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Every orbiting of a satellite around a mass is nothing else but a constant fall - and therefore acceleration - towards this mass. In a way it is a "falling around" that mass.

My question
Is it possible to measure this acceleration on earth due to its "falling around" the sun?

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Related to this question:… –  sum1stolemyname Apr 5 '11 at 14:24
We can measure the effects of differential accelleration, as tidal forces. –  Omega Centauri Apr 5 '11 at 15:32
I can't wait to see someone answer something along the lines "because of the equivalence principle, one can measure only local accelerations relative to an inertial frame" and then someone mumble something back about Mach principle... –  lurscher Apr 5 '11 at 17:19

2 Answers 2

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The answer depends in part on what you mean by "measure". You can certainly calculate the acceleration using the laws of kinematics and dynamics, as Lawrence B. Crowell points out in his answer. Does that, in your mind, count as "measuring the acceleration"?

Here's a much more direct recipe that I think would uncontroversially count as measuring the acceleration. Pick a set of fixed stars, and measure the velocity of the Earth relative to them using the Doppler effect. Any one measurement only gives you one component of ${\bf v}$, but if you measure a few, you can get the full vector. Do this twice, at two different times, and calculate $\Delta{\bf v}/\Delta t$.

With current technology it would be very easy to do this to the required accuracy. In fact, astronomers are making precisely these required measurements all the time -- not with the purpose of measuring the acceleration, but for a wide variety of other reasons. In fact, astronomers often deliberately subtract out the time-varying Doppler shifts due to Earth's changing velocities, because what they're interested in are (often much smaller) changes due to the actual motions of the stars.

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Thank you. Could you also measure the acceleration itself on earth? Acceleration is always absolute and can be felt (like in an accelerating car), so it should be possible with very sensitive instruments, shouldn't it? –  vonjd Apr 5 '11 at 18:19
No -- that's precisely what you can't do, because of the equivalence principle. The equivalence principle says, among other things, that objects in free fall "feel like" they're not accelerating. Since the Earth's motion around the Sun is a free-fall motion, experiments done locally on the Earth can't reveal this acceleration. –  Ted Bunn Apr 5 '11 at 18:25
Thank you, yeah - that clarifies the matter! –  vonjd Apr 5 '11 at 19:49

The acceleration can be found by a number of means. $F~=~ma$ with the Newtonian law of gravity and the centripetal acceleration $a~=~v^2/r$, $$ m\frac{v^2}{r}~=~-\frac{GMm}{r^2}. $$ This gives $v~=~\sqrt{GM/r}$, which is $v~=~29.5km/s$ or $v~=~2.95\times 10^4m/s$, for the mass of the sun and $r~=~1.5\times 10^11m$. The acceleration is then $0.0058m/s^2$

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