How Does Firing a Photon Torpedo Affect the Speed of a Star-Wars Satellite?

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The discussion focuses on calculating the speed of a Star-Wars satellite after it fires a photon torpedo while orbiting Earth. The satellite has a mass of 5000 kg and initially travels at 10.0 km/s. The photon torpedo contains 10^3^3 photons, each with a wavelength of 200 nm, leading to a calculated momentum change. Using conservation of momentum, the resulting speed of the satellite after the launch is approximately 9337.4 m/s, confirming the calculations align with physics principles. The final speed is rounded to 9300 m/s for significant figures.
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Homework Statement


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?


Homework Equations


Conservation Of Momentum(?)(COM)
K_m_a_x = hf - \varphi
KE = \gamma mc^2 - mc^2
p = E/c
v = f\lambda

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 into 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
 
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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. =)
 
Well i have calculated i out here's 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?
 
What you do seems correct, but the answer should be to 2 sf, so
\delta v=660 and v_sat=9,300.
 
Thanks for the help
 
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