Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 3 down vote favorite
2

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.

share|improve this question

1 Answer

up vote 3 down vote accepted

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 ...

share|improve this answer
Just for completeness (and not because there is anything wrong with John's answer) the relation $F_{drag} \propto v^2$ is only good in some regimes, though these include most of the ones that human beings experience in everyday life, and many that engineers get to play around it. It is also almost certainly the relation that the textbook expects you to apply. – dmckee♦ Oct 31 '12 at 21:32
Is $F_{drag}$ the same as $m\frac{dV}{dT}$? – Imray Nov 2 '12 at 14:07
Also, you say I have everything I need to solve $A$, but how do I get $F_{drag}$? – Imray Nov 2 '12 at 14:12
If the probe (and parachute) are travelling at constant speed, i.e. neither accelerating nor decelerating, then the drag on the parachure must equal the weight of the probe. So $F_{drag} = mg$ where $g$ is the acceleration of the Martian gravity and $m$ is the mass of the probe (and parachute, though you're probably intended to ignore the mass of the parachute). – John Rennie Nov 2 '12 at 14:50
So $F_{drag}$ is equal to zero? That wouldn't make sense, because then A would = 0. – Imray Nov 4 '12 at 19:52
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.