Solving a Second-Order Nonlinear Differential Equation

[bob14]

Homework Statement

Hi, I'm trying to calculate the formula for the position vs. time of a rocket landing from an altitude of 100km. I'm neglecting a lot of forces for simplification but basically, I want to solve $F_{net} = Drag - mg$.

Homework Equations

Drag Force: D = $\frac {C_dAρv^2} {2}$

Air Density at height h: ρ(h) = b * h
(I'm just assuming this to simplify the calculation. In the real derivation I would find an equation for each layer of the atmosphere since they differ).

Simplified D = $kρv^2$ where k = $\frac {C_dA} {2}$

The Attempt at a Solution

$F_{net} = Drag - mg$
$m \frac {dv} {dt} = kρv^2- mg$
$m \frac {dv} {dt} = kbhv^2- mg$

Now I can write v as $\frac {dh} {dt}$:

$m \frac {dh^2} {d^2t} = kbh{ ( \frac {dh} {dt})}^2- mg$

Here's where I can't solve it. I've never really taken advanced differential equations so I would just like some tips to go in the right direction. I want to get an h(t) type of equation at the end. If that isn't possible, I would still like to simplify the differential to maybe just solve for a numerical answer.

 

[BvU]

Hi,

google terminal velocity differential equation and find e.g. eq 11 here

$\rho$ and $v$ vary slowly with altitude

 

[bob14]

Hi, the website that you gave doesn't account for ρ. I have already solved the equation in the website when just velocity changes. But when I add the ρ, the equation becomes harder for me since h(t) and h'(t) are multiplied.

 

[Chestermiller]

Your equation for the air density is not correct. A more appropriate approximation would be $\rho = \rho_0 e^{-bh}$
To solve the differential equation, write dt = dh/v.

 

[bob14]

Oh ok. So I would get:

$m \frac {dv} {dt} = kp_0e^{-bh}v^2 - mg$

Since $\frac {dh} {dt} = v$ → $dh = v dt$ → $h = \frac {v^2} {2} + C$
Substituting:

$m \frac {dv} {dt} = kp_0e^{\frac {-bv^2} {2} + C}v^2 - mg$

⇒ $m \frac {dv} {dt} = Ckp_0e^{\frac {-bv^2} {2}}v^2 - mg$

⇒$\frac {dv} {Ckp_0e^{\frac {-bv^2} {2}}v^2 - mg} = \frac {dt} {m}$

 

[Chestermiller]

Oh ok. So I would get:

$m \frac {dv} {dt} = kp_0e^{-bh}v^2 - mg$

Since $\frac {dh} {dt} = v$ → $dh = v dt$ → $h = \frac {v^2} {2} + C$
Substituting:

$m \frac {dv} {dt} = kp_0e^{\frac {-bv^2} {2} + C}v^2 - mg$

⇒ $m \frac {dv} {dt} = Ckp_0e^{\frac {-bv^2} {2}}v^2 - mg$

⇒$\frac {dv} {Ckp_0e^{\frac {-bv^2} {2}}v^2 - mg} = \frac {dt} {m}$

This is not what I said. Substitute dt = dh/v in dv/dt, and integrate with respect to h.