Solar System Moons

Scroll to the bottom for a celestial mechanics tutorial.

This page displays a partial listing of the Natural Satellites in the Solar System.
Please note that we have only listed the satellites that have actual names.
So, don't expect to see S/2004 S 37 (a moon of Saturn) to be listed.
Luckily, satellites with names such as that are usually 2 kilometers or smaller.

Also note that the units used in the satellite list particularly days and kilometers are not the ones usually used.
We have also included a short tutorial about celestial mechanics, with formulas that require input units to be in kilometers, kilograms and days - as opposed to the typical equations whose required units are meters, kilograms and seconds.
You may use these equations in order to verify the natural satellite data we have listed.
You may also use these equations to answer hypothetical questions such as 'for a satellite of Jupiter to have a period of 1 year (365.25 days), at what distance must it orbit Jupiter?'
Then again, we have a celestial mechanics calculator located here.

P L A N E T   M A S S E S
 Planet Mass (kg) Earth 5.97237 x 1024 Mars 6.41710 x 1023 Jupiter 1.89820 x 1027 Saturn 5.68340 x 1026 Uranus 8.68100 x 1025 Neptune 1.02413 x 1026 Pluto 1.30300 x 1022

C E L E S T I A L   M E C H A N I C S

 Inputting the orbital 'radius' in meters and 'time' in seconds, this equation will solve for 'mass' in kilograms: mass = 4 • p ² • r ³ ÷ G • t ² Calculating 4 • p² ÷ G = 591,525,585,921 produces a more compact equation: mass = 5.91526 x 1011 • r ³ ÷ t ² To input kilometers for 'radius', and days for time, we need to recalculate that constant of 5.91526 x 1011. One kilometer = 1,000 meters but since the equation requires "r" to be cubed, then the constant needs to be increased by one billon! One day = 86,400 seconds but the equation requires time (t) to be squared and it also requires t² to be a divisor. So, now we must adjust the constant by multiplying it by 1 x 109 and dividing it by 86,400² and it equals 7.92403 x 1010 making the equation: mass = 7.92403 x 1010 • r ³ ÷ t ² By using algebra we can create the celestial mechanics equation solved for time (in days) and for inputting units of kilometers and kilograms of mass. time = Square Root (7.92403 x 1010 • r ³ ÷ mass) Here is that equation solved for radius of orbit (km) and for inputting units of time (days) and mass (kg). radius = Cube root (mass • t ² ÷ 7.92403 x 1010) or radius = Cube root (mass • t ² • 1.26198 x 10-11)