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How to Determine the Mass of a Planet

Planets on a scale

By NASA/JPL-Caltech [Public domain], via Wikimedia Commons

In my post on calculating planetary orbital periods, I mentioned that I might do a post on determining the mass of the planet.  Well, here it is.  And like that process, it sounds a lot harder than it really is.  So let’s dive in.

What’s Needed

In order to calculate the mass of your planet you again only need two things: the surface gravity and the size of the planet.  The relationship between these quantities is give by

surface gravity is equal to the gravitational constant times the mass divided by the radius squared

where g is the surface gravity, G is our friend the gravitational constant (6.67384×10-11 m3 kg-1 s-2), M is the mass of the planet, and r is the radius of the planet.  The units here are SI (meters, kg, seconds).  If you want to work in multiples of Earth masses, radii, and surface gravities you can drop the constant G.

Calculating the Mass

So the calculation is straight forward.  To get the mass, you just take the surface gravity, multiply it by the radius squared and divide by the gravitational constant. i.e.

Mass equals surface gravity times radius squared divided by the gravitational constant

So for example, the surface gravity of Earth is 9.807 ms-2 and it has a radius of 6,371,000 m.  Plugging this in gives us a mass of 5.9645×1024 kg which is really close to the actual value of 5.9726×1024 kg.

For fun, let’s do Venus as well.  Venus’s surface gravity is 8.87 m s-2 and it has a radius of  6,052,000 m.  Plugging that in gives us a mass of 4.8679×1024 kg which again is really close to the actual value of 4.8676×1024 kg.  Or doing the calculation in terms of Earth ratios we have that Venus’s surface gravity is 0.905 times that of Earth and it’s radius is 0.950 times that of Earth.  Using the equation without the G you get M = 0.905 x 0.950 x 0.950 = 0.817 Earth masses which (unsurprisingly) is the same 4.8679×1024 kg value we computed before.

And that’s all it takes.

Comparison of planetary sizes

Kepler-62 System and the Solar System

Time for a Reality Check

That’s all well and good for planets we know about, but what about a random planet you’ve made up for your game (or found in a supplement).

Let’s say I want a big planet, say 50% bigger than the Earth but with a surface gravity of only 0.8g.  So we do our calculation and we get that M = 0.8 x 1.5 x 1.5 = 1.8 Earth masses or about 1.075×1024 kg which sounds reasonable.  It’s a larger planet and has more mass.

But let’s look at the density.  The density of an object is just its mass divided by its volume.  So we have the mass above and the volume of a sphere is just 4/3 × π r3 which gives us a volume of 3.66×1021 m3. Dividing the two gives us a density of 2940 kg/m3.

You may not realize it yet but that’s awfully low.  At least for a terrestrial planet.  In fact, it’s a really weird density for object we know about.  Let’s look at some comparisons (remember that the density of water is 1000 kg/m3).

Object Density (kg/m3)
Sun  1408
Mercury  5427
Venus  5243
Earth  5514
  Moon  3346
Mars  3934
Ceres  2170
Jupiter  1326
  Io  3528
  Europa  3013
  Ganymede  1428
  Callisto  1236
Saturn  687
  Titan  1880
Uranus  1270
Neptune  1638
  Triton  2061
Pluto  2030
  Charon  1650
Halley’s Comet  600

From this we see that the terrestrial planets have densities mostly around 5500 kg/m3, although the Moon and Mars are lighter,while the gas giant planets have densities around 1500 kg/m3, and the ice moons and planets are in the 1500 – 2000 kg/m3 range.  So our hypothetical planet doesn’t really fit any of those models.  The closest match is Europa, one of Jupiter’s moons that is mostly made of silicate rock and water ice.

So what does this mean?  Well it means that the planet we just created is probably very deficient in metals.  It is a large abundance of nickel and iron in the cores of the planets that make Mercury, Venus, and Earth so dense.  Mars and the Moon have a lower iron content and so are a lower density and the outer planets and moons are even more iron poor (relatively.  There is probably more iron in Jupiter than the entire mass of the Earth).

Could such a planet exist?  Sure.  But you might want to think about the implications.  Similarly if you found that the density of your planet is really high, you might have a problem in the other direction.  It is just something to think about when designing your world and your system.

 


Categorised as: World Building


One Comment

  1. [...] “Hold on!” I hear you shouting. “You said we only needed the distance between the planet and the star and the mass of the star.  Now we need the mass of the planet too?”  Well, not really.  If you’re calculating the orbital period of an Earth-like planet around a star, the truth is that the mass of the planet doesn’t really matter.  The mass of the sun is about 2×1030 kg (that’s 2,000,000,000,000,000,000,000,000,000,000) while the mass of the earth is about 6×1024 kg (that’s a six with 24 zeros after it, you can write it out yourself if you really want to). Adding those together is 2.000006×1030 which isn’t much different from just the mass of the sun. If the star is much smaller than the sun or the planet is much larger, there might be a small effect but those combinations don’t typically result in habitable planets so we’re not going to worry about it. If it ever is important you have the full equation (and I may just do a post on determining the mass of your planet someday.  Edit:  Look I did!). [...]

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