Can a spaceship traveling close to light speed be knocked off course by a gamma ray burst?

Let's first do some math, the first part taken from this pdf regarding (solar) radiation pressure (the formulas should be applicable from any source of electromagnetic radiation).

The intensity $I$ depends on the power $P$ and the distance from the source $r$. We can write the expression $$I=\frac{P}{4 \pi r^2}$$

The force $F$ depends on the intensity and the area $A$ of the object being buffeted. It is $$F=\frac{2I}{c}A$$ and, substituting, we get $$F=\frac{2P}{4 \pi r^2c}A$$ which is $$F=\frac{P}{2 \pi r^2c}A$$ Given the mass of the ship $M$, we find that the acceleration $a$ is $$a=\frac{P}{2 M\pi r^2c}A$$

How much power does a gamma-ray burst emit? From Wikipedia:

Because their energy is strongly focused, the gamma rays emitted by most bursts are expected to miss the Earth and never be detected. When a gamma-ray burst is pointed towards Earth, the focusing of its energy along a relatively narrow beam causes the burst to appear much brighter than it would have been were its energy emitted spherically. When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 10^44 J, or about 1/2000 of a Solar mass energy equivalent"”which is still many times the mass energy equivalent of the Earth (about 5.5 × 1041 J).

We have to divide this number by two to account for the fact that only one of the two beams is hitting the ship - and this is still a little inaccurate, as it assumes the ship is hit by the whole beam. Anyway, gamma-ray bursts, on average, can last from anywhere between less than a second to 30 seconds. Let's say ours lasts for 10 seconds. Because the definition of power is $\frac{E}{t}$, where $E$ is work and $t$ is time, we can say that the power here is $$P=\frac{E}{t}$$ $$P=\frac{5 \times 10^{43}}{10}$$ $$P= 5 \times 10^{42} \text{ Watts}$$ Plugging into our earlier equation for acceleration, we get $$a=\frac{\frac{E}{t}}{2 M\pi r^2c}A$$ $$a=\frac{ 5 \times 10^{42}}{2 M\pi r^2c}A$$ Assuming a mass of the ship similar to the Space Shuttle Orbiter (109,000 kilograms) which is admittedly an unlikely comparison, we make this $$a=\frac{5 \times 10^{42}}{2 \times 1.09 \times 10^5 \pi r^2c}A$$ $$a=\frac{5 \times 10^{37}}{2.18 \pi r^2c}A$$ If you want, you can plug in the area of the underside of the Space Shuttle Orbiter (a stat I can't find, at the moment) and discover that the space shuttle would take quite a hit if it was near a gamma-ray burst.

Note that this is only valid for a ship traveling at a slow speed. At near-light speeds, the relativistic mass would increase (although I don't know if this is valid perpendicular to the direction of its initial motion). This would impact your calculations; I'll try to figure out the corrections later.

Belated Conclusion

The answer is a definitive yes. A ship moving at "normal" speed (i.e. something we could make today - think a successor to the space shuttle) would meet some severe buffeting if it was anywhere near a gamma-ray burst and was hit by one of the beams emitted from the progenitor. If it was hit full-on, it would be severely pushed back; if it was hit partially on, it could be sent spinning. Either way, things wouldn't turn out well.

Your ship, though, is a bit more advanced and is traveling at a speed pretty close to $c$. This means that it is highly unlikely that it would get hit with the beam for an extended period of time if it was anywhere close to the source. If it was further out, the cross-section of the beam would be a lot larger, though (eventually, on the order of hundreds of thousands of miles), and the ship could continue to travel through it for the duration of the burst. The downside is that the energy would be greatly dissipated over the beam's cross-section.

But the answer is yes, the sip would be buffeted if it was reasonably close (i.e. about an AU away, though that's an estimate) to the source, and would most likely be impacted in some way if it was further out.

SJuan76 and Oldcat pointed out that Lorentz contraction would impact the area of the side of the ship receiving radiation pressure if the ship was moving tangentially to the beam. At speeds nearing $c$, this phenomenon would have huge implications for the pressure on the ship. This answer is already math-heavy, so I figure adding a few more equation can't hurt. Haters of algebra beware.

The length of an object due to Lorentz contraction can be found by $$L=L_0 \sqrt{1-v^2/c^2}$$ This means that the area (previously the height times the length, $A=H \times L$) is now written as $$A=H \times L_0 \sqrt{1-v^2/c^2}$$ and so the original equation becomes $$a=\frac{5 \times 10^{37}}{2.18 \pi r^2c} \times H \times L_0 \sqrt{1-v^2/c^2}$$

posted over 9 years ago

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