# Homework Help: 2 astronauts near a space station (no rotation)

Tags:
1. Jan 11, 2017

### redivider

1. The problem statement, all variables and given/known data
You have a space station in space far from any planets or stars in form of a hollow cylinder with inner radius R1 outer R2 length L and density Rho. On a symetric axis z are 2 astronauts, 1 at the middle and the second at distance H=2L from the center of the bottom of the station. They both have mass m. http://puu.sh/tcfBz/001aed5f96.png

a) with what force does the space station act on the first and the second astronaut? b) Second astronaut lets go (initial velocity = 0). With what speed does he hit the first astronaut?

Constants are given to get numerical answers.
m=80kg, rho=50kg/m^3, R_1=1km, R_2=2km, L=1km

Masses of the astronauts are negligible compared to the mass of the station.

2. Relevant equations
F=GmM/r^2
U=GmM/r

3. The attempt at a solution
I solved the first part using newtons grav law, the second part would be too ugly I thought with the line integral of the force over the whole path, so I tried attacking it with potential. I dont have the solution for this so Im wondering what you think, how "correct" this is.

(theres some variable swaps/using different/same variables for different/same stuff, would just like to know how correct this sounds like/thought process)
This is the axial force from 1(ONE) disk, radius: "R" height: "h", mass point height over its highest center point: "a" http://puu.sh/tcoq2/0a837d4f7b.jpg [Broken] . The lower equation (F=) has to be integrated over all disks (*da right side, and /h right side) from a=a to a+h to get the cylinder, then you have to subtract the force of the smaller one from the bigger one to get the actual hollow cylinder of the space station. Force on the astro in the center is 0.

For the second part the relevant thing from the following pics is the circled equation http://puu.sh/tiShP/2375fde8a0.jpg [Broken] , which is supposed to be my grav potential of 1(ONE) cylinder at the point "a" above it. I tried some mumbo jumbo stuff and changed to b=h/2+a so that its in terms of the distance from the middle of the cylinder (being b), since I realized fast that its going to change differently, depending if the point is INSIDE or OUTSIDE of the cylinder.

But after a while, I *think* I only need the outside version so I can then say all the potential went to kinetic energy mv^2/2=U. So after that integration from a=a to a+h you get that long expression at the end with arsinh-s and all that. Then you subtract the same integral but for the smaller cylinder from the bigger one and equate with mv^2/2. http://puu.sh/tiQwf/41ae9748d2.jpg Putting that into mathematica because my calc runs out of buffer space got: (v in m/s=cca. 0.09) http://puu.sh/tiPxU/672f14734d.png [Broken] how correct/wrong does that seem like? Help appreciated in advance. (oh and also I swapped R_1 and R_2; R_2 being the smaller instead of the bigger radius, just notation; also L=h)

Last edited by a moderator: May 8, 2017
2. Jan 11, 2017

### Staff: Mentor

Did you check the calculations with cases where you can find other approximations, e. g. the potential difference between two points far away from the cylinder (so its deviation from a sphere doesn't matter)?
Does the derivative of your potential agree with the force calculated before?