Having approached an unknown planet, a spaceship went in a low circular orbit. Would astronauts be able to determine the average density of the planets using only a watch

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    $\begingroup$ Why would the astronauts only have a watch? $\endgroup$
    – Ingolifs
    Apr 23, 2019 at 3:20
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    $\begingroup$ It is a question i came across.....even i cant figure that out $\endgroup$
    – Danish
    Apr 23, 2019 at 3:28
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    $\begingroup$ This is a perfectly clear question. *Welcome to Space! @Danish ;-) $\endgroup$
    – uhoh
    Apr 23, 2019 at 4:14
  • $\begingroup$ It raises a second question: how would you be able to calculate an orbital trajectory without knowing the density of the planet? Or are we assuming the spaceship crew lucked out? :) $\endgroup$ Apr 23, 2019 at 10:33
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    $\begingroup$ @RobinWhittleton Assuming the planet had a basically normal density (i.e. somewhere in the 1.5 (dirty-ice) to 6 (iron core) g/cc range) it wouldn't be hard to pick a trajectory that would lead to a safe orbit, circular if the planet turns out to be dense, elliptical with a very high apoapsis if it's light. $\endgroup$ Apr 23, 2019 at 16:01

1 Answer 1


Planets are usually rotating, and so you can't time two passes over a single geographical feature to time an orbit. But you can time successive sunrises to get a good approximation of the orbital period, or some other astronomical coordination between the planet's limb and the celestial sphere.

From this answer:

Starting with

$$T = 2\pi\sqrt{\frac{a^3}{\mu}} = 2\pi\sqrt{\frac{a^3}{Gm}}$$


$$\rho = \frac{m}{\frac{4}{3} \pi R^3}$$

$$\frac{1}{m} = \frac{1}{\frac{4}{3} \rho \pi R^3}$$

where $\rho$ is the density and $R$ is the body's radius. Then:

$$T = 2\pi\sqrt{\frac{a^3}{G\frac{4}{3} \rho \pi R^3}},$$

and if you set $a=R$, you get:

$$T = 2\pi\sqrt{\frac{3}{4G\pi \rho}},$$

and finally

$$T = \sqrt{\frac{3 \pi}{G\rho}}.$$

And finally, flipping that around, we get:

$$\rho = \frac{3 \pi} {G T^2}.$$

The catch is the step where you set the radius of the circular orbit equal to the radius of the planet; $a=R$.

On an airless world you can get pretty close for a little while, but bodies are usually nonuniform (lumpy) and so an orbit that skims near the surface will quickly get perturbed and crash.

So to do the experiment safely you need to orbit a safe distance above the surface, and this makes $a=R$ a source of error.

One solution is to use a hand-held laser altimeter (if such a thing exists) but probably that won't be found in an astronaut's watch, not even a fancy Scott Kelly watch.

With a conventional watch, one way to estimate altitude would be to measure the time of more celestial events, like a sunrise plus a sunset, as long as you can work out the 3D geometry to convert that to an altitude. The higher the altitude, the longer the day and shorter the night is.

Once you have altitude $h$, you can solve for both the radii of the planet $R$ and of the orbit $a$.

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    $\begingroup$ If $a/R$ is fixed then a similar calculation gives$\rho$ In terms of T. That ratio can be measured by measuring the angle the planet subtends in your field of view. Fur instance of it is 120 degrees (from 4 to 8 on the face of your watch, then the ratio is $\sqrt{3}/2$ $\endgroup$ Apr 23, 2019 at 6:32
  • $\begingroup$ If we have only clock - we can measure durations of shadow passes and starshine passes during orbit. Make much enough orbits, take maximum t_shadow/t_light, and if we are sure our orbit is circular than we can estimate a/R ratio. $\endgroup$
    – Heopps
    Apr 23, 2019 at 6:40
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    $\begingroup$ In other words, no. You cannot determine the density of the planet with just a watch. $\endgroup$
    – GdD
    Apr 23, 2019 at 8:13
  • $\begingroup$ @GdD Not to unlimited precision, no, but you can't do that no matter what instruments you have. You can get a pretty decent approximation (enough to distinguish Venus density 5.24 from Earth density 5.5) by entering the lowest stable orbit that you can, and then using the watch to time the interval between successive rises of a suitable star over the horizon. $\endgroup$ Apr 23, 2019 at 8:45
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    $\begingroup$ @GdD No, I think yes! I believe you can by measuring complementary intervals to get both. Just for a simple illustration, $t_1, t_2, t_3$ = sunrise, sunset, sunrise. Two equations, two unknowns. Two time intervals ($t_2-t_1$ and $t_3-t_1$) and two radii ($r$ and $R$). Of course it would be more complicated due to (at least) two inclinations in the problem, but I think the only instrument you need (besides your eye and a window) is a watch. $\endgroup$
    – uhoh
    Apr 23, 2019 at 15:56

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