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

  1. Mar 21, 2016 #1
    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. jcsd
  3. Mar 22, 2016 #2
    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.

    If I am wrong about this, I apologize.

    ADDED

    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
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