truck like a baseball, but you could hold it up, against a gravity that low easily. The astronauts in "weightless orbit" likely feel significantly more g-forces than that, and they float around like nothing.

Small point to add to this, even those imperceptibly tiny g forces, the largest planetary one being Jupiter, 3.25E-7 *9.81 m/s^2, if you roughly calculate distance traveled using d = 1/2 a t^2, Jupiter does measurably move Earth every orbit, at least, the distance of a few earth diameters. That's not much at all compared to 93 million miles, but it's still measurable. That movement roughly, but not entirely, balances out every Jupiter orbit, 11 years and it's responsible for the orbital eccentricity variation which is one of the Milankovich cycles.

really an useless answer for astroLOGISTs coming here to figure out how planets affect their lves...

The force any planet produces on a 75kg person is given by formula:

F = G * M * m / r^2

Hence:

`body minimum distance (km) Mass (kg) Force (Newton) % earth`

Earth 6.341,00 6.000.000.000.000.000.000.000.000,00 746,488018227

Moon 384.000,00 73.000.000.000.000.000.000.000,00 0,002476552 0,000331760%

Sun 150.000.000,00 1.990.000.000.000.000.000.000.000.000.000,00 0,442443333 0,059269985%

Venus 42.000.000,00 4.900.000.000.000.000.000.000.000,00 0,000013896 0,000001861%

Mars 78.000.000,00 640.000.000.000.000.000.000.000,00 0,000000526 0,000000070%

Juipter 629.000.000,00 1.900.000.000.000.000.000.000.000.000,00 0,000024024 0,000003218%

Saturn 1.280.000.000,00 570.000.000.000.000.000.000.000.000,00 0,000001740 0,000000233%

Uranus 2.730.000.000,00 87.000.000.000.000.000.000.000.000,00 0,000000058 0,000000008%

Neptune 4.150.000.000,00 100.000.000.000.000.000.000.000.000,00 0,000000029 0,000000004%Consider using scientific notation or a Markdown table.

no, I intentionally did not use scientific notation because the answer is for people not acquainted to it and which is not able to figure out that 2E06 is NOT the double of 1E09 (see comments to the other answer, indeed).

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Content dated before 7/24/2021 11:53 AM

Stan Liou 6 years ago

Well, one could use $GM/r^2$, where $GM$ is the standard gravitational parameter and $r$ is some typical distance. So the question is basically equivalent to asking for a typical distance between Earth and the body in question. For Earth-Sun or Earth-Moon, it's sensible to use the semi-major axis of the relevant orbit, but... how do you want to measure the rest? It's essentially easy to get a rough figure, but potentially hard if you want some spatial or temporal average, etc.