space elevator


by ian2012
Tags: elevator, space
ian2012
ian2012 is offline
#1
Apr3-10, 10:50 AM
P: 80
1. The problem statement, all variables and given/known data

I've worked through part (iv) of this problem, but I am not sure if I shown enough calculation. The most important parts of the question, to put it all in context:

(i) The gravitational force on a body of mass m at a distance r from the earth with mass mE. The time it takes to complete one circular orbit is given as

[tex]T=2 \pi \sqrt{\frac{r^{3}}{Gm_{E}}}[/tex]

(ii) The radius rG of the orbit for a satellite that has a period of 1 day is 4.22x10^7 metres.

(iii) A mass m1 is to be kept in a larger circular orbit with radius rG+delta r, delta r << rG, and the same period as in (ii), a force of magnitude

[tex]F= \frac{3 \Delta r Gm_{E} m_{1}}{r^{3}_{G}}[/tex]

has to be applied on the mass in addition to the gravitational force.

(iv) Now a thin rod with mass per unit length rho is connected to m1 and reaches all the way down to the surface of the earth at radius rE. The rod is not attached to the surface of the earth. Show that the relation

[tex]m_{1} = \frac{\rho r^{3}_{G}}{3 \Delta r r_{E}}[/tex]

applies if the whole construction is to keep the same orbital period of 1 day.

3. The attempt at a solution

I think you might have to integrate the mass elements along the length of the rod, but what I have done is said that the force on the rod is

[tex]F = \frac{G m_{E}}{r_{E}} \frac{dm}{dr}[/tex]

The dm/dr term = the mass per unit length rho and so is constant. Therefore the force on the rod is constant through out. And so since m1 is attached to the rod, the force on the rod = the 'additional' force on m1. Equating part (iii) with the last expression, gives the answer. Is there a more elegant way to do this?
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