Rocket subject to linear resistive force -- two methods

[TaylorLL]

Homework Statement

Consider a rocket subject to a linear resistive force, $$f = -bv$$, but no other external forces. Use Equation (3.29) in Problem 3.11 to show that if the rocket starts from rest and ejects mass at a constant rate $$k = -\dot{m}$$, then its speed is given by:
$$v = \frac{k}{b}v_{ex}\left[1-\left(\frac{m}{m_{o}}\right)^{b/k}\right]$$

Homework Equations

Equation (3.29) $$m\dot{v} = -\dot{m}v_{ex} + F^{ext}$$

The Attempt at a Solution

[/B]
So I had no issues solving this equation when I used the chain rule and took time out of the equation. Basically, since $$\dot{m} = -k$$, we can plug that into equation 3.29, and then we can find through chain rule that $$\dot{v} = \frac{dv}{dm}\frac{dm}{dt} = -k\frac{dv}{dm}$$. This leads to the expected answer. However, it should also work by finding an equation for mass as a function of time and integrating with respect to time. This is where I'm having issues.

$$m(t) = m_{o} - kt$$
$$\left(m_{o}-kt\right)\frac{dv}{dt} = kv_{ex} - bv$$
$$\int_{0}^{t}\frac{dt'}{m_{o}-kt'} = \int_{0}^{v}\frac{dv'}{kv_{ex} - bv'}$$

When I go through this integral, I almost get the solution with the exception of a flipped exponential term.

$$v = \frac{k}{b}v_{ex}\left[1-\left(\frac{m}{m_{o}}\right)^{k/b}\right]$$

I would appreciate any help, thanks!

 

[TSny]

$$\int_{0}^{t}\frac{dt'}{m_{o}-kt'} = \int_{0}^{v}\frac{dv'}{kv_{ex} - bv'}$$
When I go through this integral, I almost get the solution with the exception of a flipped exponential term.

I don't end up with the exponent flipped. If you want to show your steps of integration, we can check it.

 

[TaylorLL]

I don't end up with the exponent flipped. If you want to show your steps of integration, we can check it.

Okay, so here's my integration:
$$\int_{0}^{t} \frac{dt'}{m_{o} - kt'} = \int_{0}^{v}\frac{dv'}{kv_{ex}-bv'}$$
$$-kln\left(m_{o} - kt'\right)|_{0}^{t} = -bln\left(kv_{ex}-bv'\right)|_{0}^{v}$$
$$-kln\left(\frac{m}{m_{o}}\right) = -bln\left(1 - \frac{bv}{kv_{ex}}\right)$$
$$\frac{k}{b}ln\left(\frac{m}{m_{o}}\right) = ln\left(1 - \frac{bv}{kv_{ex}}\right)$$
$$ln\left(\frac{m}{m_{o}}\right)^{k/b} = ln\left(1 - \frac{bv}{kv_{ex}}\right)$$
$$\left(\frac{m}{m_{o}}\right)^{k/b} = 1 - \frac{bv}{kv_{ex}}$$
$$v = \frac{kv_{ex}}{b}\left[1 - \left(\frac{m}{m_{o}}\right)^{k/b}\right]$$

As you can see, I end up with an exponential term of $$k/b$$ rather than $$b/k$$

 

[TaylorLL]

Okay, so here's my integration:
$$\int_{0}^{t} \frac{dt'}{m_{o} - kt'} = \int_{0}^{v}\frac{dv'}{kv_{ex}-bv'}$$
$$-kln\left(m_{o} - kt'\right)|_{0}^{t} = -bln\left(kv_{ex}-bv'\right)|_{0}^{v}$$
$$-kln\left(\frac{m}{m_{o}}\right) = -bln\left(1 - \frac{bv}{kv_{ex}}\right)$$
$$\frac{k}{b}ln\left(\frac{m}{m_{o}}\right) = ln\left(1 - \frac{bv}{kv_{ex}}\right)$$
$$ln\left(\frac{m}{m_{o}}\right)^{k/b} = ln\left(1 - \frac{bv}{kv_{ex}}\right)$$
$$\left(\frac{m}{m_{o}}\right)^{k/b} = 1 - \frac{bv}{kv_{ex}}$$
$$v = \frac{kv_{ex}}{b}\left[1 - \left(\frac{m}{m_{o}}\right)^{k/b}\right]$$

As you can see, I end up with an exponential term of $$k/b$$ rather than $$b/k$$

WAIT, wow, that was a stupid integration mistake.
This
$$-kln\left(m_{o} - kt'\right)|_{0}^{t} = -bln\left(kv_{ex}-bv'\right)|_{0}^{v}$$
should have been
$$-\frac{1}{k}ln\left(m_{o} - kt'\right)|_{0}^{t} = -\frac{1}{b}ln\left(kv_{ex}-bv'\right)|_{0}^{v}$$

 

[TSny]

This
$$-kln\left(m_{o} - kt'\right)|_{0}^{t} = -bln\left(kv_{ex}-bv'\right)|_{0}^{v}$$
should have been
$$-\frac{1}{k}ln\left(m_{o} - kt'\right)|_{0}^{t} = -\frac{1}{b}ln\left(kv_{ex}-bv'\right)|_{0}^{v}$$

Yes. That should fix it.