Section 11.6 found an equation for vmax 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? Referring to Section 11.6, you can write an equation for ∆Py, the change of momentum in the vertical direction, in terms of dm and dvy. ∆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.
a. 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.
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?
We need to find out :
a) vmax 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
Now dividing equation1 by dt both sides,
Mass of rocket with fuel, mo = 330,000 kg
Mass of fuel, mf = 280,000 kg
Mass of rocket, mr = 330,000 - 280,000 = 50,000 kg
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.
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at speed . The third piece has twice the mass as the other two. What are the speed and direction of the third piece? Give the direction as an angle east of north
A package of mass m is released from rest at a warehouse loading dock and slides down the -high, frictionless chute of FIGURE EX11.25 to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute.
a. Suppose the packages stick together. What is their common speed after the collision?
b. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
94% of StudySmarter users get better grades.Sign up for free