As we know, something released from around 36000 km height from a space elevator will remain in geostationary orbit around the Earth.

Things released from lower will go into an elliptical orbit around the center of the earth. If the pericenter of its orbit is at least so far from the center of the Earth, as the radius of the Earth, plus around 500 km, I think, its orbit will be highly elliptic, but relative stable.

How high do we need to release something from a space elevator to reach that?

To clarify, what is the height at which you could release an object from a space elevator and achieve an orbit with perigee >= Re + 500 km (where Re is the earth radius)?


1 Answer 1


I think the question is asking what the release altitude from a space elevator would result in an orbit with a perigee of $R_E + 500 km$ ($R_E$ is the earth radius — 6378.135 km)

The velocity of the space elevator varies linearly along its length, with the velocity at GEO altitude ($R_{GEO}$) equal to GEO orbital velocity ($V_{GEO} = \sqrt{\mu/R_g}$). So for a position $r$ on the elevator:

$$V_{elev} = {r\over R}\sqrt{\mu/R_{GEO}}$$

This velocity vector is perpendicular to the position vector, as is the orbital velocity vector at apogee (and perigee).

Orbital velocity as a function of radius $r$ is given by:

$$V_r = \sqrt{\mu(1/r-1/a)}$$

where $a$ is the semi-major axis

$$a = {R_{apogee} + R_{perigee}\over 2}$$

The release point will be the apogee of the achieved orbit, so we can re-write the equations as such:

$$V_{elev} = {R_{apogee}\over R_{GEO}}\sqrt{\mu/R_{GEO}}$$ $$V_{apogee} = \sqrt{\mu(1/R_{apogee} - 2/(R_{apogee} + R_{perigee}))}$$

We can now set $V_{elev} = V_{apogee}$

$${R_{apogee}\over R_{GEO}}\sqrt{\mu/R_{GEO}} = \sqrt{\mu(1/R_{apogee} - 2/(R_{apogee} + R_{perigee}))}$$

After some algebra, you can solve for $R_{perigee}$ as a function of $R_{apogee}$

$$R_{perigee} = (R_{apogee}^3/(2 R_{GEO}^3))[1/(1/R_{apogee} - R_{apogee}^2/(2 R_{GEO}^3))]$$

Solving for $R_{apogee}$ would be a bit messier I think, so I plugged this into Excel and used solver to numerically find the solution for $R_{perigee} = R_E + 500 = 6878.135$

$$R_{apogee} = 30276.54 km$$

Note that this is the radius from the earth center, not the height, as was used in the original question. Other quantities used:

$\mu = 398601$

$R_{GEO} = 42164.2 km$

Graph showing rising curve of release perigee crossing the fixed 500km threshold at an apogee of about 30000km


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.