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Rocket Propulsion is based on

Newton’s third law & Law of conservation of linear momentum

Rocket Propulsion is based on Newton’s third law and Law of Conservation of Linear Momentum. It is an example of Variable Mass System.

Let

Rate of ejection of gases from the rocket (r) = $\displaystyle \frac{dm}{dt}$

m₀ = initial mass of rocket (i.e., mass at t = 0)

m = (m₀ – rt) = mass of rocket at any time t

u = velocity of ejected gases w.r.t. rocket

Now consider the positions of rocket at two points P and Q. At position P, mass of rocket is m and its velocity is v and after time Δt its position is Q, at which the mass of rocket decreases to (m – Δm) and it’s velocity increases to (v + Δv). Here Δm is the mass of burnt gases which are ejected in time Δt and moving at a velocity “u” with respect to rocket.

Since Rocket Propulsion is based on Law of Conservation of Linear Momentum, so let us apply this law between points P and Q. To apply conservation of linear momentum we need velocities with resect to earth.

Linear momentum of mass m at time t = linear momentum of mass (m – Δm) & Δm at time t+Δt

m × v_{rocket w.r.t. earth at time t}=[ (m-Δm) × v_{rocket w.r.t. earth at time t+Δt} ] + Δm × v_{gases w.r.t. earth …..(1)}

Now we know that

V_AB = V_A – V_B

∴ v_{gases w.r.t. rocket} = v_{gases w.r.t. earth}– v_{rocket w.r.t. earth}

or v_gr = v_ge– v_re

⇒ v_ge= v_gr+ v_re

Here v_gr = – u (because the ejected gases seems to be going downward w.r.t. rocket) &

v_re = + v (because the rocket seems to be going upward w.r.t. earth)

∴ v_ge= – u + v

⇒ v_ge= (v – u)

Its direction is in the direction of motion of rocket.

Now using equation (1)

mv = (m – Δm)×(v + Δv) + Δm×(v – u)

mv = mv + mΔv – (Δm)v – (Δm)(Δv) + (Δm)v – (Δm)u

⇒ mΔv – (Δm)(Δv) = (Δm)u

⇒ Δv(m – Δm ) = (Δm)u

$\displaystyle \Rightarrow \Delta v=\frac{(\Delta m)u}{m-\Delta m}$

$\displaystyle \Rightarrow \frac{\Delta v}{\Delta t}=\frac{\Delta m}{\Delta t}\frac{u}{m-\Delta m}$

Here Δm = rΔt

$\displaystyle \therefore \frac{\Delta v}{\Delta t}=\frac{\Delta m}{\Delta t}\frac{u}{m-r\Delta t}$

Taking limit, Δt →0 both sides,

$\displaystyle \underset{\Delta t\to 0}{\mathop{Lim}}\,\frac{\Delta v}{\Delta t}=\underset{\Delta t\to 0}{\mathop{Lim}}\,\left[ \frac{\Delta m}{\Delta t}\frac{u}{m-r\Delta t} \right]$

$\displaystyle \Rightarrow \frac{dv}{dt}=\frac{dm}{dt}\frac{u}{m}$

$\displaystyle \Rightarrow \frac{dv}{dt}=\frac{ru}{m}$

$\displaystyle \Rightarrow \frac{dv}{dt}=\frac{ru}{{{m}_{0}}-rt}$

$\displaystyle \Rightarrow a=\frac{dv}{dt}=\frac{ru}{{{m}_{0}}-rt}$ _…..(2)

Equation (7) gives acceleration of the rocket. From equation (7) it is clear that with increase in time, acceleration also increases.

Now net force acting on the rocket in the upward direction,

F_net = ma – mg

$\displaystyle \Rightarrow m\frac{dv}{dt}=m\left( \frac{ru}{{{m}_{0}}-rt} \right)-mg$

$\displaystyle \Rightarrow \frac{dv}{dt}=\left( \frac{ru}{{{m}_{0}}-rt} \right)-g$

$\displaystyle \Rightarrow dv=\left( \frac{ru}{{{m}_{0}}-rt} \right)dt-gdt$

Integrating above expression,

$\displaystyle \int\limits_{{{v}_{i}}}^{v}{dv}=\int\limits_{0}^{t}{\left( \frac{ru}{{{m}_{0}}-rt} \right)dt}-\int\limits_{0}^{t}{gdt}$

$\displaystyle \int\limits_{{{v}_{i}}}^{v}{dv}=ru\int\limits_{0}^{t}{\left( \frac{1}{{{m}_{0}}-rt} \right)dt}-g\int\limits_{0}^{t}{dt}$

$\displaystyle \left[ v \right]_{{{v}_{i}}}^{v}=ru\left[ {{\log }_{e}}({{m}_{0}}-rt)\left( -\frac{1}{r} \right) \right]_{0}^{t}-g\left[ t \right]_{0}^{t}$

$\displaystyle v-{{v}_{i}}=u\left[ {{\log }_{e}}\left( \frac{{{m}_{0}}}{{{m}_{0}}-rt} \right) \right]-gt$

$\displaystyle v=\left( {{v}_{i}}-gt \right)+u{{\log }_{e}}\left( \frac{{{m}_{0}}}{{{m}_{0}}-rt} \right)$ _…..(3)

Here v_i is initial velocity of rocket.

Equation (8) gives velocity of rocket at any time t. We note that as time ‘t’ increases, the velocity of rocket ‘v’ also increases.

# If initial velocity v_i is zero, then

$\displaystyle v=u{{\log }_{e}}\left( \frac{{{m}_{0}}}{{{m}_{0}}-rt} \right)-gt$ _…..(4)

# If the effect of gravity is neglected(g = 0) then,

$\displaystyle v=u{{\log }_{e}}\left( \frac{{{m}_{0}}}{{{m}_{0}}-rt} \right)$ _…..(5)

Rocket Propulsion is based on Newton’s third law & Law of conservation of linear momentum

Rocket Propulsion is based on

Newton’s third law & Law of conservation of linear momentum

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