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A relativistic rocket emits a gas jet with non-relativistic velocity u constant relative to the rocket. Find how the velocity v of the rocket depends on its rest mass m if the initial rest mass of the rocket equals to `m_0`. |
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Answer» The velocity of ejected gases is u relative to the rocket. In an earth centred frame it is `(v-u)/(1-(vu)/(c^2))` in the direction of the rocket. The momentum conservation equation then reads `(m+dm)(v+dv)+(v-u)/(1-(uv)/(c^2))(-dm)=mv` or `mdv-((v-u)/(1-(uv)/(c^2))-v)dm=0` Here `-dm` is the mass of the ejected gases. so `mdv-(-u+(uv^2)/(c^2))/(1-(uv)/(c^2))dm=0`, or `mdv+u(1-v^2/c^2)dm=0` (neglecting `1-(uv)/(c^2)` since u is non-relativistic.) Integrating `(beta=v/c), int (dbeta)/(1-beta^2)+u/cint(dm)/(m)=0, 1n(1+beta)/(1-beta)+u/c1nm=const ant` The constant `=u/c1nm_0` since `beta=0` initially. Thus `(1-beta)/(1+beta)=(m/m_0)^(u//c)` or `beta=(1-(m/m_0)^(u//c))/(1+(m/m_0)^(u//c))` |
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