# Problem of photon torpedo

1. Jan 29, 2007

### Delzac

1. The problem statement, all variables and given/known data
A star-wars laser satellite, whose mass is 5000 kg, is orbiting Earth at a speed of 10.0 km/s. It launches a photon torpedo at an enemy missile which is directly in front of it. The photon torpedo contains $$10^3^3$$ photons, each having a wavelength of 200nm. What is the speed of the weapons satellite just after the launch?

2. Relevant equations
Conservation Of Momentum(?)(COM)
$$K_m_a_x = hf - \varphi$$
$$KE = \gamma mc^2 - mc^2$$
$$p = E/c$$
$$v = f\lambda$$
3. The attempt at a solution

Since we have $$\lambda = 200nm$$ we can then obtain f = 1.5*10^15 Hz.

However, from here on, i am lost. How do we use the number of photons and equate it in to $$p = E/c$$ and finally COM. And then there is also a problem of E which i am unable to find since i don't have h, $$K_m_a_x = hf - \varphi$$.

Any help will be appreciated. Thanks

2. Jan 29, 2007

### koofle

This feels like the typical momentum problem involving light, but using sci-fi scale numbers. In these cases, there are a few things to take into consideration:
-conservation of momentum
-conservation of energy

As for the relevant equations, there may have been some confusion on what $$K_m_a_x = hf - \varphi$$, the photoelectric work function, is used for.
There is also another equation that can be used to find the energy of the photon, not involving $$p$$. And h is Plank's constant. Take another look at the possibilities. =)

3. Feb 2, 2007

### Delzac

Well i have calculated i out heres what i have obtain :

$$p = \frac{h}{\lambda} * 10^3^3$$ Since no. of photon is $$10^3^3$$

Then using Conservation of momentum, i get :

$$mv = p_p_h_o_t_o_n$$
Which works v out to about 662.6 m/s
The speed of satellite is therefore 10,000 - 662.6 = 9337.4 m/s

Can anyone confirm my result?

4. Feb 2, 2007

### Meir Achuz

What you do seems correct, but the answer should be to 2 sf, so
\delta v=660 and v_sat=9,300.

5. Feb 2, 2007

### Delzac

Thanks for the help

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