Question

How long would it take a 'space elevator' to rise from the earth's surface to a geosynchronous orbit if it could rise at a constant velocity of `3.0` `ms^-1`?

Answer 1

It would take `138` days for this elevator to rise to a geosynchronous (but not geostationary) orbit.

Explanation:

I imagine there is information around this question that specifies the height of the specific kind of synchronous orbit being discussed, since there is more than one kind. Assuming we mean geosynchronous, not geostationary orbit, the orbit height is `35,786` `km` = `3.58xx10^7` `m`.

`v=d/t`

Rearranging,

`t=d/v=(3.58xx10^7)/3=11,933,33` `s`

Let's change that to more manageable units:

There are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, so:

`(11,933,33)/(60xx60xx24)=138` `days`

Answer 2

That will take a little while ...

Explanation:

In a synchronous orbit the elevator will reach a point (or satellite) that takes 24 hrs to complete 1 “lap”. One “lap” is just the circumference of a circle = `2pi.r` and the velocity can be defined as `(2pi.r)/T = sqrt(G.m/r)`

Thus `r^3 = (T^2.Gm)/(4pi^2) = (24xx60^2xx6.67xx10^-11xx6xx10^24)/(4pi^2)`

I get `r = 4.2xx10^7`m but we started at the earth’s surface (r = 6370km) so the altitude is `3.6xx10^7`m

Given we move upwards at 3 m/s and `t = d/v = (3.6xx10^7)/3` so `t = 1.2 xx 10^7`s

In more convenient units 12 million seconds = 138.9 days.

A long, slow rise to your satellite!