A small spaceship whose mass is 1.5 × 10^3 kg (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a 10 kW laser beam, what speed will the ship attain in 1.0 day because of the momentum carried away by the beam?

Question

samigupta8 · Answer

We can do it in this way ..
The total energy carried by light beam in a day is P*t 
T in seconds and P in  Watt..
Now this energy carried would be imparted to the ship as its final velocity
So equate it with 1/2 mv^2

michele_laino · Answer

hint:
the momentum transferred, is:
$$\huge P = \frac{E}{c} = \frac{{W\Delta t}}{c}$$
wherein $W,\;E$ are the power and energy respectively of the laser, so the speed $v$ of the astronaut + spaceship, is such that:
$$\huge \frac{{W\Delta t}}{c} = Mv$$
therefore:
$$\huge  v = \frac{{W\Delta t}}{{Mc}}$$
where $\Delta t=24 \times 3600= 86,400$ seconds

michele_laino · Answer

and, of course, $c$ is the speed of light

rsadhvika · Answer

$$v = \dfrac{10000*86400}{1.5*3*10^{11}} = 1.9 mm/s$$

rsadhvika · Answer

Thank you both !

michele_laino · Answer

:)

samigupta8 · Answer

@rsadhvika i were wrong for i didn't see the last line of your question...
The momentum one...

samigupta8 · Answer

And i really don't deserve the medal...so pls..give it to michele sir