# Does gravity stack?

If I am standing on a planet on the opposite side from its sun will I feel the downward pull of both the planet's gravity and the star's gravity? In other words, does gravity "stack" (Any D&D players out there?)

• Of course, otherwise you wouldn't feel any gravitational force, since it is so weak. Commented Jul 22, 2014 at 9:01
• Something that no one has pointed out (yet really should) is that because the planet in orbit around its sun (therefore in free fall) it doesn't feel that gravitational pull. The only thing you will feel are the tidal forces, which is actually why in the setup you proposed you will feel pushed away from the star, not towards it. Commented Jul 22, 2014 at 10:48
• The simple answer is yes, sure gravity "stacks". You needn't say anything as tricky as "stacks" - quite simply, gravity adds up. At this instant, you feel gravity from every single atom in the universe. Does that do it for you? :) Note that at any moment there would be many other forces involved also - for example the "centripetal force" of spinning around the star/planet/galaxy/etc in question. Commented Jul 22, 2014 at 10:51
• Regarding "D&D Players": At this instant, you feel a gravitational attraction to every D&D Player in the universe. It does not matter where they are or what they are doing. You also feel a gravitational attraction to every star in the universe, every wine glass, and every gelato. Commented Jul 22, 2014 at 11:01
• Joe did blow my mind. Commented Jul 25, 2014 at 15:18

Technically, yes, however in the case of you standing on a planet:

You feel the gravity of the planet and star, so you accelerate accordingly.

However the planet also feels the gravity of the star, so accelerates towards it. So you only notice your acceleration towards the planet.

Well given Newtons famous equation the force of gravity you feel is equal to

$$F_{\textrm{gravity}}=G\frac{mm'}{r^2}.$$

It increases with the mass of the two objects being measured and is inversely proportional to the distance between the two objects in question. Strangely enough however, this technically stretches out nearly infinitely so the gravitational force of a black hole a billion light years away is technically exerting an almost completely negligible but still (hardly) measurable force upon you right this instant. So yes, gravity does "stack" but in a more equilibrium fashion. Large masses you are closer to exert more force than those farther away.

It might not be best to say you'd feel it, as the oscillations of the sun's gravity (as due to rotations and revolutions) are relatively subtle here on earth. The moon's gravity has twice as much influence in terms of tidal force, which is probably the most prominent consequence of gravitational "stacking" in the macroscopic sense you had in mind.* Here's an illustration of a planet's tidal field by Krishnavedala:

In the case of the earth and the moon, the oceans are pulled toward and away from the moon, causing high tides directly below the moon and on the opposite side of the earth. Meanwhile, low tides occur at points halfway across the surface of the earth from where the moon hangs directly overhead. Other consequences of tidal forces include the preservation of some planetary rings (i.e., prevents them from coalescing into moons) and the splitting of Shoemaker–Levy 9 before it hit Jupiter. These suggest other scenarios in which tidal forces could be well beyond perceptible, and even deadly.

* In a more microscopic sense, which I think @pfnuesel may have had in mind, gravity would not be perceptible at all if it didn't "stack". If only the single subatomic particle nearest you had any gravitational effect and the rest of the earth didn't, none of us would be here!

• "It might not be best to say you'd feel it, as the oscillations of the sun's gravity (as due to rotations and revolutions) are relatively subtle here on earth". Do you mean that changes of Sun's gravity are very subtle? If these are subtle, then we can consider them negligible, and therefore Sun's gravity should just be felt on Earth. And it must be, because Earth is revolving around the Sun due to its centrifugal force. Or you meant something else, which I did not understand? Commented Jul 22, 2014 at 9:51
• Yes, just the oscillations, not the unchanging portion. Within the scope of my answer, I suppose I have neglected them by focusing on the moon's effect...but as I said, the sun's tidal force is (almost) half as large as the moon's, which isn't quite negligible. See Wikipedia on tidal range variation. Commented Jul 22, 2014 at 9:53
• OK, I'm just wondering, how is that possible (that Sun's tidal force is much weaker than Moon's), since Earth is revolving around the Sun, and not around the Moon (which means the Sun must exert larger force on Earth than Moon does). Commented Jul 22, 2014 at 10:04
• Yes, a larger gravitational force, but not a larger tidal force. The sun is much further away, so (to paraphrase Wikipedia once more) the sun has a smaller field gradient at our position. Hence its gravitational force doesn't change as much depending on whether you're on the near or far side of the earth, especially in proportion to its overall force. Commented Jul 22, 2014 at 10:11
• Sure, but gradient explains only the difference in the attraction (and therefore different movement of the oceans) nearer and further away from the Moon. It doesn't explain, why there is no Sun's tide at all (which should be equal throughout). Commented Jul 22, 2014 at 10:49