7
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Stan Liou
    Dec 30, 2015 at 3:58
  • $\begingroup$ 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 :) $\endgroup$
    – Marcus Quinn Rodriguez Tenes
    Dec 30, 2015 at 5:00
  • $\begingroup$ @MarcusQuinnRodriguezTenes : Please post your results if you decide to do the calculations yourself. I think I might be a little lazy... :p $\endgroup$
    – Nico
    Dec 30, 2015 at 8:07
  • $\begingroup$ @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 :) $\endgroup$
    – V-J
    Jan 2, 2016 at 14:24

2 Answers 2

Reset to default
9
$\begingroup$

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: enter image description here

Improve this answer
$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$
    – Dave
    Dec 30, 2015 at 23:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – Conrad Turner
    Dec 31, 2015 at 8:26
  • $\begingroup$ 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). $\endgroup$
    – RononDex
    Dec 31, 2015 at 12:34
  • 1
    $\begingroup$ @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. $\endgroup$
    – userLTK
    Feb 7, 2016 at 7:32
  • 2
    $\begingroup$ 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. $\endgroup$
    – userLTK
    Feb 7, 2016 at 8:01
0
$\begingroup$

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

F = G * M * m / r^2

Hence:

table of values

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%
Improve this answer
$\endgroup$
2
  • $\begingroup$ Consider using scientific notation or a Markdown table. $\endgroup$
    – Mike G
    Apr 22, 2021 at 16:28
  • $\begingroup$ 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). $\endgroup$
    – jumpjack
    Apr 23, 2021 at 7:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .