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 ΔPy, the change of momentum in the vertical direction, in terms of dm and vy. ΔPy 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.What is vmax for a vertical launch from an airless planet? Your answer will be in terms of mR, the empty rocket mass; mF0, the initial fuel mass; vex, the exhaust speed; R, the fuel burn rate; and g Express your answer in terms of mR, mF0, vex, R, and g.
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 ΔPy, the change of momentum in the vertical direction, in terms of dm and vy. ΔPy 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.What is vmax for a vertical launch from an airless planet? Your answer will be in terms of mR, the empty rocket mass; mF0, the initial fuel mass; vex, the exhaust speed; R, the fuel burn rate; and g Express your answer in terms of mR, mF0, vex, R, and g.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter8: Momentum And Collisions
Section: Chapter Questions
Problem 46P: A rocket has total mass Mi = 360 kg, including Mfuel = 330 kg of fuel and oxidizer. In interstellar...
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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 ΔPy, the change of momentum in the vertical direction, in terms of dm and vy. ΔPy 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.What is vmax for a vertical launch from an airless planet? Your answer will be in terms of mR, the empty rocket mass; mF0, the initial fuel mass; vex, the exhaust speed; R, the fuel burn rate; and g
Express your answer in terms of mR, mF0, vex, R, and g.
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