# Are there Practical Methods to Measure the Distance between a Spacecraft and an Astronomical Body?

I came across this detailed article on NASA's JPL website about how mission teams are able to calculate the trajectories (and by extension the position) of spacecraft at faraway distances. However, I was curious if it was possible for a spacecraft equipped with reasonable instrumentation (such as those found on any previous spacecraft design, or feasible planned design) to determine the distance to a target astronomical body.

For example, suppose a spacecraft in orbit around the sun needed to determine the distance between itself and another body relatively far away (up to 5AU). Could the craft do this with, say, a radar beam aimed at the target body?

Edit: I'm going to make the assumption for simplicity that the spacecraft already knows its position and velocity vector with reference to the sun, but that the orbital elements of other known bodies are not already pre-programmed into the craft.

References:

1. Basics of Space Flight - Section II - Chapter 13 - Spacecraft Navigation | NASA Jet Propulsion Laboratory @ California Institute of Technology
2. Physics Stack Exchange - Can Radio Waves be Formed into a Pencil Beam? | Q: Theodor, A: ptomato
3. Radio Images of the Solar System | National Radio Astronomy Observatory
• Interesting question! That's quite far - is the body planet-sized so that it the reflected signal will be strong? What kind of uncertainty in distance is good enough - even if if there is a tiny received signal, it may be so noisy that after processing and averaging over an extended period of time with some velocity model, there's a very large uncertainty in the distance. Unless it's a new, unknown planet, it's position would be in an ephemeris in the spacecraft's computer to a few tens of km uncertainty (or better), so is this actually about the spacecraft finding its own position? – uhoh Dec 4 '16 at 1:55
• That's a fair point; of course known bodies' orbital parameters would be pre-loaded into the spacecraft's firmware! I'm going to clarify my question a bit more in an edit. – gate_engineer Dec 4 '16 at 14:25

Yes. The spacecraft doesn't need radar or lasers or anything active (which won't work due to the incredible distances).. A decent telescope is all that's required. The telescope will measure the targets movement against the background starfield (known as Parallax). Since the spacecrafts velocity and position is known, this movement can be used to make estimates of the distance, which can be refined over time. This is how we knew the distances to the various moons and planets before sending spacecraft there..

• Sounds interesting! Since the other body is not a known, there may be some assumptions necessary. For example, if it was an unknown object in the asteroid belt, you could assume that it is in orbit around the sun, but if you are trying to calculate a precise distance you'd need to calculate the effects of Jupiter at least. If the spacecraft doesn't have information about Jupiter's mass and orbit, then it could introduce some uncertainty. Also, it sounds like "over time" is many months or even a year, and it would have to know it's position all the time. But this does seem like a good answer! – uhoh Dec 5 '16 at 14:37
• @uhoh spaceship should be able to see jupiter too – Innovine Dec 5 '16 at 15:21
• @uhoh "over time" depends greatly on the orbits, and the power and quality of the telescope.But it won't take a full solar orbit to gather the maximum parallax data which should give some good results. You could probably fine polish the results over a long long time, true. The effects of jupiter would help in estimating the bodies mass, which is not related to its orbital elements or range (which are the parameters under discussion here) – Innovine Dec 5 '16 at 15:30
• The gravity of Jupiter changes orbits. If a body is acted on by both the sun and jupiter (and everything else), but you interpret the motion against the stars using only the gravity of the sun in your model, you will get the wrong orbit and therefore the wrong distance. The principle sounds easy while in prose, but it's messy if you do it correctly with math. To plan a spacecraft trajectory you can't use an approximate distance - you can't turn around and try again. You need a precise state vector. "...orbital elements of other known bodies are not already pre-programmed into the craft." – uhoh Dec 5 '16 at 23:49
• I couldn't help but notice today's Google Doodle - here's the GIF if you'd like to add it to your answer :) – uhoh Dec 6 '16 at 17:30

TL;DR: won't work with any spacecraft further away than one third of the larger semiaxis of the earth's path, even for fully-armed battlestation sized spacecraft.

Radar (more exactly, time-of-flight) is limited by two things:

1. speed of light: The nearest star is 4 light years away. So anything that we send into that direction could, earliest, return in 8 years. Now, this is not really the problem at hand, because
2. free space attenuation is the fact that for any wave front, the power density decreases by the same amount the sphere surface increases with radius. I.e., you get $\frac1{4\pi d^2}$ of the original power density at distance $d$.

Now, the result is the so-called Radar Equation:

$$P_r = {{P_t G_t G_r \sigma \lambda^2}\over{{(4\pi)}^3 R^4}}$$

with

\begin{align} P_r && \text{received signal power}\\ P_t && \text{transmitted signal power}\\ \lambda && \text{wavelength}\\ G_t,\,G_r && \text{directional gain of the transmit, receive antennas}\\ \sigma&&\text{radar cross section, "effective reflection area"}\\ R && \text{the distance between you and the radar target} \end{align}

Let's plug in a few numbers.

First of all, let's assume your radar spacecraft has enough power, and sends 1 MW. It's also got an excellent receiver and lots of signal processing, so that it can even detect a reflected signal at far below thermal noise level at 20°C. Let's say it can work with -180dBm of power – that's $10^{-21}$ W. Pretty much nothing. (in fact, we're getting close to action quantization here)

Then, we come to the following reasoning for our maximum distance $R$:

\begin{align} 10^{-21} \text{ W}&= \frac{10^{6} \text{ W} G_t G_r \lambda^2 \sigma}{{(4\pi)}^3 R^4}\\ 10^{-27} &= \frac{{ G_t G_r\lambda^2 \sigma}}{{(4\pi)}^3 R^4} \end{align}

Let's furthermore assume your spacecraft has something slightly smaller than the Arecibo Observatory (72dBi) as antenna – something with a gain of 60 dBi, and let's also assume you use that for both transmitting and receiving, $G_t=G_r=G$

\begin{align} 10^{-27} &= \frac{{ G_t G_r \lambda^2\sigma}}{{(4\pi)}^3 R^4}\\ &= \frac{{ G^2 \sigma \lambda^2}}{{(4\pi)}^3 R^4}\\ &= \frac{{ 10^{12} \sigma \lambda^2}}{{(4\pi)}^3 R^4}\\ 10^{-39}&= \frac{{ \sigma \lambda^2}}{{(4\pi)}^3 R^4}\\ \end{align}

The question remains: What's a good estimate for the radar cross section of your target? So, we need to pick a target.

I arbitrarily chose the Imperial Death Star. Which is nearly spherical, so we can analytically determine its RCS based on its radius $r$, assuming they have a nice, flat, metal surface freshly polished for the visit of the emperor (first Death Star had a $r=70\text{ km}$

\begin{align} \sigma &= \pi r^2\\ &=\pi {(7\cdot 10^4)}^2\text m^2\\ &\approx 3\cdot 50 \cdot 10^5 \text m^2\\ &= 1.5\cdot 10^7 \text m^2\text{ .} \end{align}

Back to our maximum distance:

\begin{align} 10^{-39}&= \frac{{ 1.5\cdot 10^7 \text m^2 \lambda^2}}{{(4\pi)}^3 R^4}\\ 6.67\cdot10^{-47}&= \frac{{m^2 \lambda^2}}{{(4\pi)}^3 R^4}\\ \end{align}

Let's assume we're doing some 1 GHz as frequency, so we have a wavelength of \begin{align} \lambda &= \frac cf\\ &=\frac{3\cdot 10^8 \frac{\text m}{\text s}}{10^9\frac1{\text s}}\\ &=3\cdot 10^{-1}\text{ m .} \end{align}

Why not a lower frequency, you ask? Simply because the size of an antenna of 60 dBi gain scales linearly with the wavelength. We need to get that antenna to space, so we can't have it being arbitrarily large (and as you've noticed, I'm overly concerned with realism).

It follows that

\begin{align} 6.67\cdot10^{-47}&= \frac{{m^2 \lambda^2}}{{(4\pi)}^3 R^4}\\ &= \frac{{9\cdot 10^{-2} m^4}}{{(4\pi)}^3 R^4}\\ 0.74\cdot10^{-46}\text{ m}^{-4}&= \frac{1}{{(4\pi)}^3 R^4}\\ {(4\pi)}^3 \cdot 0.74\cdot10^{-46}\text{ m}^{-4}&= \frac{1}{ R^4}\\ R^4 &= \frac{1}{{(4\pi)}^3 \cdot 0.74\cdot10^{-46}}\text{ m}^{4}\\ &\approx \frac{1}{2000 \cdot 0.74\cdot10^{-46}}\text{ m}^{4}\\ &\approx \frac{1}{2 \cdot 0.74\cdot10^{-43}}\text{ m}^{4}\\ &\approx \frac{1}{1.5\cdot 10^{-43}}\text{ m}^{4}\\ &= \frac{1}{1.5}10^{43}\text{ m}^{4}\\ &= \frac{2}{3}10^{43}\text{ m}^{4}\\ R&=\sqrt[4]{\frac{2}{3}10^{43}}\text{ m}\\ &=\sqrt[4]{\frac{2}{3}10^{3}}\cdot\sqrt[4]{10^{40}}\text{ m}\\ &=\sqrt[4]{\frac{2}{3}10^{3}}\cdot 10^{10}\text{ m}\\ &\approx 5\cdot 10^{10}\text{ m}\\ &\approx 0.334 \text{ AU .} \end{align}

Since from the formula we see that radius of the target only contributes to maximum range with the square root, to get a max distance of 5 AU, we'd need to increase the radius by a factor of $\left(\frac5{0.334}\right)^2\approx 15^2=225$, ie. that body would need to have a diameter of 31,500 km at least – about one fourth of the diameter of Jupiter!

• Interesting!! At 10 GHz a 60 dBi dish is about 30 meters, which is conceivable as an ultralight foldable space-origomi. Commercial (cheap) GPS receivers have tracking sensitivity of -165 dBm, so your -180 dBm is a little low but not outrageously low, and planets have radar albedo's of a few percent. So this is a real lower-limit estimate but that's fine, it means that it is worth at least the time to do a better estimate. It's not hopeless yet, but it would need to be a giant planet for sure. – uhoh Dec 4 '16 at 16:58
• Why use radar, though, and not e.g. laser/maser beams? Consider the lunar ranging laser reflectors left by several of the Apollo missions, where the return can be observed by a telescope much smaller than 30 meters. – jamesqf Dec 4 '16 at 18:59
• @jamesqf radar=electromagnetic phase coherent waves=laser. It's the same, just that light has much much smaller $\lambda$ and instead of an antenna with directivity gains, you have lenses which focus plane wave fronts in the focal point - exactly the same mechanism; you'd replace the $G$ with the appropriate formula with lens area and aberration. The same path loss due to sphere surface increase applies, just with much worse effect of $\lambda ^2$. – Marcus Müller Dec 4 '16 at 20:25
• @jamesqf the lunar range measurements are reflected from retroreflectors which were placed on the moon during Apollo missions, and which always face Earth. The lasers are large, heavy, and require cooling. The single-photon measurements are accurate only because of substantial averaging over time, while the telescope carefully tracks the movement of the moon. Optical telescope mirrors are heavy and thick to maintain accurate surface, a microwave dish can be thin and unfold and require a thousand times less figure accuracy. – uhoh Dec 4 '16 at 23:54
• Just adding lots of semi-random math to your answer may make it look impressive and get you a vote or two, but doesn't actually answer the question. – Innovine Dec 5 '16 at 11:56

Well, the classical far-distance unit is the parsec – a unit relative to the distance of something that is, perpendicular to the plane in which your own body is moving, so far away that, if you follow a circle with 1AU radius, the angle under which it's observed changes by 2 arcseconds (1 arcsecond relative to the center of that circle and any point on it).

That definition, on the other hand, can be used to estimate the distance of something if you know how large your own ellipse is – and have appropriate angle measurement.