# Measuring acceleration of earth due to its fall around the sun

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?

• Related to this question: physics.stackexchange.com/questions/5994/… – 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

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 $\Delta{\bf v}/\Delta t$.
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 $0.0058m/s^2$