I understand that we can calculate the gravitational force other planets exert on Earth, but can it be measured?

  • $\begingroup$ Great question! There's a similar but different question here. The answer is yes, but I'm not able to write a full answer now. Rocky bodies that have had landers or rovers that contained accelerometers for navigation or even seismology will likely (but not always) have made measurements of surface gravity to various levels of accuracy. But even so more accurate numbers will come by monitoring the orbits of satellites around those bodies, and calculating. $\endgroup$
    – uhoh
    Jan 14, 2019 at 14:21
  • $\begingroup$ Worth reading this paper on measuring Jupiter's gravity: google.com/amp/s/www.researchgate.net/publication/… $\endgroup$
    – Rory Alsop
    Jan 14, 2019 at 22:46

1 Answer 1


I think the questions you're asking is, "All of the planets exert a gravitation force on everything else in the Solar System, including the Earth. Can we make measurements of that force in a laboratory on the surface of the earth? (I.e., not by observing events on the other planets or their orbits.)"

If I'm right, the answer is "No, the effects are too small."

First, calculate the gravitational acceleration we feel on Earth from, say, Jupiter. As it happens, Earth is roughly 10,000 Jupiter radii away from Jupiter. The acceleration due to gravity follows an inverse square law, so the acceleration due to Jupiter on Earth is about 10-10 that at the surface of Jupiter or about 3x10-10 g. This is small, but perfectly measurable.

The problem is that we can't measure gravity directly. We're sitting on a freely-falling Earth and (as far as Jupiter is concerned) are freely falling with it. Whatever acceleration Jupiter is imparting, it imparting the same acceleration to the Earth, us and our laboratories and everything in them. So there's no force to measure.

We can measure tidal forces directly, since they are in essence the differences in acceleration due to objects being a different distances from the gravitating object, but they drop off as the inverse cube of distance and are undetectable at this distance using current technology.

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    $\begingroup$ That 10^^-1 looks like a typo? $\endgroup$
    – user1569
    Jan 15, 2019 at 13:43

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