- #1

Adoniram

- 94

- 6

- No gravity, No drag
- Gravity, No drag
- No gravity,
*linear*drag (b*v where b is a constant)

I have never seen

- Gravity, linear drag
- Gravity, quadratic drag

I took John Taylor's two examples of linear drag and gravity (without drag), and tried to combine them (since quadratic drag would add even more complexity). A couple basic assumptions at first:

The mass of the rocket is described as:

[tex]m(t)=m_o-kt[/tex]

Where [itex]m_o[/itex] is the initial mass of the fueled up rocket, and [itex]-k[/itex] is the rate at which mass leaves the rocket. Also, [itex]u[/itex] is the velocity of the exhaust (which is taken as a constant), and [itex]v[/itex] is the velocity of the rocket itself.

For linear drag + gravity, the resultant equation of motion would be:

[tex]\frac{m dv}{dt}=-\frac{u dm}{dt}-mg-bv[/tex]

At this point, for the case

*without*gravity, you can do separation of variables. However, that doesn't seem possible here:

[tex]\frac{dv}{dt}=\frac{1}{m}\left(ku-bv\right)-g[/tex]

(I've substituted -

*k*in for dm/dt)

So, putting in m->m(t):

[tex]\frac{dv}{dt}=\frac{\left(ku-bv\right)}{\left(m_o-kt\right)}-g[/tex]

Any help on solving this analytically would be greatly appreciated!