# Lift to space

1. Sep 19, 2010

### dirk_mec1

1. The problem statement, all variables and given/known data

[PLAIN]http://img194.imageshack.us/img194/2062/57916122.png [Broken]

2. Relevant equations
$$F_{gravitational}= \frac{MmG}{r^2}$$

$$F_{centrifugal}= \frac{mv^2}{r}$$

3. The attempt at a solution
I got this:

$$dF_{centr} = dF_{grav} \longrightarrow \frac{dm \cdot v^2}{R+x} = \frac{MdmG}{(R+x)^2}$$

Is this correct?

Last edited by a moderator: May 4, 2017
2. Sep 19, 2010

### Delphi51

The dF's look okay to me. I don't think they would be equal, though. There would be tension in the wire, except perhaps at one value of x.

3. Sep 20, 2010

### dirk_mec1

Tension? How can I calculate this tension and how does this equation then changes?

4. Sep 20, 2010

### physhelper301

I wouldn't worry about tension or balancing the forces if the question does not ask for it. I think your equations look correct. Here's how I think you should approach the problem.

From the definition of linear density:
rho = dm/dx.

Therefore you could substitute dm in both equations for rho*dx, so that you could actually solve for Fg and Fc as function of x. Then it's just simple calculus to get to a solution.

dFg = M*rho*G*dx/(R + x)^2
Fg = -M*rho*G/(R + x)

dFc = rho*v^2*dx/(R+x)
Fc = rho*v^2*log(R + x)

Hope this helps.

5. Sep 20, 2010

### dirk_mec1

Thanks a lot that helped!