Section 11.6 found an equation for v_max of a rocket fired in deep space. What is v_max for a rocket fired vertically from the surface of an airless planet with free-fall acceleration g? Referring to Section 11.6, you can write an equation for ∆P_y, the change of momentum in the vertical direction, in terms of dm and dv_y. ∆P_y is no longer zero because now gravity delivers an impulse. Rewrite the momentum equation by including the impulse due to gravity during the time dt during which the mass changes by dm. Pay attention to signs! Your equation will have three differentials, but two are related through the fuel burn rate R. Use this relationship—again pay attention to signs; m is decreasing—to write your equation in terms of dm and dv_y. Then integrate to find a modified expression for v_max at the instant all the fuel has been burned.

a. What is v_max for a vertical launch from an airless planet ? Your answer will be in terms of m_R, the empty rocket mass; m_F0, the initial fuel mass; v_ex, the exhaust speed; R, the fuel burn rate; and g.

b. A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of thrust. What is this rocket’s speed at the instant all the fuel has been burned if it is launched in deep space ? If it is launched vertically from the earth?

Step by step solution

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Step 1. Given Information isMass of Rocket with fuel = 330,000 kgMass of fuel = 280,000 kgTime = 250 sEngine generates 4.1 MN of thrust 

We need to find out :

a) v_max for a vertical launch from an airless planet

b) Rocket’s speed at the instant all the fuel has been burned if it is launched in deep space

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Step 2. Part a) Calculating vmax considering that gravity delivers an impulse

$$\begin{gathered} P_f=P_i+J \\ {m}_{rocket}{v}_{y_{rocket}}+m_{fuel}{v}_{y_{fuel}}=m{v}_y+Fdt \\ (m+dm)(v_y+d{v}_y)+(-dm)(v_y-v_{e}x)=m{v}_y+Fdt \\ m{v}_y+v_{y}dm+md{v}_y+dmd{v}_y-v_{y}dm+v_{ex}dm=m{v}_y-mgdt \\ (dmd{v}_{y}isnegligible) \\ md{v}_y+v_{ex}dm=-mgdt-1. \\ d{v}_y=-v_{ex}\frac{dm}{m}-gdt \\ {\int }_0^{v}d{v}_y=-v_{ex}{\int }_{m_o}^m\frac{dm}{m}-g{\int }_0^{t}dt \\ v=-v_{ex}(\ln m-\ln m_o)-gt \\ v=-v_{ex}\ln \left(\frac{m}{m_o}\right)-gt \\ v=v_{ex}\ln \left(\frac{m_o}{m}\right)-gt \\ {v}_{max}=v_{ex}\ln \left(\frac{m_R+m_{F0}}{m_R}\right)-gt \end{gathered}$$

Now dividing equation1 by dt both sides,

role="math" localid="1647806477903" $$\begin{gathered} m\frac{d{v}_y}{dt}+v_{ex}\frac{dm}{dt}=-mg \\ m{a}_y=-v_{ex}\frac{dm}{dt}-mg \\ \left(\frac{dm}{dt}=R\right) \\ Ristherateatwhichfuelisburnt \\ {F}_{thrust}=v_{ex}R-mg \end{gathered}$$

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Step 3. Part b) Rocket’s speed at the instant all the fuel has been burned if it is launched in deep space 

Mass of rocket with fuel, m_o = 330,000 kg

Mass of fuel, m_f = 280,000 kg

Mass of rocket, m_r = 330,000 - 280,000 = 50,000 kg

$$\begin{gathered} R=\frac{280,000}{250}=1,120kg/s \\ {F}_{thrust}=4.1MN=4.1\times {10}^{6}N \\ For{v}_{ex} \\ 4.1\times {10}^{6}N=v_{ex}(1,120)-(330,000)(9.81) \\ {v}_{ex}=\frac{4.1\times {10}^6+3,237,300}{1,120} \\ {v}_{ex}=6,551.16m/s \\ {v}_{max}=v_{ex}\ln \left(\frac{m_R+m_{F0}}{m_R}\right)-gt \\ {v}_{max}=(6,551.16)\ln \left(\frac{330,000}{50,000}\right)-(9.81)\left(250\right) \\ {v}_{max}=9,910m/s \end{gathered}$$

Rocket’s speed at the instant when all the fuel has been burned if it is launched in deep space is 9,910 m/s.

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