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Could the acceleration in the rate of expansion of the universe be due to the weakening of gravitational forces, as the distance between objects continues to increase?

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Modified gravity (both starting from Newtonian and Einsteinian bases) have been proposed in serious contexts and given some attention. They are not, however, very popular. – dmckee Apr 29 '13 at 23:35

I guess that you are imagining an expansive force accelerating the Universe versus gravity pulling the Universe together, and that if somehow gravity were weaker, the expansive force would win. That is not the correct picture.

In popular models, the accelerating Universe is caused by gravity, because of a vacuum energy with negative pressure (see dark energy/cosmological constant/quintessence). This statement requires a much fuller explanation.

A weakened gravity would merely dampen any gravitational contraction in a model without dark energy, or amplify any expansion in a model with dark energy. A weakened gravity wouldn't help with the underlying problem, namely, why is the Universe accelerating?

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I think the term acceleration implies a change in direction of the net forces acting on galaxies on the large scale. Gravity (regardless of magnitude) acts to collect the galaxies, while the acceleration acts in the opposite direction.

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Why did this answer get downvoted? The answer is correct. The reduction in density causes a reduction in $\ddot{a}/a$ but in the absence of dark energy it's value always remains negative and acceleration requires it's value to become positive. – John Rennie Apr 30 '13 at 6:44

The expansion of the universe is (approximately) described by the FLRW metric. The Wikipedia article I've linked gives lots of gory details, but the key result we need is the dependance on the acceleration on the density and pressure:

$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left( \rho + \frac{3p}{c^2} \right) + \frac{\Delta c^2}{3} $$

If there is no dark energy and we can ignore the pressure then this simplifies to:

$$ \ddot{a} \propto -\rho \space a $$

The parameter $a$ is the scale factor, and $\rho$ is the average energy density. The density varies as $a^{-3}$, so as the universe expands the magnitude of the acceleration $\ddot{a}$ does indeed decrease, and as you say this is because the matter is less dense and produces a lower gravity. However the acceleration always remains negative, i.e. it is always a deceleration. The expansion of the universe will never cause an acceleration.

To see how we get an acceleration note that if the cosmological constant $\Delta$ is much greater than the contribution from matter and pressure our equation simplifies toL

$$ \ddot{a} \propto \Delta \space a $$

So the sign of $\ddot{a}$ is now positive i.e. we have an acceleration.

So the answer to your question is that we do require a cosmological constant to get acceleration. It can't happen just due to the energy density decreasing.

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