As the title says, what would the universe be like if the gravitational force was 0.1% stronger but the other forces (strong and weak nuclear forces, electromagnetic force) stayed the same? Would the universe have formed the same as it is now?

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    $\begingroup$ I'm assuming you mean something along the lines of "if $G$ were greater"? $\endgroup$ – HDE 226868 Oct 19 '14 at 18:19
  • $\begingroup$ How much stronger is "slightly stronger" $\endgroup$ – Tim B Oct 19 '14 at 18:20
  • $\begingroup$ @TimB Say 0.1% stronger. $\endgroup$ – Beta Decay Oct 19 '14 at 18:20
  • $\begingroup$ 0.1% wouldn't make much difference here, but if it were more than that - say 5% - we probably couldn't get off the planet with chemical rockets. At least not with any useful payload. nasa.gov/mission_pages/station/expeditions/expedition30/… $\endgroup$ – Kevin Krumwiede Oct 20 '14 at 3:44
  • $\begingroup$ While this got good answers, in the interest of consistency, closing as too broad: worldbuilding.stackexchange.com/questions/79180/… $\endgroup$ – kingledion Apr 26 '17 at 13:52

Not a lot would be different, actually.

Let's take the orbits of the planets around the Sun. In the Newtonian world, gravity is represented by Newton's law of universal gravitation: $$F=G\frac{m_1m_2}{r^2}$$ where $F$ is force, $G$ is the universal gravitational constant, $m_1$ and $m_2$ are the masses of the objects, and $r$ is the distance between them.

Now, gravity is the centripetal force responsible for the motion of the planets. In other words, $$F_g=F_c$$ and so, because $F_c=\frac{mv^2}{r}$ (where $v$ is the tangential velocity), $$G\frac{m_1m_2}{r^2}=\frac{mv^2}{r}$$ Given that $m=m_2$ (where $m_2$ is the smaller mass - the planet), $$G\frac{m_1}{r}=v^2$$ Solving for $r$, $$r=G\frac{m_1}{v^2} \tag{1}$$ Now let's call $G$ at its current value $G_o$, and call its future value (of $1.01G_o$) $G_f$. We now know that $$(G_f)\frac{m_1}{r}=v^2$$ $$(1.01G_o)\frac{m_1}{r}=v^2$$ and, solving for $r$, $$r=(1.01G_o)\frac{m_1}{v^2} \tag{2}$$ So the radius of the planet's orbit is just a bit smaller - in fact, if we write $(1)$ as $$\frac{r_o}{G_o}=\frac{m_1}{v^2} \tag{1.1}$$ and write $(2)$ as $$\frac{r_f}{1.01G_o}=\frac{m_1}{v^2} \tag{2.1}$$ and do a bit of algebra, we find that $$r_f=1.01r_o$$ assuming that $v$ is the same for both $G$s. For Earth, $r=150 \text { million km}$, so we would be about $1.5 \text { million km}$ further away from the Sun. Would that make a huge difference? Maybe a little. As I wrote here, we could have minor seasons if Earth was 1.05 AU away from the Sun in its new "winter" and 0.95 AU from the Sun in its new "summer" (this all neglects axial tilt, as per the question). So there would be some change if we were further away from the Sun, but not a lot.

We can do this kind of approximation on larger scales, too - that is, like stellar orbits around the Galactic center.

Would this have impacted how the universe formed? I'm not sure. 0.1% isn't a very big change. If things were changed substantially, we'd be in trouble. This is the case with some of the other forces, such as the strong nuclear force. As Wikipedia says,

If, for example, the strong nuclear force were 2% stronger than it is (i.e., if the coupling constant representing its strength were 2% larger), while the other constants were left unchanged, diprotons would be stable and hydrogen would fuse into them instead of deuterium and helium. This would drastically alter the physics of stars, and presumably preclude the existence of life similar to what we observe on Earth. The existence of the di-proton would short-circuit the slow fusion of hydrogen into deuterium. Hydrogen would fuse so easily that it is likely that all of the Universe's hydrogen would be consumed in the first few minutes after the Big Bang.

So it's a good thing that all the constants are the way they are. But I'm not sure just how a small change like the one mentioned could affect a universe from the very beginning.

  • $\begingroup$ The other option being a faster orbit at the same distance from the sun as we have now. $\endgroup$ – Tim B Oct 19 '14 at 20:12
  • $\begingroup$ @TimB Yes, correct. $\endgroup$ – HDE 226868 Oct 19 '14 at 20:14
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    $\begingroup$ @HDE226868 A small change in the strength of gravity may have a dramatic effect on the fate of our universe. I read recently that our current best guess (including dark energy estimates) is that our universe is within 0.4% of being flat. It may be possible that a small change in gravity could tip the balance either way into an open or closed universe, "et voila" good-bye Euclid. $\endgroup$ – User2178 Oct 19 '14 at 21:14
  • $\begingroup$ @NickR The thing that would tip the scales one way or another is the energy density of the universe. $\endgroup$ – HDE 226868 Oct 19 '14 at 21:16
  • $\begingroup$ @HDE226868 Yup. I should have remembered that. It is the density parameter, $ \Omega $ , that determines our fate. $\endgroup$ – User2178 Oct 19 '14 at 21:18

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