Question

A small single-stage rocket is launched vertically as shown in Figure 1.3.19. Once launched, the rocket consumes its fuel, and so its total mass m(t) varies with time t > 0. If it is assumed that the positive direction is upward, air resistance is proportional to the instantaneous velocity v of the rocket, and R is the upward thrust or force generated by the propulsion system, then construct a mathematical model for the velocity v(t) of the rocket. [Hint: See (14) in Section 1.3.]

Figure 1.3.19 Single-stage rocket in problem 21

Short answer

The differential equation is $v\times \frac{dm}{dt}+m\times \frac{dv}{dt}=-mg-kv+R$

Step-by-step solution

Step 1:

Given, mass of rocket$=m\left(t\right)$

Mass varies with time$t>0$

Upward direction is positive

Upward thrust$=R$

Velocity of the rocket$=v\left(t\right)$

Air resistance role="math" localid="1667978107945" $\alpha$ instantaneous velocity(v) of rocket

Step 2:

As the rocket climbs in positive direction, it spends amount of fuel i.e., mass of fuel changes with time.

The air resistance work in opposite direction of the motion.

So, air resistance $=-kv$

and R works in same direction.

Using Newton’s law, we get

$F=ma=\frac{d}{dt}\left(mv\right)$

$\frac{d}{dt}\left(mv\right)=$weight of rocket in downward direction+ air resistance+ upward thrust

$\frac{d}{dt}\left(mv\right)=-mg-kv+R-------\left(1\right)$

Step 3:

Now mass is variable, we use product rule to the left side of the equation (1),

we get,$v\times \frac{dm}{dt}+m\times \frac{dv}{dt}=-mg-kv+R$