# Electromagnetic waves and conservation of momentum. A small spaceship, whose mass, with occupant, is `1.5 * 10^3` kg is drifting in outer space, where no gravitational field exists. If the...

Electromagnetic waves and conservation of momentum.

A small spaceship, whose mass, with occupant, is `1.5 * 10^3` kg is drifting in outer space, where no gravitational field exists. If the astronaut turns on a 10-kW laser beam, what speed would the ship attain in one day because of the reaction force associated with the momentum carried away by the beam?

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### 1 Answer

To answer this question, you have to be familiar with units and the equations that relate different values to each other.

Let's start with the equation that relates the energy of a photon (or set of photons) with momentum, considering the equation for relativistic energy (`E^2 = p^2c^2 + m^2c^4`) and the fact that for a photon, `m=0`:

`E = pc`

`E/c = p`

Here, `E` is the energy of the photon, `c` is the speed of light (`3*10^8` m/s), and `p` is the momentum of the photon.

Recognizing that we have units of power (kW), we need to tweak the equation a bit. Recognizing that c is invariate with respect to time, we can change the above equation to the following:

`P/c = (dp)/(dt)`

Here, P is Power and `(dp)/(dt)` is the time derivative of momentum.

Now, we can start plugging in values:

`(10000 J/s)/(3*10^8 m/s) = 3.33*10^-5 N = (dp)/(dt)`

We now know the constant rate of change of momentum produced by the laser. For every second the laser stays "on," the momentum changes by `3.33*10^-5 kg *m/s` .

Now, we need to multiply this value by the number of seconds in one day to determine the total change in momentum:

`Deltap = 3.33*10^-5 kg*m/s^2 * 3600 s/h * 24 h/(day) = 2.88 kg*m/s`

We must now consider that because we started at rest (`p_0 = 0`, in the reference frame of the spaceship before firing the laser), the above result is the total momentum of the spaceship. Now, because we don't seem to be moving too fast, we can use Newtonian mechanics to solve the rest of the problem.

We start with the following relationship between momentum, mass, and velocity:

`p = mv`

Now, we just fill in the momentum and mass to find the velocity:

`2.8 = 1500v`

`2.8/1500 = v`

Thus, our **final velocity** is `1.92 * 10^-3` m/s.

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