Question #186795

The star of a distant solar system explodes as a supernova .At the moment of the explosion ,an resting exploration spaceship is AU away from the shock wave . The shock wave of the explosion spaceship travels with 25000km/s towards the spaceship .To save the crew , the spacecraft makes use of a special booster that uniformly accelerates at 150m/s^2 in the opposite direction .

Determine if the crew manages to escape from the shock wave (neglect relativistic effects)


1
Expert's answer
2021-04-29T10:41:49-0400

Let the x-axis be directed from star to the starship. At the initial moment the coordinate of the ship is 1AU=150 mln km=1.51011m1\,\text{AU}= 150\text{ mln\, km}= 1.5\cdot10^{11}\,\mathrm{m} and this is the initial distance between the shock and the ship. After the beginning of the acceleration the coordinate of the ship after time t will be x=1.51011m+at2/2=1.51011m+75m/s2t2.x=1.5\cdot10^{11}\,\mathrm{m} + a t^2/2 = 1.5\cdot10^{11}\,\mathrm{m} + 75\,\mathrm{m/s^2}\cdot t^2.

The coordinate of the shock wave is xw=25106m/stx_w = 25\cdot10^6\,\mathrm{m/s}\cdot t . Let us determine if there is a moment t at which the shock wave reaches the starship: x=xw.x=x_w.

1.51011m+75m/s2t225106m/st=0.1.5\cdot10^{11}\,\mathrm{m} + 75\,\mathrm{m/s^2}\cdot t^2 - 25\cdot10^6\,\mathrm{m/s}\cdot t = 0.

We calculate the discriminant: D=(25106)24751.51011=5.81014>0.D=(25\cdot10^6)^2 - 4\cdot75\cdot 1.5\cdot10^{11} = 5.8\cdot10^{14} > 0.

The roots of the equation are

t1,2=25106±5.81014275,    t1=6.1103s,  t2=3.3105s.t_{1,2} = \dfrac{25\cdot10^6 \pm \sqrt{5.8\cdot10^{14}}}{2\cdot75}, \;\; t_1 =6.1\cdot10^3\,\mathrm{s} ,\; t_2= 3.3\cdot10^5\,\mathrm{s}.

Therefore, after approximately 6000 seconds the shock wave will reach the ship.


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