If the Earth and Moon were smaller (and everything on them too, but keeping the same average density and the same "relative" distances, I mean... the same proportions!), would the tides caused by the Moon on Earth be more or less intense? I don't know, maybe this is too vague, but I would like to know if something would change. The motivation for this question is that I am trying to write a sci-fi book, and this happens in the story, but my knowledge of physics and mathematics is too shallow to know what the difference would be (if any). Thank you!


1 Answer 1


Assuming all LENGHTHS (like the radii of earth $r_e$ and moon $r_m$ as well as the distance $d$ between their centers) are scaled by the same scaling factor $s$ (which will be set to $<1$ to make things smaller) while everything maintains the same density, the volumes and therefore the MASSES $m_e$ of the earth and $m_m$ of the moon, both approximated as spheres, will be scaled by $s^3$, since the volume $V$ of a shpere is given as $$V=\frac{4}{3}\pi r^3$$

Due to this non-linear relation between "size" (in the sense of length), volume (and thereby mass) and, for that matter, surface (which scales with $s^2$), scaling down things and creatures has a variety of consequences which you might want to consider in your story by the way.

But back to the tides. They are caused by the difference in gravitational force the moon exerts (on a drop of water for example) at different points of the earth, forming two "mountains" at the places closest to and farthest away from the moon. I won't go into detail here but just point to Wikipedia.

Gravitational acceleration, here the one caused by the moon $a_g$, decreses quadratically with distance ($G$ is a natural constant): $$a_g=G\frac{m_m}{d^2}$$

Relevant to the tide is the difference $\Delta a_g$ in gravitational acceleration between different points on earth. Taking the aforementioned points closest to and farthest from the moon as an example, it would be $$\Delta a_g=Gm_m(\frac{1}{(d-r_e)^2}-\frac{1}{(d+r_e)^2})$$

Now let's see what happens when we scale everything - i.e. all lengths by $s$ and the mass by $s^3$, to get the scaled force difference $\Delta a_s$:

$$\Delta a_s=Gs^3m_m(\frac{1}{(sd-sr_e)^2}-\frac{1}{(sd+sr_e)^2})$$

We see that we can reduce the fractions, eliminating $s^2$, and are left with $$\Delta a_s=s\cdot\Delta a_g$$

So the difference in tidal acceleration at different points of the eath goes linear with $s$. Therefore, the tide would be less intense, in the sense of graviational acceleration caused by the moon, when scaling $s$ down.

If what you mean by "intensity" also includes the speed with which the wave goes around the planet, that depends on whether earth still rotates ones every 24 hours.

  • $\begingroup$ Thank you so much!! Very nice explanation. I can understand your answer with the rudiments of physics that I know from school, which is a very nice thing. ^^ $\endgroup$
    – jainemarie
    Commented Aug 16, 2022 at 22:57

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