# I Deriving an Expression of an Object in Free-Fall on Mars

1. Mar 21, 2016

### lithium123

Hi everyone. I'm trying to derive an expression of the velocity of an object in free fall on Mars, and I am having trouble. Here is what I've done so far:

Drag on Mars can be expressed as
$$F_d = \frac{C_d A \rho_M v^2}{2},$$ where C_d denotes the drag coefficient, A the reference area, and rho_M the air density on Mars. Let $$k = \frac{C_d A \rho_M}{2}$$.

Suppose the lander is in free-fall. Considering its two main forces, the lander hence has an equation of motion
\begin{align*}
F_d - mg &= m \hspace{1mm}\frac{dv}{dt} \\
\frac{k}{m}v^2 - g &= \frac{dv}{dt} \\
\int {dt} &= \int{\frac{dv}{\frac{k}{m}v^2 - g}}
\end{align*}
Applying the substitution $$v = u \sqrt{\strut\frac{gm}{k}}$$, the equation can then be written as

\begin{align*}
t &= \int \frac{1}{\left(u^2-1\right)\sqrt{\frac{gk}{m}}} du \\
&= -{\sqrt{\frac{m}{gk}}}\int \frac{1}{-u^2+1} \hspace{1mm} du \\
&= -{\sqrt{\frac{m}{gk}}} \hspace{1mm} \text{arctanh} \hspace{1mm} u
\end{align*}
And substituting back, $$u = v \sqrt{\strut\frac{gm}{k}}$$ and expressing this as a function $v(t)$ gives us $$v(t) = \sqrt{\frac{gm}{k}} \hspace{1mm} \text{tanh} \hspace{1mm} ( \hspace{0.5mm}\sqrt{\frac{gk}{m}} (C-t)).$$ Now, we must find the values of the constants in our above expression.

From information about our object, we have $$k = \big(\frac{1}{2}\big) (0.67)(\pi r^2)(0.0155) \approx 0.0163 r^2 \hspace{1mm} \frac{\text{kg}}{\text{m}}.$$
The force of $g$ on Mars is expressed as $$g = 3.711 \frac{\text {m}}{\text{s}^2}.$$
Note m and r are not constants and vary. Our velocity expression can then be written as $$v(m,t,r) = \frac{15.083}{r} \sqrt m \hspace{1mm}\text{ tanh} \hspace{1mm}\frac{0.285211r (C-t)}{\sqrt m}.$$ We can determine the constant C by noting that our object has a mass of 600 kg, begins the landing phase at $t = 0$ at a speed of 250 m/s with a cross sectional area of 81pi, or $$v(600, 0, 9) = 250$$.

I have run into a problem here. If I use v(600, 0, 9) = 250, the C value turns out to not be real. Why is this, and is it still possible to get an expression for v(m,t,r) with this special condition?

2. Mar 22, 2016

### Buzz Bloom

Hi @lithium123:

I don't think I can offer much help, but I think the integration of
t = ∫... du is flawed. You assumed that m is independent of u, and I don't think that is correct. Both m and u are functions of t.

I am not sure why you said
Note m and r are not constants and vary.​
m might vary if some fuel is used to control attitude while falling.
I think you intended r to be ρ.

In any case, gk/m is a function of t, so the integral with respect to t should have an integrand of
√m/gk

Regards,
Buzz

Last edited: Mar 22, 2016