I had read that the Sun's gravity at the distance of the Earth's orbit was 0.0006 times that of Earth's gravity. So, I was wondering what it would be further out, at the distance of Jupiter's orbit.
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Acceleration due to gravity is given by
$$ a = \frac{GM}{r^2} $$ where G is the universal gravitational constant, M is the mass of the central body and r is the distance between the bodies' centers.
For Jupiter at ~5.2 AU average orbital radius, this works out to 0.22 mm/s$^2$. This is $2.2\times10^{-5}$ (or 0.000022) times the gravitational acceleration due to the Earth at its surface.
answered Apr 14 '16 at 17:12
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4$\begingroup$ Alternatively, centripetal acceleration for circular motion is $r\omega^2$. Taking from Wikipedia Jupiter's semi-major axis (778.299 Gm) and angular velocity ($2\pi/$(4332.59 days)), and hand-waving that an ellipse is approximately circular motion, gives the same value 0.00022m/s$^2$. Which is a relief. $\endgroup$ – Steve Jessop Apr 15 '16 at 0:53
