6
$\begingroup$

I thought I read somewhere that everything we've ever sent into space used only Newton for navigation, because relativistic effects are orders of magnitude smaller than other causes of course deviation.

But now I can't find it. Now all I can find is articles about probes testing various aspects of relativity--but nothing about whether relativity was necessary for navigation, beyond the needs of the experiment.

Do we need relativity for navigating the solar system?

$\endgroup$
3
  • $\begingroup$ Thanks; that's about navigation of non-space vehicles on the surface of the earth. I'm wondering about the spacecraft's navigational needs in space. Do you have any thoughts on how I can make my question clearer to reflect that? $\endgroup$ Jul 2 '18 at 13:53
  • 1
    $\begingroup$ Related, but not a duplicate: Could we send a man safely to the Moon in a rocket without knowledge of general relativity? $\endgroup$ Jul 2 '18 at 13:54
  • $\begingroup$ @DavidHammen Yeah, I read that one, hoping it would give me a starting point, but no luck. Cheers $\endgroup$ Jul 2 '18 at 13:55
9
$\begingroup$

Keep in mind that interplanetary probes don't know where they are in space until they get close to the target object (and sometimes not even then). The onboard computer of a vehicle about to land on Mars doesn't need to consider third body perturbations, let alone relativistic effects. For example, the Mars Science Laboratory Entry, Descent, and Landing flight software modeled gravity using a simple J2 (equatorial bulge) gravity model of Mars. This simple model was the largest contributor to the error in the landed position; Mars exhibits significant gravitational anomalies. The errors that resulted from this simplistic model of Mars gravity field are orders of magnitude than are the errors that resulted from ignoring the Newtonian gravitational effects of the Sun and Jupiter, which in turn are orders of magnitude larger than the errors that resulted from ignoring the relativistic effects.

Navigation of a spacecraft in interplanetary space is performed by people and computers on the Earth rather than by the spacecraft. Earth-bound computers don't have the huge constraints imposed on onboard computers, so they can model relativistic effects. Those relativistic effects are important for a spacecraft that takes many months to many years to get from Earth to the target. As a result, the Deep Space Network uses a Parameterized Post-Newtonian (PPN) formulation to calculate the effects of gravitation on spacecraft; see section 4 of Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation.

$\endgroup$
4
  • $\begingroup$ Wow, it's more complex than I thought. Can you clarify for me: whoever said that we don't need relativity for navigating the solar system was just plain wrong to say so? $\endgroup$ Jul 2 '18 at 17:35
  • 1
    $\begingroup$ @GreatBigBore - It depends on where the spacecraft is going, on how accurate the planetary ephemeris and spacecraft trajectory needs to be, and how much fuel one is willing to waste on course corrections. The 43 arc second per century relativistic precession of Mercury means a non-relativistic ephemeris degrades by 120 km per year. That's not good if the goal is to hit a 2 km window at Mercury, which was the goal for Mercury MESSENGER. $\endgroup$ Jul 2 '18 at 18:02
  • $\begingroup$ Thanks again. I hate to pester, but I don't suppose there's a document somewhere I could read to find out which missions have used relativity and which haven't? $\endgroup$ Jul 2 '18 at 18:07
  • $\begingroup$ @GreatBigBore - All NASA missions that use the Deep Space Network for navigation implicitly use a relativistic gravity model because that's what the DSN uses, and has used since the 1960s. $\endgroup$ Jul 2 '18 at 18:57
0
$\begingroup$

The impact of general relativity on an orbiting spacecraft can be seen according to modifications of Newtonian mechanics. It is not that hard to compute orbits in the Schwarzschild spacetime without pounding out Christoffel connection terms. The invariant interval $ds^2~=~g_{\mu\nu}dx^\mu dx^\nu$ has zero variation $\delta\int ds~=~0$. We can then see the action is equivalent to the proper time, and there is a Lagrangian ${\cal L}~=~ds/dt$ which has zero variation.

For the Schwarzschild metric $$ ds^2~=~c^2\left(1~-~\frac{2m}{r}\right)dt^2~-~\left(1~-~\frac{2m}{r}\right)^{-1}dr^2~-~r^2d\Omega^2, $$ for $m~=~GM/c^2$. We assume there is a weak field so $(1~-~\frac{2m}{r})^{-1}~\simeq$ $1~+~\frac{2m}{r}$ and we place the dynamics in a single plane so $$ ds^2~=~c^2\left(1~-~\frac{2m}{r}\right)dt^2~-~\left(1~+~\frac{2m}{r}\right)dr^2~-~r^2d\phi^2. $$ Now get the Lagrangian with $$ \left(\frac{ds}{dt}\right)^2~=~c^2\left(1~-~\frac{2m}{r}\right)~-~\left(1~+~\frac{2m}{r}\right)\left(\frac{dr}{dt}\right)^2~-~r^2\left(\frac{d\phi}{dt}^2\right)^2. $$ While this is the square of the Lagrangian, since the left hand side of the Euler-Lagrange equation has two derivatives there will be a term with $O(c^-1)$ from the derivative of the square root. This will be associated with a “square of the Lagrangian” in the denominator. This is the reason the $c^2$ term is kept on the $dt^2$ part of the line element. We drop this “square of the Lagrangian” in a semi-relativistic case. This means we can just use $\left(\frac{ds}{dt}\right)^2$ as the Lagrangian to $O(c^{-1})$ error. We then have the following differential equations $$ \frac{\partial}{\partial t}\left(\frac{\partial{\cal L}}{\partial\dot r}\right)~-~\left(\frac{\partial{\cal L}}{\partial r}\right)~=~0~\rightarrow~\ddot r~\simeq~\left(1~-~\frac{2GM}{rc^2}\right)\left[\left(1~-~\frac{v^2}{c^2}\right)\frac{GM}{r^2}~+~\frac{1}{2}r\dot\phi\right] $$ $$ \frac{\partial}{\partial t}\left(\frac{\partial{\cal L}}{\partial\dot\phi}\right)~-~\left(\frac{\partial{\cal L}}{\partial\phi}\right)~=~0~\rightarrow~r^2\ddot\phi~+~-2r\dot r\dot\phi~=~0 $$ These are the two relevant equations of motion

The first of these may be put in standard form with the angular momentum $L^2~=~m^2r^4\dot\phi^2$ so $$ \ddot r~=~\left(1~-~\frac{2GM}{rc^2}\right)\left[\left(1~-~\frac{v^2}{c^2}\right)\frac{GM}{r^2}~+~\frac{L^2}{2mr^3}\right]. $$ Now with the elimination of the $O(c^{-2})$ terms we just have the standard Newtonian result for the force of gravitation. We then see there is the modification by the Lorentz factor $1~-~\frac{v^2}{c^2}$ and the Schwarzschild metric term $1~-~\frac{2GM}{rc^2}$.

For a spacecraft in Earth orbit with $v~=~\sqrt{GM/r}$ that $v/c~\simeq~2.3\times 10^{-5}$ and the relativistic correction terms are going to be on the order of $(v/c)^2~=~5\times 10^{-10}$. This appears small, until one considers navigation. This error is largely going to influence the relative rate that clocks tick. If in one second there is this error in timing of $5\times 10^{-10}sec$ this corresponds to a drift in distance error of $d~=~ct$ $=~16$cm. This suddenly becomes a considerable problem. For GPS tracking this is something of importance and knowledge of the gravitational field is a part of the geodesy of satellites.

$\endgroup$
0
$\begingroup$

It is even closer at home than you think. You need it to navigate on earth using satellite data.

Do you use GPS?

More sophisticated techniques, like Differential GPS (DGPS) and Real-Time Kinematic (RTK) methods, deliver centimeter-level positions with a few minutes of measurement. Such methods allow use of GPS and related satellite navigation system data to be used for high-precision surveying, autonomous driving, and other applications requiring greater real-time position accuracy than can be achieved with standard GPS receivers.

To achieve this level of precision, the clock ticks from the GPS satellites must be known to an accuracy of 20-30 nanoseconds. However, because the satellites are constantly moving relative to observers on the Earth, effects predicted by the Special and General theories of Relativity must be taken into account to achieve the desired 20-30 nanosecond accuracy.

.....

The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time.

$\endgroup$
2
  • $\begingroup$ The OP explicitly excluded GPS. He wants to know about the effects of general relativity on satellite navigation as opposed to the the effects of general relativity on a satellite's clock. These are different things. $\endgroup$ Jul 4 '18 at 22:14
  • $\begingroup$ @DavidHammen I am sorry but I do not see the exclusion . The same mathematics is used in the navigation of the satellites. The GPS gives a relevant example in numbers. $\endgroup$
    – anna v
    Jul 5 '18 at 3:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.