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At the surface of the Earth, the force of gravity is defined as 1 G.

I am wondering what gravitational force the sun excerts at the distance of 1 AU (the distance Earth is at). Imagine it like this; you are in space, at a distance of 1 AU from the sun, and you have absolutely no velocity relative to the sun. At this point you would start to fall directly towards the sun, but in this case you have a platform with rockets that counteract this so that the platform remains at the same distance from the sun. How heavy would you feel when you are standing on this platform?

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  • $\begingroup$ $g=G\frac{m}{r^2}$ or $f=\frac{G}{1au^2}m_{sun}.m_{you}$ $\endgroup$ – JCRM Jun 19 at 12:49
  • $\begingroup$ Well, that ended up with a whole lot of big numbers... Finally arrived at the result of 593 mN of force, which means that "how heavy I would feel" would be about 60 grams (assuming I'm 100 kg on Earth). That's 0.0006% of Earth's gravity... $\endgroup$ – Anju Maaka Jun 19 at 13:02
  • $\begingroup$ This question and my answer are almost identical to this, but with Earth's orbit instead of Jupiter's and actually contains the value 0.0006g. $\endgroup$ – Jack Jun 19 at 13:19
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Gravitational acceleration $a$ at some distance $r$ from a central body mass $M$ is given by:

$$a = \frac{GM}{r^2}$$

Where $G$ is the gravitational constant $\approx6.67\times10^{-11}m^3kg^{-1}s^{-1}$

For an object at 1AU ($\approx1.5\times10^{11}m$) from the Sun ($\approx2\times10^{30}kg$), this works out as:

$$a = \frac{GM}{r^2} \approx 0.006ms^{-2} \approx 0.0006g$$

So a 70kg person would 'feel like' they weigh 0.06% of their Earth weight or 42 grams.

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    $\begingroup$ I guess they would only feel that way if they were standing on a 1 AU radius Dyson Sphere. $\endgroup$ – uhoh Jun 19 at 14:08
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    $\begingroup$ @uhoh or if they had a slick new pair of rocket boots (patent pending) $\endgroup$ – Jack Jun 19 at 14:11
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At earth's distance from the sun solar gravity is about 6 millimeters/sec^2 . Earth surface gravity is 9.8 meters/sec^2, about 1,600 times greater.

Here is a graphic from an orbital mechanics coloring book I'm working on:

enter image description here

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