This question keeps coming to me as I daydream about space.

Suppose, hypothetically, I traveled to Jupiter (or any other relatively large object compared to earth) but had no knowledge of its size or the size of its moons. Would I be able to tell immediately that it is so much bigger than earth? Would it look big to me?

My thought is, without a reference point, the approach to Jupiter would look very similar to the approach to Earth.

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    $\begingroup$ Up until the point that jupiter starts filling up half the sky and looking more and more like a flat plane, the approach would lack any easy visual cues. However, if the space traveller knows anything about jupiters mass, or the approach path, then geometry can be used to give a size estimate. So passengers might not know the size, but the pilot certainly would. $\endgroup$ – Innovine Apr 22 at 13:19
  • $\begingroup$ Earth looks big, and it is, compared to you at least. There's a reason there are loonies out there who think it's flat - most of us don't often get the opportunity to fly high enough to see the curvature of its surface. If you could stand on the surface of Jupiter (assuming it had a surface rather than being a gas giant), the horizon wouldn't look all that different from Earth's, because Earth is already large enough to appear flat from the surface. The difference is you'd have to be flying much higher above the "surface" to be able to see that curvature. $\endgroup$ – Darrel Hoffman Apr 23 at 14:18
  • $\begingroup$ This is similar to how the moon looks huge when its on the horizon, but looks comparatively smaller when its in the middle of the sky (it's always the same size). With nothing else around to judge size, it's really hard to tell. $\endgroup$ – JPhi1618 Apr 23 at 17:55
  • $\begingroup$ I don't think we ever find out how big Solaris is, but it's certainly beautiful!! Perceiving scale is tough from orbit, refer to the GIF shown here where the cubesat looks like it's going to hit the surface any second. $\endgroup$ – uhoh Apr 23 at 22:43
  • $\begingroup$ Welcome new user! it's unclear if you mean if you were standing on the surface of another planet, or, if you were approaching the planet in your spaceship. $\endgroup$ – Fattie Apr 24 at 15:10

The simple answer

You are correct: it would only look big because you know it's big.

There are really three things, purely in terms of your visual system which together tell you how big something is:

  • what angle it subtends in your vision;
  • combined with focus & depth of field information;
  • combined with comparing what your two eyes see, or how what you see changes if you move around.

For something like a planet the second and third of these don't give you any useful information: if you can see the planet as a whole then it's 'at infinity' from the point of view of the optical system of your eyes, and also you're not going to get any useful information from the differing input to your eyes, or by moving your head around.

So the things that tell you it's big are then just comparing it with other objects you know the size of. But, again, for a planet, there's not much useful to compare it with.

A more complicated answer

One thing I realised based on a comment is: if you wake up one day in a spacecraft orbiting some unknown planet (this sort of thing happens to all of us, I'm sure), what can you tell about the planet by how fast it seems to move under you, if you don't know your altitude over it? The answer is that you can tell some interesting things but not its mass.

So, let the mass of some planet be $M$ and its radius $R$. And let it be spherical (good enough approximation!), and not rotating (less good, but we can sort that out below).

Kepler's third law says that

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

Where $a$ is the semimajor axis of the orbit. Let's just worry about circular orbits (see below), and define $r \doteq a$, the radius of the orbit.

So now the angular velocity of such an orbit is just $2\pi/T$:

$$\omega = \sqrt{\frac{GM}{r^3}}$$

OK, so a bit of trigonometry tells you two useful angles:

The angle subtended by the planet in your vision is $\theta = 2\sin^{-1}(R/r)$, and the angle through which you need to orbit for a feature on the planet's surface to move between the horizons as seen from orbit is $\phi = 2\cos^{-1}(R/r) = 2\cos^{-1}(\sin(\theta/2))$. Let's say it takes some time $t$ for this to happen.

So the things you can know, just by looking are:

  • $\omega$, or equivalently $T$, by measuring the positions of the distant stars with your distant-star-o-meter (all spacecraft come with these);
  • $\theta$, by looking at the planet (with a planet-o-meter, of course);
  • $t$, the time it takes for features on the planet to move between horizons, with your trusty watch.
  • $G$ which is engraved on a plaque on all spacecraft.

From $\theta$ you can work out $R/r$ and hence $\phi$, and in fact this and $t$ tells you $\omega$ so all that star-peering was not necessary. If you suspect the planet may be rotating, then you still want to measure $\omega$ using the star-o-meter because then you can use that value for it to compute the angular velocity of the planet by measuring how much $t$ differs from what you think it should be.

But you don't know $r$, or $R$: you just know $R/r$.

Well, looking at the Kepler expression for $\omega$ again: the mass of the planet is

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

where $\rho$ is the average density. Plugging this into the Kepler expression you get

$$\omega = \sqrt{\frac{4\pi G\rho}{3}\left(\frac{R}{r}\right)^3}$$


$$\rho = \frac{3\omega^2}{4\pi G}\left(\frac{r}{R}\right)^3$$

So you can measure how dense the planet is, as well as its rotational velocity, by timing things and looking at its subtended angle, but you can't know how big it is.

Although I've shown this just for circular orbits, it's true for any orbit at all in Newtonian gravity: this is reasonably obvious from looking at Kepler's third law: because $a^3$ occurs in this, and because $M\sim R^3$, everything works out so that all you can know is $M/R^3$ if you only know $R/r$. A similar thing works for masses: if all you know is $m/M$ you can't use this to find either $m$ or $M$.

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    $\begingroup$ It would subtend a much, much larger angle than earth would when in a low orbit. If the trajectory is even approximately known, then Jupiter would start to look rather large as it fills all the windows while you're still nowhere near your periapsis. The mass is probably well known to a space traveller on approach, and can also be used to understand the size, assuming jupiter is not made of exotic material. $\endgroup$ – Innovine Apr 22 at 13:16
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    $\begingroup$ @Innovine: Well, if you know how low an orbit you're in then you know how big it is. But your eyes don't tell you that. However you've raised an interesting question: does how fast it seems to move under you tell you useful things? The answer seems to be 'yes, but not what you think it might': I'm going to add to my answer... $\endgroup$ – user21103 Apr 22 at 14:44
  • $\begingroup$ Uh, it was better without the maths, which only complicate the answer unecessarily in an attempt to sound authorative. Furthermore, Jupiter rotates very rapidly, and is also composed of gas, which rotates at different speeds at different latitudes. Movement of surface phenomenon will not give useful data. The maths behind orbit determination of static surface features is much more complex than what's presented here. $\endgroup$ – Innovine Apr 22 at 16:01
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    $\begingroup$ Ah, today I learned. Now that I think some more on it, I wonder if there's a way to determine your orbital elements if you throw something overboard with a known delta-v and watch how it moves relative to your ship... $\endgroup$ – Innovine Apr 22 at 18:47
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    $\begingroup$ @Innovine: yes, I think that would tell you things! In fact what it would tell you is things about the tidal effects, or in other words the spacetime curvature near you. And that tells you absolute distances I think. That's a clever trick! $\endgroup$ – user21103 Apr 23 at 9:47

No, you won't be able to tell immediately, and yes, it will look big.

First, I think it's extremely unlikely that a spacecraft could travel to Jupiter without knowing Jupiter's size.

However, maybe a passenger aboard the spaceship is unaware of the size of Jupiter, and completely unaware of orbital mechanics. To this person, taking one instant look out the window, they will have no idea what size Jupiter is compared to Earth. Even an experienced space pilot, who has not seen Jupiter before (!) will not be able to tell. The human eye will not get any of its usual size and depth cues. Parallax from moving your head around are useless at space distances, and familiar surface shading (like rivers and mountains) provide no information here.

If the spaceship was in a low Jupiter orbit, then Jupiter would start to fill quite a bit more of the sky than Earth would in an low Earth orbit. This would start to give an experienced astronaut info that the planet is bigger than Earth, but someone not used to any view from orbit may be quite confused. Jupiter will look very big, perhaps much bigger than Earth, but the viewer will have no possibility of really determining HOW much bigger, since they can't tell how far they are from the surface, or how much of the surface they are seeing. Plus they will die of radiation poisoning a few minutes later, anyway. Jupiter is very radioactive.

In fact, just by looking out the window once, there's no way to tell. Some clues may be gained by making observations over time, as the spacecraft moves. If some information is known about the spacecrafts current orbit, then perhaps the size can be derived from that, and Jupiters horizons motion against background stars, but this is extremely difficult to do and would require familiarity with some very advanced maths, and assumes the spaceship is in a closed and unchanging orbit already. If someone doesn't know the relative sizes of the planets, they won't know how to do any of this.

Will it look big? Absolutely yes. I can add a personal note to this bit, since I have been building a space flight simulator with VR support for some years now. I am very aware of the depth and distance limitations of the VR systems compared to real life. Even with these limitations, in absolutely every case, Earth, Moon, Jupiter, Jupiters moons, Ceres, a large rock, the feeling has always, always been one of "wow, these things are HUGE". Even floating alongside a rock which is a few kilometers in size gives an awe-inspiring sense of immense scale. There is only ONE feeling when looking out the window at something, and it is that it is huge. If you are in a 'low' orbit, such that the planet fills up a good portion of the sky outside, you are only able to look at a small part of the surface at one time, not half of it like you might expect. So you have really no clue how big it is. It looks like a flat disk, with some shadowing sometimes suggesting a curve, and you just boggle at how huge it all is (even Ceres, which is pretty small compared to Earth, feels mindblowingly gigantic). It takes a full 90 mins to do an orbit, with seemingly endless mountains and craters just slowly scrolling by below). Having spent many, many hours orbiting Ceres, I can now estimate my altitude to within a few kilometers just by looking, but this is impossible if you are not familiar with the sight.

Nothing in space looks small, apart from you.

  • $\begingroup$ How well does cloud structure on Jupiter come through in VR? I imagine that's one of the things that's difficult to get realistic from source data. $\endgroup$ – Erin Anne Apr 22 at 22:36
  • $\begingroup$ "I have been building a space flight simulator with VR support for some years now" - does "Elite Dangerous" not scratch your itch? $\endgroup$ – iwarv Apr 23 at 8:58
  • $\begingroup$ @iwarv not really. It's pretty, but not realistic, and kind of a dull game imho. $\endgroup$ – Innovine Apr 23 at 20:54
  • $\begingroup$ @ErinAnne I haven't tried elaborate texturing of Jupiter, I just stuck some googled images on it. Most of my action is centered around Ceres, and there I use some NASA height data at ~20m per pixel, and rocky textures which I create myself but are designed to look like the Dawn mission photography. I did a few megapixel texturing of Earth as well, and that was pretty nice, with roughly 1 texture pixel to 1 headset pixel, so it's as good as you can see in VR, but I didn't work much on an atmosphere shader. That's pretty important for a good looking Earth, especially sunrise/sunsets $\endgroup$ – Innovine Apr 23 at 20:56
  • $\begingroup$ We've had to do Earth From The Space Station in VR and it's a complete pain in the ass. One major problem is that viewers are used to the unbelievably good / beautiful / unrealistic synthetic images of that, in features. Tough! $\endgroup$ – Fattie Apr 24 at 15:13

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