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Assuming a relativistic rocket travelling at 0.95 times the speed of light (c), what would be the drag force on the cross-section area $(\pi500^2)$ of the ship facing the direction of travel assuming here that drag coefficient is 0.25. The equation for force drag in classical mechanics is: $$ F_D = \frac12\ pv^2C_DA $$ where $p$ is density of interstellar space; which should be about 2 million protons per $m^3$.

However I am not sure if in relativistic mechanics it is as such: $$ F_D = pv^2\gamma^2 $$ where $\gamma$ is the lorentz factor (gamma factor).

Both of these equations will give different results.

Furthermore, what is the kinetic energy when matter within the vacuum of space impacts on cross-section area $(\pi500^2)$ of the ship facing the direction of travel?

The equation is as follows: $$ K_E = (\gamma-1)pc^2 $$

However the results from these equations are about 26.6 N, 0.0028 N and $6.6e^-4$ J respectively. The latter is not much! However in many articles on the web and as per certain experts, the KE should have been greater and as explosive as nuclear bombs?! Where am I wrong?

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  • $\begingroup$ You might find the following useful: https://what-if.xkcd.com/1/. I think the drag force will not be anywhere as simple as just adding a Lorentz factor. $\endgroup$ – honeste_vivere Mar 29 '17 at 13:32
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Calculations of air drag are always an approximation. And the equations you are trying to use are certainly not going to be valid under such extreme situations—huge speed and tiny density. Furthermore, when these protons collide with the ship, they will very likely penetrate into the metal unlike the conditions for which equations like you are using are applicable where the gas molecules just bounce off the object. This will cause radiation damage to the ship's hull. Furthermore, at such high speeds interaction with photons will become important, in particular the cosmic microwave background, causing additional drag.

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