A rocket is fired from rest into an outer space where the gravity can be neglect

Question

A rocket is fired from rest into an outer space where the gravity can be
neglected. What is the ratio of the total mass of the rocket when the speed
of the rocket reaches 80% of the exhaust speed with respect to the rocket?

My teachers answer was:

v = -v exhaust ln(m final/m initial) --> m final/m initial = exp(-v/ v exhaust) = exp(-0.80) = 0.45 = 45%

But I have no clue how he got there. What is exp? and why does it make .80 equal .45. So if someone could please explain, that would be awesome!

Explanation / Answer

this problem is of reduced mass concept

where

d/dt(MV)=dM/dt(V-v)

V=V0 + v loge(1-ct)-gt

if gravity is neglected then g=0

V=V0 + v loge(1-ct)

if initialy rocket was in rest V0=0

V/v =  loge(1-ct)

1-ct=exp(V/v)

1-ct= ratio of masses of the rocket=M/M0

M/M0=exp(-0.80)

M/M0=0.449

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