Can I have some help with this problem?

[srakocy]

Our professor gave us this monster of a problem in Differential Equations. I'm not even sure where to start. Can you possibly give me a bit of help on this?


A lunar lander is free-falling toward the moon's surface at a speed of 1000 mi/h. Its retrorockets, when fired in free space, provide a deceleration of 33,000 mi/h^2. At WHAT HEIGHT above the lunar surface should the retrorockets be activated to insure a "soft" touchdown (v=0) on impact?

Hints: Use units of "kilomiles" and hours and derive by y(t) the lunar lander's distance from the center of the moon at time t.
The moon's radius is about 1.08 "kilomiles," so we want v=0 when y=1.08.
The moon's SURFACE gravitational acceleration is around 5.3 ft/sec^2 which is aproximately 13 "kilomiles"/h^2.
Since gravitational acceleration is inversely proportional to the square of the distance, the lunar gravitational acceleration at distance y (from the center) is ((1.08/y)^2)(13), which is about (15.16/(y^2)) "Kilomiles"/h^2

Hence, subtracting this from retrorocket acceleration of 33 "kilomiles"/h^2 we obtain dv/dt = 33 - (15.16/(y^2))


I appreciate any help!

 

[stapel]

Does this PowerPoint help?

 

[HallsofIvy]

I cannot understand why you would call that a "monster"! This is hardly even a "differential equation". You just integrate.

You are told that there is a constant deceleration of 33000 mi/h^2 so your differential equation is just dv/dt= -330000 or, using the suggestion of units of "kilo-miles", dv/dt= -33. You are told that that initil speed is v(0)= 1000 mi./h or v(0)= 1 using "kilo-miles". All you need to do is integrate. Find the time, t, when v(t)= 0 and then integrat h= dv/dt to find the distance traveled in that time.