3
$\begingroup$

If I’m travelling in space near the speed of light, I would like to use radar to give me advance warning of approaching hazards. However, will I have enough time to be able to react?

From a perspective on the spaceship, the radar beam travels away from me at the speed of light, so I might imagine that I would have ample warning of an object positioned at rest one light-hour away. However, observed from a position of rest, the radar beam is just slowly pulling away ahead of the spaceship, and the radar pulse would arrive at the obstruction only shortly before the spaceship.

I suspect therefore that I would not have enough time to react, but can someone shed some light on this for me please?

$\endgroup$
  • 6
    $\begingroup$ think about it this way: your resting spaceship is subject to a debris storm with near-speed-of-light velocity. $\endgroup$ – Bort Jul 29 '15 at 11:51
  • $\begingroup$ So I would not have enough notice, because the returning pulse would arrive shortly before the debris, but would "claim" that the debris was 30 minutes out? $\endgroup$ – Richard Corfield Jul 29 '15 at 11:54
  • 2
    $\begingroup$ The distance to the object of interest is different in the two frames. In your frame, the object is much closer than in the global frame, so although your radar signal travels at the speed of light in your frame as well, the distance to the object may be too small for you to react. $\endgroup$ – pela Jul 29 '15 at 12:11
3
$\begingroup$

From a perspective on the spaceship, the radar beam travels away from me at the speed of light, so I might imagine that I would have ample warning of an object positioned at rest one light-hour away.

One light-hour away in which frame? If that one light-hour away is from the perspective of the rest frame of that object, you don't have much time at all. From your perspective, that one light-hour is length contracted to 508 light-seconds. You have five seconds to react before the object hits you at 0.99 c.

You also need to be able to know if the object is going to hit the spaceship and, if it is, which way the spaceship should move to avoid it. You need to assess the trajectory of the object. You aren't going to get this with one ping. This takes multiple pings, and time. Suppose it takes 20 seconds total to determine the trajectory and get out of the way. You need to multiply 20 seconds by 99 because the object is coming toward you at 0.99 c. That's 2000 light seconds, more or less, or about four astronomical units.

Note well:

  • That 20 seconds is a very generous supposition. It currently takes hours of sightings to develop a reasonably accurate estimate of the trajectory of a hitherto unknown near-Earth object, and that's using multiple radars spaced continents apart.
  • A one centimeter radius rock with a density of 3.6 grams/cc hitting you at 0.99 c is equivalent to 100 Fat Man nuclear bombs, all going off at once.
  • You are looking for objects smaller than a centimeter across, 4 AUs away.
  • That 4 AUs is in your frame. In the rest frame of the objects that you are about to hit, that's a distance of about 28 AUs, or roughly the distance between the Sun and Neptune.
  • There are lots of sub-centimeter sized objects between Neptune and the Sun.
$\endgroup$
4
$\begingroup$

Per Bort's comment, this is easier to think about from the spaceship's frame rather than the debris's frame. A rock is traveling toward your spaceship at .99 times the speed of light. You send out a pulse of light that intercepts the rock when it's one light-minute away from the ship. The pulse bounces back and arrives at the ship 1 minute later. The rock arrives another $1/.99-1\approx .01$ minutes after that. That's how much warning you get.

$\endgroup$
  • $\begingroup$ Ok, this might be a question about how radar works, but would you know that you had 0.01 minute (by knowing the approaching speed by doppler shift?) or would you think you had the full minute to react? $\endgroup$ – Richard Corfield Jul 29 '15 at 15:22
  • $\begingroup$ @rdc: If you know the speed of the rocks, you know how long you have to react. If not, not. $\endgroup$ – WillO Jul 29 '15 at 16:55
  • $\begingroup$ Doppler will only give you range rate (radial component of velocity). You also have to assess the transverse component to determine if the rock is going to hit you. Moving the spacecraft into the path of a rock moving at 0.99 c is a bad idea. $\endgroup$ – David Hammen Jul 29 '15 at 17:36
  • $\begingroup$ @DavidHammen I think we're fundamentally assuming here that the rocks were "stationary" when the spaceship was at speed 0. In other words, the only real velocity the rocks have is relative to the spaceship. $\endgroup$ – barrycarter Mar 17 '16 at 3:28
  • $\begingroup$ @barrycarter -- There is no such thing as absolute velocity. Everything is relative. From the perspective of the spacecraft, the spacecraft isn't moving. From this perspective, it's the rocks that are moving, and doing so at 0.99 c. $\endgroup$ – David Hammen Mar 17 '16 at 12:51
0
$\begingroup$

As stated in the comments:

Even though the radar signal may come back at the speed of light at any frame, it would require an enormous distance for you to react to any resting object. If your computer could re-calculate the routes in $\frac{1}{c} \textrm{ s} $ (s.i system) it would take you at least $1\textrm{ m}$ to have enough space to manouver, since the best computer operates with speeds in the order of $10^{-3}\textrm{s}$ (being optimistic) you would need at least about $10^5 \textrm{ m}$ to manouver. If you consider that in your frame there is the space contraction, your situation would be even worse.

$\endgroup$
0
$\begingroup$

You can easily understand this if you read about relativity of simultaneous events. Events which occur to be simultaneous in one frame of reference are not simultaneous in any other frame of reference.

You can not calculate the time left for you to react based on your intuitions. You should carry out the calculations strictly using the relativistic equations of motion. Mark the events as event A, event B and so on along with their co ordinates including time as one of the coordinate. Choose a particular reference frame and strictly stick to it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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