Will it be possible to see BFR approaching the moon from earth, with naked eye?

The title pretty much sums it up. I'm not old enough to have experienced any of the previous manned moon landings. Given that Big Falcon Rocket will go around the moon in 2023, I'm wondering if one will be able to see the whole rocket or at least its second stage from earth when it's approaching the moon - but with my plain eyes (I know camera technology's good).

I'm not expecting to see it like a plane or something, but maybe reflections, thrusters, ...? A tiny dot getting closer and closer to the moon, knowing there are humans in there - that'd be cool...

• Hi new user! The simple answer is no, not a chance. Not even close. It would be like trying to see a cellphone from about 100 miles away. Nov 17, 2018 at 16:47
• You may enjoy this, OP .. what-if.xkcd.com/13 Nov 17, 2018 at 16:51
• Hi @Fatti - yes, that'd be a tiny, tiny cellphone... that's why I thought maybe reflections or the thrusters would make a difference... That "what if" page is indeed awesome! Nov 17, 2018 at 17:32

tl;dr: at an apparent magnitude of about +18.5 you need a several meter telescope and a dark sky. Hubble can do it too. So by reflected sunlight, definitely not by eye. The exhaust from a Methalox (CH4 + LOX) engine barely makes any light in the visible, so no help there.

Starting with the math from this answer:

I'm going to characterize the 55 x 9 meter white 2nd stage as a 22 meter diffuse sphere(ical cow) with albedo 0.3 for order of magnitude estimation.

The expression for absolute magnitude $$M_{Abs}$$ by rearranging the equation here is:

$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right)$$

For the "spherical cow" spacecraft that turns out to be an absolute magnitude of +25.

Knowing the absolute magnitude of an object, you calculate the apparent magnitude $$m$$ using:

$$m = M_{Abs} + 2.5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right),$$

where $$d_{SR}$$ and $$d_{RE}$$ are the Sun-Roadster and Roadster-Earth distances, each normalized by 1 AU, and the factor $$O(1)$$ is the phase integral, of order unity, taking into account the angular difference between the direction of illumination and the direction of viewing. In an order of magnitude calculation, this only becomes really significant when the body moves between the Sun and the viewer. See https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H).

In this case, replace Roadster with BFR. Since $$d_{SR}$$ and $$d_{RE}$$ are 1.0 and 0.0025 AU respectively, and ignoring the phase integral, the apparent magnitude is about -6.5 lower or +18.5 magnitude.

I think that can be seen with the Hubble Space Telescope above the atmosphere in a fairly short exposure. On the ground you'd need a several meter telescope and a dark night, but you can do it. If you want to use your eye, better find one of these telescopes that has an active eyepiece!

BFR's Raptor engines burn "methalox" (methane and liquid oxygen) CH4 + O2. The products are CO2 and H2O both gases. The brightness from the exhaust from many launches are from carbon soot glowing when kerosene (RP-1) is burned, not methane.

Here is a screenshot from the video Blue Origin BE-4 Engine Compilation during the daytime, so you can have a reference to brightness. I haven't found an outdoor firing of the Raptor yet:

• Thanks for the detailed answer, especially your addition about the engine chemistry! I'm not a native english speaker, but after I look up some of the physics vocabulary used I'll have a more in depth look on the formulas! Nov 17, 2018 at 12:47
• Feel free to ask me to make something clearer here, or you can post a new question about the calculations after checking the links. Welcome to Space!
– uhoh
Nov 17, 2018 at 12:51
• Doesn't New Shepard use the BE-3, which is hydrolox? Did they test it with the BE-4? Nov 17, 2018 at 18:21
• Agreed, there's no methane in that picture. Methalox != LH2/LOX - that is hydrolox. Nov 17, 2018 at 18:29
• Example methalox engine firing video, if you wish to make a replacement. Though, I think it's easy to say the vehicle's engine firings will be visible while in or near Earth orbit due to the size of the plume, not the brightness of the exhaust. youtube.com/watch?v=e7kqFt3nID4 Nov 17, 2018 at 18:30

If you want an easy way to think about it, imagine how bright it might be in low Earth orbit, 240 miles up (which is just a bit lower than the International Space Station). However bright that is, it will be only a millionth as bright when it’s near the Moon, 240,000 miles away (and so a thousand times the distance). Is it likely that it will be bright enough in low orbit that you can still see it when it’s a millionth as bright as that?

• Good one Mike. I figure you can "see a cellphone" from about 1/10th of a mile. So I said it would be like trying to see a cellphone at 100 miles. Sounds about right? Nov 17, 2018 at 16:48
• @Fattie It’s about right, but the crucial point is that something a thousand times further away isn’t a thousand times dimmer, it’s a million times dimmer. Inverse square law. Nov 17, 2018 at 17:20
• You got the same answer as I did with about a thousand times less work, congratulations! One million times dimmer is 15 magnitudes dimmer. Start with 3rd magnitude and add 15 and you get 18. +1
– uhoh
Nov 17, 2018 at 18:04
• @linusg I think that this is a better answer than mine, because it's just as correct, but a lot more accessible.
– uhoh
Nov 17, 2018 at 18:06

Rocket engines burning in vacuum produce large exhaust plumes that rapidly spread outward to 'fill' the vacuum. This means that for a distance of many dozens of meters - or more - they produce a visible cloud of gas that can catch sunlight and produce very large, visible artifacts in a clear night sky. So - engine burns near Earth (departure, or arrivals burning into orbit) will be quite easily visible to those who are in night and under the burns.

For example, this GLONASS launch trail taken from a thread full of rocket plume images.

Mid course burns may be visible as well, though much less dramatic. Arrival and departure burns happening in the vicinity of the moon are not likely to be visible at all to the naked eye due to great distance.

• I am not sure that CO2 + H2O exhaust in vacuum would ever form a visible plume. Without a buffer gas (i.e. low pressure atmosphere) to slow it quickly, wouldn't the molecules expand to large separation distances and have no chance to recombine into droplets?
– uhoh
Nov 18, 2018 at 2:00
• I can't find suitable evidence for my hunch they will remain visible. Gaseous nitrogen thrusters produce visible plumes. Fuel dumps are visible, though that contains liquid fuel. I believe water vapor will produce visible ice crystals, but this is just my hunch. Nov 18, 2018 at 15:26
• @uhoh Wouldn't they flash-freeze into tiny crystals with pressure drop in vacuum?
– Eth
Nov 19, 2018 at 19:05
• @Eth It's a race. If the molecules bump into each other frequently enough, they can stick and start forming particles, but if their relative velocity moves them far enough apart fast enough so that collisions are infrequent, then they won't. They won't slow down in vacuum. Watch launch videos and especially the views behind 2nd stages where you see the red-hot nozzle and the path back to Earth, RP-1/LOX engines don't seem to leave any trail behind themselves at all, and that's very similar chemistry to methalox.
– uhoh
Nov 20, 2018 at 0:02
• RP-1/LOX engines don't produce a plume visible from the vehicle's POV, but they produce plumes that are exceptionally visible when lit by sunlight against a dark sky (see any SpaceX sunset/sunrise launch). Nov 20, 2018 at 20:57