I would like to use PlanetData to get µ, the product of the Gravitational constant G and a planet's mass m.

This property is listed in the documentation (Mathematica v10.1.0.0) for PlanetData as the property "GravitationalConstantMassProduct".

Unfortunately, the property appears unknown upon execution:

PlanetData["Earth", "GravitationalConstantMassProduct"]

Missing["UnknownProperty", {"Planet", "GravitationalConstantMassProduct"}]

In addition, the property also does not appear to be in the list of properties when queried as follows:


Given that this is a highly useful constant (known to far greater accuracy than a planet's mass FWIW), I'd like to know what has happened to it.

  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ Jun 3, 2015 at 19:25
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    $\begingroup$ To expand on @belisarius comment - it seems to be a recurring feature of Wolframs Curated Data that the Mathematica version of it tends to be the poor relation of the similar dataset held in W|A. There are similar issues with CountryData $\endgroup$ Jun 3, 2015 at 19:53
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    $\begingroup$ In case anyone's wondering why we don't just multiply EntityProperty["Planet", "Mass"] by Quantity["GravitationalConstant"]: we can measure $\mu=MG$ with much greater accuracy (~9 decimal places) than we can measure $G$, or by extension, $M$ (~4 decimal places). $\endgroup$ Jun 5, 2015 at 1:37
  • $\begingroup$ @2012rcampion Yes, thank you; this is exactly the reason I'd prefer to have µ. $\endgroup$ Jun 6, 2015 at 12:18

1 Answer 1

WolframAlpha["mars Gravitational Constant  mass product", {{"Result", 1}, "NumberData"}]



WolframAlpha["mars Gravitational Constant  mass product", {{"Result", 1}, "ComputableData"}]

Quantity[4.28*10^13, ("Meters")^3/("Seconds")^2]

  • $\begingroup$ I guess, WA or example data isn't going to be too useful here. I couldn't force WA to give answers to more than two decimal places. $\endgroup$
    – LLlAMnYP
    Jun 5, 2015 at 11:09
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    $\begingroup$ Perhaps data accuracy doesn't warrant more digits. Wikipedia has two more digits for Mars. $\endgroup$ Jun 5, 2015 at 11:44

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