Hi I'm trying to solve this textbook example but I don't know where to begin;

NASA has decided to send an experimental probe to Mars. Its weight on earth is $40 kg_f$. When the probe is near the planet it will be

attracted by the its gravitational field ($g_{Mars} = 3.75\ m/s^2$). Determine the parachute’s diameter so that the probe will touch Mars’ surface with a velocity of $3 m/s$ (which is equivalent of dropping the probe from 0.5 m height on Earth). ($A_{chute} = π \frac{D^2}{4}, C_D = 1.4$, the density of Mars’ atmosphere is $\frac{2}{3}$ of Earth’s).

[$C_D$ is the drag coefficient of the parachute $C_D = 1.4$ (no dimensions, dimensionless number, or clear number,) and where $A$ is the projected area of the solid on a plane perpendicular to the body’s motion.]

My knowledge of phyics is still at a beginner-intermediate level. I've been working on this problem for more than an hour and understand the concepts of drag coefficient, and force of a falling object.

What I don't understand, and couldn't find good resources for, is how to measure the speed of a falling object attached to a parachute (aka calculate the drag the parachute has on the fall), how to figure the density of Earth's atmosphere (isn't it different everywhere?) and how to figure out what the speed of the probe would be without a parachute.

## Best Answer

You probably know the equation for the drag, but just for record it's:

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

and rearranging this gives:

$$ \frac{2F_{drag}}{\rho \space C_d \space v^2} = A $$

and you're given $C_d$ (1.4) and $v$ (3 m/s). I would guess you're meant to take the density of Earth's atmosphere at sea level, and a quick Google gives this as 1.2754 kg/m$^3$ (is the density of the Martian atmosphere really 2/3 that of Earth? I thought it was more like 1%). The force is just the weight of the probe (40kg) multiplied by the acceleration due to gravity at the surface of Mars ($g_{Mars} = 3.75\ m/s^2$). So you have everything you need to calculate $A$.

It's not obvious to me why you need the speed of the probe without the parachute ...