Difference between revisions of "Talk:Gravitational parameter"
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I would like to point out that mu also equals g*r^2. Gravity, g, = G * m<sub>1</sub>/r^2. So G = g * r^2 /m<sub>1</sub>. Substituting that into mu = GM, we get mu = g*r^2, QED. This is the form of mu that I have been in the habit of using since gravity and radius are usually more available to me than G and m<sub>1</sub>. I have done paper missions with less accurate values but they were sufficient for my purposes. See [[People from Earth to Mars in 30 days]] - [[User:Farred|Farred]] 04:35, 19 February 2013 (UTC) | I would like to point out that mu also equals g*r^2. Gravity, g, = G * m<sub>1</sub>/r^2. So G = g * r^2 /m<sub>1</sub>. Substituting that into mu = GM, we get mu = g*r^2, QED. This is the form of mu that I have been in the habit of using since gravity and radius are usually more available to me than G and m<sub>1</sub>. I have done paper missions with less accurate values but they were sufficient for my purposes. See [[People from Earth to Mars in 30 days]] - [[User:Farred|Farred]] 04:35, 19 February 2013 (UTC) | ||
:I'm not sure why you would want to do that. For heavenly bodies, the average surface gravity g is calculated from mu using the very formula that you just gave (mu and r being determined from astronomical observation). Why would you want to work backwards from this to get mu? I agree that inaccurate values are fine for most ballpark feasibility studies, but finding accurate values takes little time and offers several advantages. A small force over an Earth-Mars transit orbit makes a big difference and inaccurate values are compounded when you work backwards from the fuel needed for the last leg of a mission to determine the fuel needed at the beginning. [[User:ChristiaanK|ChristiaanK]] 17:01, 20 February 2013 (UTC) | :I'm not sure why you would want to do that. For heavenly bodies, the average surface gravity g is calculated from mu using the very formula that you just gave (mu and r being determined from astronomical observation). Why would you want to work backwards from this to get mu? I agree that inaccurate values are fine for most ballpark feasibility studies, but finding accurate values takes little time and offers several advantages. A small force over an Earth-Mars transit orbit makes a big difference and inaccurate values are compounded when you work backwards from the fuel needed for the last leg of a mission to determine the fuel needed at the beginning. [[User:ChristiaanK|ChristiaanK]] 17:01, 20 February 2013 (UTC) | ||
| + | ::However g is calculated, it is the value that I find in some references. If I do not have mu but have g and r from a reference naturally that is what I would use. Actually I have seldom found mu tabulated in the references that I have used. Where it was tabulated, it was referred to as g*r<sup>2</sup>. I just do not have the sort of references that are available to you. - [[User:Farred|Farred]] 18:38, 20 February 2013 (UTC) | ||
Revision as of 11:38, 20 February 2013
Alternate value of mu
I would like to point out that mu also equals g*r^2. Gravity, g, = G * m1/r^2. So G = g * r^2 /m1. Substituting that into mu = GM, we get mu = g*r^2, QED. This is the form of mu that I have been in the habit of using since gravity and radius are usually more available to me than G and m1. I have done paper missions with less accurate values but they were sufficient for my purposes. See People from Earth to Mars in 30 days - Farred 04:35, 19 February 2013 (UTC)
- I'm not sure why you would want to do that. For heavenly bodies, the average surface gravity g is calculated from mu using the very formula that you just gave (mu and r being determined from astronomical observation). Why would you want to work backwards from this to get mu? I agree that inaccurate values are fine for most ballpark feasibility studies, but finding accurate values takes little time and offers several advantages. A small force over an Earth-Mars transit orbit makes a big difference and inaccurate values are compounded when you work backwards from the fuel needed for the last leg of a mission to determine the fuel needed at the beginning. ChristiaanK 17:01, 20 February 2013 (UTC)
- However g is calculated, it is the value that I find in some references. If I do not have mu but have g and r from a reference naturally that is what I would use. Actually I have seldom found mu tabulated in the references that I have used. Where it was tabulated, it was referred to as g*r2. I just do not have the sort of references that are available to you. - Farred 18:38, 20 February 2013 (UTC)
