## Saturday, October 27, 2018

### [davcwnxt] Standard gravitational parameter from units

The 'units' program does not have built in the standard gravitational parameter, mu=G*M, for various solar system bodies.  However, we can get it from gauss_k, the Gaussian gravitational constant.  Here is GM for the sun:

You have: gauss_k^2 au^3 day^-2
You want:
Definition: 1.3271244e+20 m^3 / s^2

Both au and day are exact.  gauss_k is known to 10 significant digits.  In contrast, if we were to multiply G and solarmass directly, the universal gravitational constant suffers from precision of only 4 significant digits, so the product would too.

Here is GM for the Earth and Jupiter:

You have: gauss_k^2 au^3 day^-2 solarmass^-1 earthmass
You want:
Definition: 3.9860044e+14 m^3 / s^2

You have: gauss_k^2 au^3 day^-2 solarmass^-1 jupitermass
You want:
Definition: 1.2671277e+17 m^3 / s^2

Although solarmass and planetary masses aren't known to high precision, their ratios are.  Planetary masses are defined in 'units' as ratios to solarmass, so the low precision of solarmass cancels out in the expressions above.

You have: jupitermass
You want:
Definition: solarmass / 1047.3486 = 1.8991766e+27 kg

Here is the escape velocity of earth calculated using the above standard gravitational parameter for earth:

You have: sqrt(2 gauss_k^2 au^3 day^-2 solarmass^-1 earthmass earthradius^-1)
You want:
Definition: 11186.127 m / s

Here is a pretty bad way to get gravitational acceleration g, though it is within the range seen on earth (9.76 - 9.83 m/s^2).  What causes the discrepancy?  I'm guessing it's because the earth is an ellipsoid not a sphere.

You have: gauss_k^2 au^3 day^-2 solarmass^-1 earthmass earthradius^-2
You want:
Definition: 9.8202195 m / s^2

You have: gravity
You want:
Definition: 9.80665 m/s^2 = 9.80665 m / s^2