- #1
Adoniram
- 94
- 6
I have seen many examples of the EOM for a rocket derived for the following cases:
I have never seen
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!
- 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!