Monday, August 15, 2016

How efficient can a rocket engine be?

Let's say we have some mass of propellant m. We can release ALL the energy of this mass as according to mc^2, in some kind of perfect fusion or antimatter reaction. The burned mass is m_b and the unburned mass, into which we dump all the energy from the burned mass as kinetic energy, is m_u.

So
m_u + m_b = m

Relativistic kinetic energy is
KE = (γ-1) m_u c^2

We put all the burned mass's energy into this, giving
KE = m_b c^2.

We have the expression for relativistic momentum,
p = γ m_u v
where v is the exhaust velocity of the fuel.

The Lorentz factor γ is
γ = (1 - (v/c)^2)^-(1/2).

We want to find the maximum momentum for a given amount of fuel.
The above reduces to
p/m = c sqrt(1 - (m_u/m)^2)



So in other words, you get the most momentum out when you convert all of your fuel to pure energy.