physics question, centripetal acceleration?

Question

The Mars Pathfinder is designed to drop the vehicle's instrument package from a height of 20 meters above the Martian surface, after the speed of the probe has been brought to zero by a combination parachute-rocket system at that height. To cushion the landing, giant air bags surround the package. The mass of mars is 0.1074 times that of earth and the radius of mars is 0.5282 that of earth. FInd a) the accerleration due to gravity at the surface of Mars, b) and how long it takes for the instrument package to fall the last 20 meters.

help please!

the mass of mars is 6.42252 X 10^18

the radius is 3369916 m

I don't really need the answers, just the formulas used; [if they're derived from other equations; what they're derived from would be helpful] and the process

ps- why is the equation sqrt [2d/g]? shouldn't it be sqrt [d/g]; derived from v(f) = v(i) + at

nevermind; i figured it out! thanks so much for the help! i got 3.26 seconds for the answer

Answer 1

Ok first part. Mars's gravity. You could always look it up, but let's use Newton's law of gravity;

F=G m1m2/r^2

We will also use F=ma

This means that, for gravity

F/m = a = G m(planet)/r^2

therefore

a (earth)= g = G m(earth)/r(earth)^2

a(mars)= G m(mars)/r(mars)^2

therefore

a(mars)/a(earth)= a(mars)/g=[G m(mars)/r(mars)^2]/[G m(earth)/r(earth)^2]

Canceling everything out and moving g over, we get

a(mars)=g* m(mars)/m(earth) *[r(earth)/r(mars)]^2

a(mars)= g * .1074/ (.5282)^2

=g * 0.3850

= 3.77 m/sec^2

(This is pretty close to what Wikipedia says Mars's gravity is)

Part B

t=sqrt (2 * h/a(mars))

I'll let you do this part, but you should be a little bit above 3 secs.

Answer 2

I was wondering the same question myself yesterday