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An example on a physics assignment asks this question:

NASA is preparing a probe to send to Mars. The probe weighs $40kg_f$ on Earth. As it approaches Mars, the gravitational field of Mars, which is given as $g_{mars}=3.711 m/s^2$, will pull it down. The question is, what should the diameter of the parachute be so that the probe touches the surface with a speed of $3m/s$?

Here's what's given:

  • $A_{chute} = \pi \frac{D^2}{4}$
  • $C_D$ (the parachute's drag coefficient) $=1.4$
  • Density of Mars' atmosphere = $0.8167kg/m^3$
  • Landing the probe on Mars at $3m/s$ is equivalent to dropping it from a height of $0.5 $ meters (sans parachute) on Earth.
  • The problem involves a differential equation.

Here's where I'm stuck: I'm using the formula $$ m \frac{dv}{dt} = \frac{1}{2} \rho_{air} \space C_d A \space v^2$$ to solve for $A$, area. The problem is, that when the parachute is drifting down through the Martian atmosphere with the parachute deployed, it quickly meets its terminal velocity and $\frac{dv}{dt}=0$, which makes it impossible to solve for $A$. I can't find another way/formula to use to solve for $A$.

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2 Answers 2

up vote 1 down vote accepted

The probe reaches terminal velocity when the drag force from the parachute equals the gravitational force of the planet.

You have the drag force:

$$F_{drag} = \frac{1}{2} \rho_{air} \space C_d A \space v^2$$

And you have the gravitational force:

$$F_g = m_{probe} \space g_{mars}$$

You just need to solve the equation:

$$F_{drag} = F_g$$

Where the only unknown is $A$ (you already know that you want $v = 3.0 m/s$).

P.S.: You would have probably gotten a faster response if you had placed this question in

P.P.S.: Reading your question again it seems that your doubt came from the expression the drag force as $m \frac{d_v}{d_t}$. This expression is correct in the absence of gravity, i.e. $\frac{d_v}{d_t}$ refers to the change of velocity due to the drag force only when no other forces are acting.

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Where do differential equations come into play in the answer though? –  Imray Nov 4 '12 at 20:52
You don't need to solve any differential equation in this problem. You just need to justify that $g_{mars}$ = $\frac{d_v}{d_t}$ of your differential equation when the probe touches the surface. –  pedrosorio Nov 4 '12 at 20:55
I'm struggling to understand that.. what do you mean $g_{mars} = \frac{dv}{dt}$ when it touches the surface? Why not the whole time? –  Imray Nov 4 '12 at 22:20
I say "when it touches the surface", but I mean "after it reaches terminal velocity". This includes the moment of touching the surface which is the interesting one for you. –  pedrosorio Nov 4 '12 at 22:32
Does this make sense: $A = \frac{\sqrt(112)}{\sqrt(\pi)}$? –  Imray Nov 4 '12 at 22:34

Continuing from the previous answer

$m_{probe} \space g_{mars}= \frac{1}{2} \rho_{air} \space C_d A \space v^2$

You could put $v=dy/dt$ to get differential equation

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