Talk:Standard gravitational parameter

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Info[edit]

Thanks for this great info. I'm trying to use it to calculate the orbital distance between two hypothetical stars. But I have become quite lost in the formulas. I am no math wiz - obviously - so that may expain why. I'm having trouble figuring out how to connect the formulas together. As well as determining what values to use for some of the items such as Mass (M). Is it in solar masses? kg? Plus, what is "<<"? It can't mean less-less than <grin>. Sorry about all this -- I guess I'm just one of those lost souls who need an example after the formulas. BY THE WAY: These pages are really great! They helped me develop a table that tells what the Apparent Magnitude of any object is at any given distance! Only took me all night to write it, so my own apparent magnitude was pretty dim! Tesseract 501@aol.com

I am glad you like Wikipedia. "<<" means "much less". M can be taken in kg. Note the difference between m^3 and km^3.--Patrick 12:27, 19 August 2005 (UTC)[reply]
Thank you Patrick. "Much less" being a relative term, I suppose. When I only get two slices of pizza instead of four, I consider it <<. I'm still struggling with the Gravitational Constant and the SGParameter. It all started when I just tried to calculate Orbital Periods for hypothetical objects. The various websites seem to give info along the lines of "if planet was .68 AU from the star, its Orbital Period would be 228 days. If it was 0.01 AU, its Orbital Period would be 18 days. Etc." So, I tried to plug in value into the T = 2\pi\sqrt{a^3/\mu} formula for determining orbital period (small body around central body) -- and the bit-more involved formula for two bodies orbiting each other. It's straightforward enough to get the Semi-Major Axis values and the masses. But, the plugging the Gravitation Constant into the equation is where I'm a bit stuck. I'm not sure how to incorporate the kg-1, m3 and s-2 aspects.
I wish the page had at least one example using hypothetical objects (one for the small-body equation, and one for the two-body equation).
Thanks again -- I've seen your work on other pages. It is really great stuff. Inspiring.
Tesseract501@aol.com (August 27, 2005)
Thanks. The "kg-1, m3 and s-2" mean that distances should be in meters, not AU, and times in seconds, not days. Anyway, you may not need G, you need the sum of the two μ-values; for those you can e.g. multiply the number of solar masses by the μ-value for the Sun.--Patrick 00:13, 28 August 2005 (UTC)[reply]

Unit Conversion[edit]

This might go better in the units of measure area, but since there were already people talking about it in this area, I'll stay here for the moment. :)

Patrick, could you give an example of converting, for example, km3/s-2 to m3/s2? I understand how to convert km3 to m3 (multiply the value by 1000, since there are 1000 meters in a kilometer), but I don't understand what unit of time s-2 is referring to, or how to convert between different exponentations of the same time unit, which is what it sounds like it's asking me to do. --Kintar 15:04 CST, 06 October, 2005

1 km3s-2 = 1 km3/s2 = 1,000,000,000 m3/s2.--Patrick 22:17, 6 October 2005 (UTC)[reply]
I'm not really sure how km3 is the same thing as m3. I understand that GM is m3s-2, but wouldn't that only make the equation kms-2? So is the equation on the front page correct?--oscabat 18:13 CST, 06 January 2005.
Perhaps you're misunderstanding the difference between km and m^3. 1 km = 1000m . 1 km^3 = 1 km*km*km = (1000m)*(1000m)*(1000m) = 1'000'000'000 m^3 = 1*10^9 m^3. The units on the front page are correct. I would prefer that the units of GM be changed to m^3/s^2. SI units all the way. DavidMcKenzie 20:45, 1 August 2007 (UTC)[reply]

The units are not correct right now. On the table (top right corner) the unit is (m3s−2) and the value for Earth is 398600.4418(9). In the article, the value given is 398600.4418±0.0008 km3s−2. Same value, different unit. The correct value should be 398600000000000 m3/s2 =3,986*10^14 (rounded), from the values of G (6,67*10^-11m3/kgs2) and M(6*10^24kg). That is 398600 km3/s2 so the value in the text is correct, the table is not. Hoemaco (talk) 10:40, 8 December 2014 (UTC)[reply]

Standardized table to m3.- Elmidae 12:46, 19 November 2015 (UTC)[reply]

Moon GM parameter[edit]

I thought that the Moon should have a listing in that table, especially if Ceres has! referenced from here: [1]

Precision[edit]

From the article: "The value for the Sun is called heliocentric gravitational constant and equals 1.32712440018 × 1020 m3s-2." How did we get that many places of precision? RJFJR (talk) 21:18, 18 July 2009 (UTC)[reply]

All of the values in the table are shown as having an uncertainty (9) in the last figure. That seems quite unlikely, and in the text the value for Earth is given as 8 in the last place, so that would contradict. Were these just assumed because the true uncertainties were unknown? And if so, is that a commonly accepted practice? Sethhoyt (talk) 19:55, 1 February 2016 (UTC)[reply]

Cite this definition?[edit]

The first paragraph currently defines the "standard gravitational parameter" as based on the sum of the two objects' masses and states the mass of one object times the gravitational constant (GxM) is an approximation of the "standard gravitational parameter". If this is the "true" meaning of the phrase, a source should be cited. The notion seems wrong to me: while I see some logic in such a notion, it means a "standard gravitational parameter" does not apply to a single body (e.g., the Sun's standard gravitational parameter), but actually applies to a pair, e.g., the Sun-Earth standard gravitational parameter. This is radically different than the manner in which the term is used. While "G times the sum of the masses" is a very useful quantity, I would say that the "standard gravitational parameter" of a particular body could be described as a good approximation of this useful quantity, for smaller objects orbiting around it. 74.69.160.254 (talk) 17:48, 29 September 2023 (UTC)[reply]