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Does cumulative mass of Mercury, Venus, Earth's moon and Mars total 99% of Earth's mass ?

Reid
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    Show your sources and your work. That'll make a better question. – RonJohn Apr 23 '23 at 05:10
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    it would certainly be close - Venus is a bit lighter than Earth and the rest are basically loose change. Before getting too golden ration on the topic you need to work out how the moon's formation as an ejected chunk of earth fits into things. – GremlinWranger Apr 23 '23 at 05:28
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    This page https://ssd.jpl.nasa.gov/astro_par.html has planetary system masses in the form of GM (gravitational parameter), which is known to much higher precision than the mass. – PM 2Ring Apr 23 '23 at 13:05
  • FWIW, using that JPL data, I get a ratio of just over 0.990018; that includes Phobos & Deimos. GM (or mass) values for other Solar System bodies can be found in Horizons body data pages. https://ssd.jpl.nasa.gov/horizons/ – PM 2Ring Apr 23 '23 at 13:32
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    Interesting related fact: If you just look at planets, each one is larger than all the smaller ones combined, and Earth is actually the one where that comes closest to being false. This is because Venus is so massive relative to Earth. When you look at all bodies ordered by mass, that pattern no longer holds, as other answers and your own calculation indicate. – Mark Foskey Apr 23 '23 at 18:59
  • Why does it matter? Add up some arbitrarily-selected masses of almost anything of the same type and you'll be able to make that sum to the mass of another object of that type. In other words there's nothing at all notable about what you've posted, except for the human desire to find meaning in coincidence. – Ian Kemp Apr 23 '23 at 19:14
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    I, for one, welcome the human desire to find meaning in coincidence. I hadn't considered this kind of question before and had fun my writing my sort-of-related answer, so I've upvoted. Welcome to Space SE! – Erin Anne Apr 24 '23 at 02:45
  • @IanKemp It might not be pure coincidence, though. Sure, there's a lot of randomness involved in the formation of planetesimals, but there are also clear patterns in the resulting size distribution, which we can see in the Main Belt asteroids. Asteroid sizes roughly follow a power law. I have some info about that here: https://astronomy.stackexchange.com/a/49425/16685 So it's not unreasonable for the dynamic processes to lead to some patterns in the partitioning of masses of the larger Solar System bodies, IMHO. – PM 2Ring Apr 24 '23 at 07:37
  • I found it surprising, if for no other reason than I ignorantly overestimated how massive Mars was. (I knew it was smaller than Earth and Venus, but didn't realize its mass was an order of magnitude less.) – chepner Apr 25 '23 at 13:27
  • You can do even better, Venus + Mercury + Mars + Titan is within about .02% – AKHolland May 17 '23 at 15:04

3 Answers3

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According to Wolfram Alpha, your calculation is correct. (The masses of Phobos and Deimos are negligible here.)

It's not a particularly remarkable coincidence; glancing at a list of solar system bodies ordered by mass, it didn't take me long to find that the mass of Mercury is 99.8% of the combined masses of Ganymede, Titan, and Europa.

Screen shot of Wolfram Alpha, showing the sum of the masses of Mercury, Venus, the Moon, and Mars is 5.91 x 10^24 kg, or 0.99 Earth masses

Russell Borogove
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    "(The masses of Phobos and Deimos are negligible here.)": in fact, adding the entire asteroid belt makes virtually no difference. Occasionally someone suggests dropping asteroids on Mars to add to its gravity, but all the asteroids together are only a few percent of the mass of the moon. – Christopher James Huff Apr 23 '23 at 17:34
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Russell Borogove's answer got me wondering how one might best successively approximate the mass of the Earth by using other solar system bodies (say, if you wanted to balance a scale). My every interaction with Wolfram Alpha causes it to choke so I've instead referred to Wikipedia's List of Solar System objects by size.

Venus, Mars, and Mercury are good first choices; they're the next three-most-massive bodies in the solar system. Those alone get you 97% of the way there.

The next choices are critical.

  • Adding Titan seems to put you slightly over (don't make the same mistake I did; the precision of the table in Earth masses is slightly lower, so stick with the $10^{21} \text{kg}$ column).

  • Adding the Moon instead leaves you at about 0.99 of Earth's mass, as the OP indicates.

  • Instead adding Callisto, Triton, and Haumea seems to put you just under an Earth mass, with the error bars overlapping those of Earth's mass as given in the table.

Link to the spreadsheet where I played with the numbers

Erin Anne
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Using the gravitational parameters of JPL: DE430/431 (the most accurate I could find), the discrepancy between the cumulative sum and the mass of the Earth is about five magnitudes higher than the uncertainty, so this clearly checks out.

 22.03178E+12     Mercury
324.858592E+12     Venus
  4.902800066E+12  Moon
 42.828375214E+12  Mars

394.621547280E+12  Sum


398.600435436E+12  Earth

The ratio is thus indeed very close to 99%, 0.9900179

The similarity is a bit less striking when one considers that the missing mass is about 80% of a Moon mass.

SE - stop firing the good guys
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    Pluto is about 17.7 % of a Moon mass, the total mass of the asteroid belt is calculated to be 3% that of the Moon. The mass of Eris is 22.4 %, Haumea is about 5.5 %, Makemake 4.2 %, Gonggong 2.4 %, Quaoar 1.9 % and Sedna about 1 %. Together about 58 % of the Moon. So still 22 % missing. – Uwe Apr 24 '23 at 09:29
  • The GM values I linked in my 1st comment on the question come from DE 440. Horizons body data has GM for numerous bodies, including many of the larger asteroids, eg, here's Vesta, but they aren't of high precision, and the Horizons docs are rather vague on the sources for the body data pages. – PM 2Ring Apr 24 '23 at 09:55
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    But if we also look at the moons of the gas giants, we get Ganymede 202 %, Titan 183%, Callisto 146 % and Io 122 % of the Earth's Moon. – Uwe Apr 24 '23 at 10:53
  • What units are these masses in? – Jason Goemaat Apr 25 '23 at 18:54
  • @JasonGoemaat They're GM values, in units of $m^3/s^2$. See https://en.wikipedia.org/wiki/Standard_gravitational_parameter – PM 2Ring Apr 25 '23 at 19:41
  • Ah, I think I get it. It's a combination of M (the mass which was asked in the question) and G (the gravitational constant). We don't know M or G as accurately as we've calculated GM for these bodies, but since G is the same for all of them, GM can be used to compare their relative masses, right? – Jason Goemaat Apr 25 '23 at 23:19
  • Could you explain the "five magnitudes higher than the uncertainty" statement, maybe include the uncertainty explicitly? – user2705196 Apr 26 '23 at 12:00