# Calculating the Angular Momentum of a planet

I want to calculate the Angular Momentum of any planet at any point on the orbit around the Sun. Example: I want to calculate the Angular Momentum of Pluto today, on Jan 27, 2021 00:00:00 Hrs

Where can I get the actual data to calculate this? I will need Mass, Velocity and Radius to calculate. Do I need the θ too? Is it readily calculated by NASA?

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html

• Why do you think the angular momentum will change?! – ProfRob Jan 27 at 22:18
• One could want to calculate the angular momentum at many points in time to demonstrate that it does not change. – Phil Frost Jan 28 at 15:13
• @PhilFrost To demonstrate that it varies, I'd say. – Cristiano Jan 28 at 17:09

## 2 Answers

Where can I get the actual data to calculate this?

If you are a programmer, you could use the SPICE library: https://naif.jpl.nasa.gov/naif/toolkit.html and the data files available here: https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/ or you could use https://wgc.jpl.nasa.gov:8443/webgeocalc/#StateVector or HORIZONS: https://ssd.jpl.nasa.gov/horizons.cgi

The masses are listed here: https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/ file "gm_de431.tpc".

For example, the state of Pluto (here I'm using Pluto system: Pluto + satellites) on 2021-01-27 00:00 UTC in the mean ecliptic and equinox of J2000 reference frame is:

r = (2.11488 · 1012, -4.65878 · 1012, -1.13045 · 1011) m
v = (5088.6; 1090.04; -1584.49) m/s

GM = 977 · 109 m3/s2. If we use G = 6.6743 · 10-11 m3 kg-1 s-2 (https://physics.nist.gov/cgi-bin/cuu/Value?bg), M= 1.4638239 · 1022 kg.

L = r x Mv = (1.09861 · 1038, 4.06324 · 1037, 3.80771 · 1038) kgm2/s, the magnitude is 3.9838 · 1038 kgm2/s.

If we do the calculations for 20 years ago (2001-01-27 00:00 UTC) we get:

L = (1.09685 · 1038, 4.04601 · 1037, 3.78485 · 1038) kgm2/s, the magnitude is 3.9613 · 1038 kgm2/s.

EDIT

For Mars, 1970-08-25 00:00 UTC (the notation 123e+011 means 123 · 1011):

r = (-1.9051451e+011; 1.58990113e+011; 8.02428218e+009) m
v = (-14604.5348; -16539.1841; 13.5952337) m/s
L = (8.65493375e+037; -7.35385224e+037; 3.51194063e+039) kgm2/s
||L|| = 3.51377655e+039 kgm2/s.

For today (Jan 28) I get:

r = (4.11669129e+010; 2.27295989e+011; 3.75334592e+009) m
v = (-22924.868; 6375.0766; 695.965252) m/s
L = (8.61551162e+037; -7.35994071e+037; 3.51209564e+039) kgm2/s
||L|| = 3.51392307e+039 kgm2/s.

WARNING! Do not to assume that all the digits are significant. I wrote the numbers with an unrealistic accuracy just to verify the calculations.

EDIT #2

It could be interesting to see the variation of L against the time.

The following graph shows the value of (L - Lmin) / Lmin, where Lmin is the smallest calculated angular momentum and L is the instantaneous angular momentum.
The calculations are done with the SPICE library and the DE430, jup310 and mar097 ephemerides. The independent variable is the time wrt the orbital period of each planet. The starting epoch is 2021-01-28 00:00 UTC. The graph is as expected. Mercury, well inside the Sun gravity well, is only slightly perturbed by the other bodies.
The Earth is heavily perturbed by the Moon and also Jupiter is significantly perturbed by its moons.

• A BIG thank you! This helps so much. I loved the State Vector calculator. :) Would you mind editing your answer with one more example for another planet and another random date, let's say for Mars on date 25 Aug 1970. – Majoris Jan 28 at 15:53
• Oh, that's fair enough. Even Horizons does that in their generated ephemeris data, and they tell you not to assume all the digits are significant. – PM 2Ring Jan 28 at 18:36
• @PM2Ring I wrote the numbers with unrealistic accuracy, so Majoris can verify his calculations. For Pluto, I already wrote "here I'm using Pluto system: Pluto + satellites". – Cristiano Jan 28 at 19:30
• @Cristiano Awesome! That graph is what I want to do. We should connect. How do I contact you? – Majoris Jan 30 at 20:05
• @Majoris If it's not a marriage proposal, I guess you can write here. :-) What's the problem? – Cristiano Jan 30 at 21:03

The big essential fact about momentum is that it is conserved (in the presence of a central force, as is the case here). So the Angular momentum of Pluto today is the same as it was yesterday, and the same as last year and (excepting perturbations) the same as it has ever been.

It is easiest to calculate for a body that is moving perpendicular to its position vector. This is always true for circular motion, about the centre of the circle. It is not true for elliptical motion, except at apoapsis and periapsis.

At these times $$L = mvr$$. For Pluto the periapsis speed is $$v = 6.10\, km/s$$ the distance is 4.44 billion km and the mass is $$1.31 × 10^{22}$$ kg. To get the angular momentum you've got to multiply them together. If you want SI units, convert those km to m first.

The angular momentum today is the same.

Alternately you can use the relationship

$$L= \sqrt{\mu \ell}$$ where $$\mu = GM =1.33×10^{20}$$ and $$\ell$$ is the semi latus rectum or $$\ell=a(1+e^2)$$, and you have to plug in the semimajor axis for Pluto and eccentricity of its orbit.

Or you can look it up, since it is constant $$L=3.6\times 10^{38}$$

• How important are satellites in calculation of angular momentum of a planet? Do the 79 satellites of Jupiter change/affect angular momentum of Jupiter? – Majoris Feb 19 at 2:17