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According to another answer on this site, the time it takes for two bodies in space to collide can be computed with the following formula:

$$\frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}}$$

However, if you plug in the distance between the earth and the sun and their two masses the value comes out to 64 years. Why haven't they collided yet?

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    $\begingroup$ In the text of the question you link: "Starting from rest, they both begin to accelerate towards each other.". Preconditions matter. $\endgroup$ – dmckee May 8 '16 at 16:45
  • $\begingroup$ There exists the law of conservation of angular momentum. The planets are in stable orbits because of their angular momentum about the barycenter of the system , which they got during the creation of the planetary system. $\endgroup$ – anna v May 8 '16 at 16:53
  • $\begingroup$ The most surprising thing to me is the result of 64 years. I'm trying to work out how the Earth can orbit the sun once every year with such a small acceleration. $\endgroup$ – Asher May 9 '16 at 5:55
  • $\begingroup$ 64 years?The earth and Sun are not in rest. $\endgroup$ – user167288 Aug 30 '17 at 14:11
  • $\begingroup$ @Asher - The OP did the math wrong. A test mass at a distance of one AU from the Sun and initially at rest with respect to the Sun would take 64.5 days rather 64 years to fall into the Sun. $\endgroup$ – David Hammen Aug 22 '18 at 17:10
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You sound disappointed that the earth hasn't yet collided with the sun.

Applying random formulas to a physical situation is never a good idea. I suspect that the formula you are having issues with is for two bodies in space which are not in orbit around one another, as unfortunately the earth is with the sun.

When one body orbits another, the motion of the orbiting body is such that as it falls toward the central object, it moves laterally with respect to that object just enough to avoid colliding with it.

Isaac Newton dreamed up a thought experiment to explain how orbiting bodies do not collide with one another, only he used the earth and a hypothetical cannonball fired from the top of a tall mountain in his explanation, which is sometimes referred to as Newton's Cannonball:

Newton's cannonball

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    $\begingroup$ I thought the time for the cannonball to hit the ground was independent of its tangential speed? $\endgroup$ – Ambrose Swasey May 8 '16 at 17:11
  • $\begingroup$ Not sure who down voted this, as it's basically correct. +10 from me. $\endgroup$ – Gert May 8 '16 at 17:27
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    $\begingroup$ @TylerDurden The working that show you that in a Physics 101 class assume the ground is planar and that gravity has constant magnitude and direction ($9.81 \,\mathrm{m/s}$ down), which is fine at human scale but incorrect on distances scales comparable to the radius of the planet. $\endgroup$ – dmckee May 8 '16 at 17:42
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As stated in the answer you link, that formula is for 2 bodies starting from rest. The Earth and Sun are not at rest relative to each other, they are in orbit at a relative tangential speed of nearly $30\,{\rm km}\,{\rm s}^{-1}$.

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The answer you refer to is to a question where both bodies are initially motionless.

That's not the case for the Earth/Sun system. The Earth orbits the Sun. The gravitational force the Sun exerts on the Earth provides the centripetal force needed to keep the Earth in a stable orbit.

In a sense the Earth is constantly free-falling towards the Sun but because of the orbital velocity (the velocity along the path of the orbit, aka tangential speed) it constantly misses the Sun and stays at a more or less constant distance (idealised for a perfectly circular orbit).

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