Bearing in mind the moon's perfect fit when eclipsing the sun, were the two orbs to have the same density, would they not each exert the same gravitational pull on the Earth? (Much like a dim star in close proximity appears just as bright as a distant star of high luminosity.) Although the nature of the solar interior remains in the realm of speculation, it is presumed to be of high temperature and gaseous. Conversely the interior of the moon is assumed to be cold (no active volcanoes) and of a solid nature. As gases are less dense than liquids and solids, this would imply that (if the moon isn't hollow) the gravitational attraction of the moon on the Earth should exceed the gravitational pull of the sun? Pray tell me, is there a flaw in this logic? Or does the Nobel Prize beckon?

  • $\begingroup$ Even if the sun if gaseous, it is about 99% mass of the Solar System. Now, compare that to the size of the moon. $\endgroup$ – Yashbhatt Nov 12 '14 at 17:14
  • $\begingroup$ Have you computed $F=GmM/r^2$ for the two cases? $\endgroup$ – Kyle Kanos Nov 12 '14 at 17:20
  • $\begingroup$ I don't understand the downvote. $\endgroup$ – garyp Nov 12 '14 at 17:27
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    $\begingroup$ @garyp: I do. The question is asking for a simple calculation, but wanders in many ways off topic. $\endgroup$ – Ross Millikan Nov 12 '14 at 19:10

The apparent diameter of an object like the Sun or Moon varies as $r^{-1}$, that is an object twice as far away looks half the size. So the nearly perfect fit during an eclipse tells us that the Earth-Moon distance $d_M$ and the Earth-Sun distance, $d_S$, are related to the radius of the Moon, $r_M$, and the radius of the Sun, $r_S$, by:

$$ \frac{d_M}{d_S} = \frac{r_M}{r_S} \tag{1} $$

If the density of the Sun and Moon were the same then the mass would be proportional to the radius cubed, and the gravitational force would be proportional to $d^{-2}$, so the ratio of the gravitational forces would be:

$$ \frac{F_M}{F_S} = \frac{\frac{r_M^3}{d_M^2}}{\frac{r_S^3}{d_S^2}} = \left(\frac{r_M}{r_S}\right)^3 \left(\frac{d_S}{d_M}\right)^2 \tag{2} $$

A a quick rearrangement of equation (1) tells us that:

$$ \left(\frac{r_M}{r_S}\right) \left(\frac{d_S}{d_M}\right) = 1 $$

So equation (2) simplifies to:

$$ \frac{F_M}{F_S} = \frac{r_M}{r_S} $$

The gravitation forces from the Moon and Sun are not the same just because they have the same angular diameter even assuming equal density.


Assuming they have the same density (the Sun's average density is not much smaller than that of the moon) , if they had the same apparent size in the sky, then the mass M of the object will grow as $r^3$ (because $M=4/3\rho \pi R^3$ and $R=\theta r$), so the force actually grows linearly with $r$.


The responses so far seem to suggest that, assuming known density, we can tell how far away an object is by the strength of its gravitational attraction. I'm quite sure this is wrong. Since we are in free-fall about the sun (as we are about the moon) I don't think there is any direct way to measure the force of its attraction.

It is however possible to measure the quadrupole component of that force. That is what we measure when we observe the height of the tides. And in that case, the tidal force of the moon is indeed about three times that of the sun, in line with their differences in density. If they were both made of rock, then we would be unable to distinguish them based on their tidal force.

  • $\begingroup$ free fall or not, the gravitational attraction is the same, right? $\endgroup$ – Wolphram jonny Nov 12 '14 at 18:25
  • $\begingroup$ The force you caluculate for the sun is much greater. I'm just not sure it's a force you can actually FEEL. $\endgroup$ – Marty Green Nov 12 '14 at 18:33
  • $\begingroup$ yes you can! that is why the moon is in orbit around Earth. I mean, you could measure it if you had a sensible enough gravimeter (but the Earth's own gravity would be too strong, I guess $\endgroup$ – Wolphram jonny Nov 12 '14 at 18:34
  • $\begingroup$ What I meant in my first comment is that regardless of free fall towards the Sun or not, the force from the Moon doesn't change $\endgroup$ – Wolphram jonny Nov 12 '14 at 18:40
  • $\begingroup$ +1 I like your second paragraph, for same apparent size and density the tidal force is independent of distance. $\endgroup$ – Wolphram jonny Nov 12 '14 at 18:45

For the two orbs, we use Newtons universal law of Gravitation: $$ F_{1\to2}=G\frac{m_1m_2}{r_{12}^2}\tag{1} $$ which does not require any information about densities, purely the total mass of the objects.

The data we need are:

  • $m_1=5.972\times10^{24}$ kg
  • $m_2=7.34767309 \times 10^{22}$ kg (moon)
  • $m_2=1.99\times10^{30}$ kg (sun)
  • $r_{12}=384,400$ km (earth-moon)
  • $r_{12}=149,600,000$ km
  • $G=6.674\times10^{-17}$ N$\cdot$km$^2$/kg$^2$

These return \begin{align} F_{e-m}&=1.98\times10^{20}\,{\rm N}\\ F_{e-s}&=3.54\times10^{22}\,{\rm N} \end{align} so the earth-sun force is about 100 times stronger than the earth-moon force.


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