You know the equation v max= v e x ln( m R+ m F0 m R) for the maximum speed of a

Question

You know the equation vmax=vexln(mR+mF0mR) for the maximum speed of a rocket fired in deep space. What is vmax for a rocket fired vertically from the surface of an airless planet with free-fall acceleration g? Assume that free-fall acceleration g is the same at every location of the rocket. You can write an equation for Py, the change of momentum in the vertical direction, in terms of dm and vy. Py 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 dvy. Then integrate to find a modified expression for vmax at the instant all the fuel has been burned.

Vmax= vexln(mR+mF0/mR)g/R*mF0

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?

Explanation / Answer

for a rockjet launched form the planet ingnoring acceleration due to gravity
Vmax = Vex*ln((mRo + mFo)/mRo)

now, intiial KE of rocket = 0
final KE = 0.5(mRo)Vmax^2

so for the rocket fired from a planet with acceleration due to gravity g ( earth)
g = GM/R^2 ( where M is mass of earth and R is radius of earth)
also, initial PE = -GM(mRo + mFo)/R = Rg(mRo + mFo)
so final KE = 0.5(mRo)Vmax^2 - Rg(mRo + mFo) = 0.5(mRo)u^2 ( where u is the max speed in this case)
u = sqroot((Vex*ln((mRo + mFo)/(mRo)))^2 - 2Rg(mRo + mFo)/(mRo))

also for engine thrust T
T = mFo *vex/ t
t = 250 s
T = 4.1*10^6
mRo = 50,000 kg
mFo = 280,000 kg
so Vex = 3660.71428 m/s
R = radius of earth = 6371,000 m
hence
Vmax = 27879.6117 m/s

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