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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 $A$.

2 Answers 2

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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 https://physics.stackexchange.com/

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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    I used $A = 28$ as an estimate2012-11-07
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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