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It is clear that the gravitational pull diminishes with increasing distance, but I can't find the answer to whether there is a point where it becomes "exactly zero". Is there such distance? Is it calculable?

Qmechanic
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gunakkoc
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3 Answers3

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Is there such distance?

No. Gravity is a fundamental force whose range is infinite. The same applies to the electromagnetic force. For more info see

http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html

Hope this helps.

Bob D
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Yes, in a way. Gravitational force decreases inside a planet. At the center it is 0 because all parts of the planet pull outward symmetrically. The formula you used only applies outside.

Also at an infinite distance, the force would be 0. That is of course a bit of a cheat. The limit is 0, but you can't find a point where the value is the same as the limit.

mmesser314
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Not in Newtonian gravity, but in relativistic cosmology. The most suitable metric to study this problem is the Schwarzschild De Sitter metric. By setting $\dot{r}=\dot{\theta}=\dot{\phi}=0$ we get

$$\ddot{r}=\frac{\lambda c^2 r}{3}-\frac{G M}{r^2}$$

where $\lambda$ is the cosmological constant. By setting $\ddot{r}=0$ and solve for $r$ we get

$$r=\sqrt[3]{3 G M / \lambda / c^2}$$

which is the coordinate radius at which the test particle is neither attracted nor repelled and keeps a constant distance to the dominant mass $M$.

Another approach is to set the recessional velocity equal to the Newtonian orbital velocity

$$H r = \sqrt{G M/r}$$

and solve for $r$, then we get

$$r=\sqrt[3]{G M / H^2}$$

Since in a De Sitter universe the Hubble and the cosmological constants are related by

$$H=\sqrt{\lambda c^2/3}$$

this is the same expression for $r$ as obtained above.

Yukterez
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