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

In summary, the conversation revolves around deriving an expression for the velocity of an object in free fall on Mars. The main equation of motion is given, but there is difficulty in finding the right constants to substitute. The integration method used is questioned and it is mentioned that the mass and radius of the object may vary. There is also some confusion about the units used in the calculations.
  • #1
lithium123
2
0
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?
 
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  • #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:
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1. What is free-fall on Mars?

Free-fall on Mars refers to the motion of an object when it is falling towards the surface of Mars, under the influence of gravity. In free-fall, the object is only affected by the gravitational force and there is no other external force acting on it.

2. How is free-fall on Mars different from free-fall on Earth?

Free-fall on Mars is different from free-fall on Earth because Mars has a lower gravitational acceleration (3.7 m/s²) compared to Earth (9.8 m/s²). This means that objects will fall slower on Mars compared to Earth. Additionally, the atmosphere on Mars is much thinner than Earth's, so there is less air resistance during free-fall.

3. How do you derive an expression for an object in free-fall on Mars?

To derive an expression for an object in free-fall on Mars, you can use the equation for free-fall motion: d = 1/2 * g * t^2, where d is the distance traveled, g is the gravitational acceleration on Mars, and t is the time in seconds. By plugging in the values for g on Mars and solving for d, you can derive an expression for the distance an object will fall on Mars in a given amount of time.

4. How does the mass of the object affect its free-fall on Mars?

The mass of an object does not affect its free-fall on Mars. According to Newton's Law of Gravitation, the gravitational force between two objects is directly proportional to their masses. However, in free-fall, the object is only affected by the gravitational force of Mars, which remains constant regardless of the object's mass. Therefore, the mass of the object does not affect its free-fall on Mars.

5. Can an object in free-fall on Mars ever reach a terminal velocity?

Yes, an object in free-fall on Mars can reach a terminal velocity. Terminal velocity is the maximum velocity an object can reach when falling due to the balance between the gravitational force and air resistance. While the air resistance on Mars is significantly less than on Earth, it still exists and will eventually balance out the gravitational force, causing the object to reach a constant terminal velocity during free-fall on Mars.

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