The sun is strong enough to keep gas giants close, but why not people?

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    $\begingroup$ For your scenario to be true the gravitational parameter between the sun-human system would have to be much greater than for the sun-earth system. Instead it seems to be a universal constant. $\endgroup$ Nov 29, 2016 at 15:37
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    $\begingroup$ Last time I checked, people are a lot closer to the Sun than the gas giants. Did I miss some manned mission to Uranus or something? :P $\endgroup$
    – Luaan
    Nov 30, 2016 at 14:06

4 Answers 4


The Sun is keeping you close. After all, you are orbiting it just like the Earth. You don't fly off into space because the Earth and you experience the same acceleration due to the Sun's gravitational force, so you orbit together; this is sometimes called the equivalence principle.

If, however, you were floating near Earth but closer to the Sun, you would experience stronger gravity. You would be in a smaller orbit which would make you drift away from the Earth. You wouldn't fall into the Sun, though.

Edit: I forgot to say something about the outer planets, something which the other answers touch on but I think get wrong. First, we should speak of acceleration rather than force, because like I said earlier all objects at a given distance from the Sun experience different forces but the same acceleration.

You ask "how come the Sun is strong enough to keep the distant planets in orbit but I don't fall into it?". The important point is that you don't need such a huge acceleration to keep the planets in orbit, because they are far away and move very slowly.

But, the smallness of the acceleration isn't the reason you don't feel it. The reason is that you're in free fall around the Sun; even if you were zipping around kilometers from the Sun's surface, you would not feel the huge gravitational force, because it affects everything around you in exactly the same way (disregarding tidal effects).

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    $\begingroup$ If you were "floating near Earth but closer to the Sun", it is definitely not true that you would necessarily drift towards the Sun. It depends completely on how close you are to the Earth. (It also depends upon your velocity. There are plenty of things, such as satellites and the moon, which are often between the Sun and Earth but do not drift closer to the Sun.) $\endgroup$ Nov 30, 2016 at 9:05
  • $\begingroup$ @GregMartin: fair enough, I took a simple example to illustrate tidal forces. You can interpret "floating" to mean "standing still in the rotating frame", in which case you will drift towards the Sun and then out again in an elliptical orbit. $\endgroup$
    – Javier
    Nov 30, 2016 at 10:55
  • $\begingroup$ Isn't the earth free falling toward the sun, just like astronauts are to Earth. So the Earth, and therefor Everyone on Earth, is experiencing the suns Gravity just like the Astronauts are for Earths gravity, we kind of ignore it since our angular momentum counters it. But unless we are in space orbiting earth, we are not also countering Earths gravity, so its the only one that matters (on earth) in relation to this question. (in that we ignore the Suns gravity not because its actually negligible, but it is negligible in this case) $\endgroup$
    – Ryan
    Nov 30, 2016 at 18:46
  • $\begingroup$ @Ryan: That's one way to think about it, yes. $\endgroup$
    – Javier
    Nov 30, 2016 at 19:15
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    $\begingroup$ While we don't feel the effects on our body (a lot of answers explain why effect is so "small" compared to the earth gravity), the Sun gravity has a real impact on what is on earth. A visible example of this is how the tide effect may be amplified with the right alignment of the Sun, the Moon and the Earth. $\endgroup$
    – dotpush
    Dec 1, 2016 at 3:42

First of all, planets are much, much more massive than you are and so the sun's gravity exerts a much larger force on the planets than it does on you.

Secondly (and more importantly) you are much closer to the Earth than you are to the sun. Gravity follows what's called an inverse square law which in simple terms means that if you double the distance between the two objects, the strength of the force gets 4 times weaker (because 2 squared is 4). If you triple the distance, the force gets 9 times weaker.

The sun is around $150\,000\,000$ kilometres away whereas the centre of the Earth is only about $6\,300$ kilometres away from you. That equates to a distance increase factor of $24\,000$... So the force would be $24\,000^2 = 576\,000\,000$ times weaker. The sun's mass is only around $300\,000$ times greater than the mass of the Earth. So it's not anywhere near enough to make up for the weakness caused by the huge distance.

  • $\begingroup$ If I'm calculating correctly, the Sun's gravitational acceleration at Earth orbit is merely 0.0059 m/s² (compared to Earth's 9.8 m/s² on the surface). $\endgroup$
    – Jan Hudec
    Nov 29, 2016 at 21:41

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The sun is pulling you up (towards its center) with a force $$\rm F_{sun}=G \frac{M_{sun} × m_{you}}{R^2}$$ and Earth is pulling you down (towards its own center) with a force $$\rm F_{earth}=G \frac{M_{earth} × m_{you}}{r^2}$$

If we find the ration between Earth’s down force and the sun’s up force, we get

$$\rm α = \frac{M_{earth}}{M_{sun}}\frac{R^2}{r^2}$$

$\rm α $ bigger than 1 means that earth is pulling you down with a bigger force. If smaller than 1, the sun is pulling you up with a bigger force. Since

$\rm M_{earth}=6×10^{24} kg$; $\rm M_{sun}=2×10^{30} kg$; $\rm r=6.4×10^3 km$; $\rm R=1.5×10^8 km$, then


So, Earth is pulling you down with a force about 1600 times bigger than the sun. That’s why the sun cannot rip you off the earth.

For the planets no matter how big the force between them and the sun is, they would still be trapped (actually you, me and the whole earth too) since there is no other force pulling them away from the sun just like the Earth acted opposing sun’s gravity on you (forget the other stars and the whole universe, ok).


The force of gravity is generated between all objects in our universe. you can calculate that force using this equation:

$$F=\frac{G \cdot M \cdot m}{r^2}$$

where $G$ is the gravitational constant $6.67\times {10^{-11}}$,

$M$ is the mass of one object and $m$ is the mass of the other ( in kg)

and $r$ is the distance between the objects (in meters).

If you calculate the force applied between you and the sun, you will see that its much smaller then the force between you and the earth.

So earth is actually pulling you harder.

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    $\begingroup$ This is wrong; it has nothing to do with the Earth's gravitational field, as surprising as that may seem. $\endgroup$ Dec 1, 2016 at 21:22

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