Calculating Rocket Velocity with External Forces

[MathematicalPhysicist]

A rocket ascends from rest in a uniform gravitational field by ejecting exhaust with constant speed u.
Assume that the rate at which mass is expelled is given by dm/dt=ms, where m is is the instantaneous mass of the rocket and s is a cosntant, and that the rocket is retarded by air resistance with a force bv where b is a constant. find the velocity of the rocket as a function of time.
in my text they have proven the case when the total external force is F and it equals: F=mdv/dt+udm/dt, i used this in the calculation, and got:
mg-bv=mdv/dt+udm/dt, but i don't think this correct, your help as always is appreciated.

 

[NotMrX]

A rocket ascends from rest in a uniform gravitational field by ejecting exhaust with constant speed u.
Assume that the rate at which mass is expelled is given by dm/dt=ms, where m is is the instantaneous mass of the rocket and s is a cosntant, and that the rocket is retarded by air resistance with a force bv where b is a constant. find the velocity of the rocket as a function of time.
in my text they have proven the case when the total external force is F and it equals: F=mdv/dt+udm/dt, i used this in the calculation, and got:
mg-bv=mdv/dt+udm/dt, but i don't think this correct, your help as always is appreciated.

Newtons third law applies, you know, the equal but oppisite force one. The force that the mass leaving excerts on the rocket the rocket exerts on the mass. Then we are left with:
$$m*\frac{dv}{dt}=-v\frac{dm}{dt}$$
Since:
$$\frac{dm}{dt}=ms$$
Then:
$$m*\frac{dv}{dt}=-vms$$
But I don't really know what I am doing and something seems wrong.
Maybe:
$$F=ma=m\frac{dv}{dt}-mg-bv=v\frac{dm}{dt}-mg-bv=vms-mg-bv$$

Edit: on the last line on the last equal sign there should be a v by the ms, don't know why I am incompatible with latex sometimes

 

[mezarashi]

It looks like you were on the right track except some of the signs may have been reversed around. There are three forces involved, namely: F_thrust, F_grav, and F_air. Together, they produce a net force in the upward direction. Draw a free-body diagram if you cannot clearly visualize the direction in which these forces are acting.

$$F_{net} = m\frac{dv}{dt} = F_{thrust} - F_{grav} - F_{air}$$

Substituting the values, you will get a linear differential equation you are probably expected to solve. Solving this, you will get v(t). Grab a math textbook, your physics text's appendix, or do a quick google if you don't know how to do differentials.

 

[MathematicalPhysicist]

so mezarashi it should be:
mdv/dt=udm/dt-mg-bv
right?

 

[MathematicalPhysicist]

just one more question, i get this equation:
v(t)=(us-g--bv/m)t and according to the text the teminal velocity is:
(us-g)/b, but i don't see how can i infer that from this equation?

 

[mezarashi]

Your differential equation looks correct, but your solution is clearly wrong. The solution to most differentials of this kind are inverse exponential, i.e. exp(-at). Since there is no (t) in the differential, the solution will most definitely have a 1/exp(t), thus as t approaches infinity, that factor will disappear leaving some other portion.

As I've mentioned, if the mathematics is a problem, you may want to consult a good introductory text that will lead you through step by step. To assist you there, I found a good scan online:

http://college.hmco.com/mathematics/larson/calculus_analytic/7e/shared/downloads/clc7eap1502.pdf

 

[MathematicalPhysicist]

what is wrong in my integration?, I am given that:
mdv/dt=ums-mg-bv
dv/dt=us-g-bv/m
v=dx/dt
dv=(us-g-b/mdx/dt)dt
integrating gives:
$$\int_{0}^{v(t)}dv=\int_{0}^{t}(us-g)dt+\int_{0}^{x(t)}b/mdx$$
what is wrong here?

 

[MathematicalPhysicist]

dv/dt=us-g-dx/dtb/m
d^2x/dt^2=us-g-(b/v)dx/dt
shouldnt you point me to a second order differential equation text?

 

[MathematicalPhysicist]

mezarashi, are you still here?