# What happens to an orbiting body moving with the Earth's angular speed at various altitudes?

I was learning about inertial frames amongst other things, and I somehow came up with this question. Imagine we have a pole, made of imaginary very strong material that will never bend, placed at the equator and extending into space for a great length. I then climb this pole to certain heights, and release. The question is, at different heights, what exactly does my motion look a) in the perspective of Earth and b) in perspective of the pole.

My hypothesis is that there should be a point at which I would have enough speed (since the pole is rotating with the Earth at one rotation per day) that if I let go of the pole, I would simply enter geostationary orbit, and stay right beside the pole. So from the earth's perspective, I would stay right beside the pole, and from the pole's perspective I would just hang right beside where I let go.

Below this point however, I should drop towards earth, but the first question is whether I would actually make contact with earth, or whether at a lower altitude I would enter elliptical orbit instead? I coded this to test what happens at altitudes below the geostationary point (if it exists)(red is my path, blue is the path of the pole), and apparently it enters elliptical orbit. Then again my code isn't that great. Another point to note is that in this simulation, I go forward, past where the pole is at.

Much above the 'geostationary' point, there will be a region where you will simply have enough velocity to have escape velocity, and escape the gravitational field. I'm 98% sure on this point. Just above the 'geostat' point but below the 'escape' point (which should be root2 times the geostat point?) I'm not too sure what happens here. One conjecture is that the circular orbit deforms into other orbits, such as elliptical. Just pulling at straws, but perhaps just above and just below the 'geostat' point, are both elliptical orbits, but earth is different focus for both (as ellipse has two foci? maybe)

My main question is, how do we determine using actual physics and not computers the path? It seems to me that below the 'geostat' point I will go faster than the pole in respect to the earth, but above the 'geostat' point I will go slower than the pole in respect to the earth. (just from running dodgy simulations.) I had the idea of using angular momentum and velocity to try and attempt this question (as radius decreases, moment of Inertia decreases, angular velocity goes up? not sure if applicable) but I still had no clue how to start to attempt. Also, since gravity ONLY acts toward the centre of the planet, how exactly do we get this deviation to the front or back of the pole? This is my current model:

• I tried to edit your question to clarify it, but you seem to touch on a bunch of different questions here and I'm not sure what you're actually asking. Any way you could edit your post to clarify what your one question is? – David Z Nov 26 '14 at 10:24

You've pretty much covered everything, except that below the geostationary orbit the orbit is elliptical - there is no transition to a hyperbolic orbit at small $r$.

Let's draw a picture of the pole and get some preliminary formulae out of the way:

If you are on the pole at a distance $r$ from the centre of the Earth (NB from the centre of the Earth not from the Earth's surface) then your tangential velocity is:

$$v = r\omega \tag{1}$$

where $\omega$ is the angular velocity of the Earth: $\omega \approx 7.2921 \times 10^{-5}$ radians/sec.

The orbital velocity at a radius $r$ is given by:

$$v_o = \sqrt{\frac{GM}{r}} \tag{2}$$

and the escape velocity at a radius $r$ is given by:

$$v_e = \sqrt{\frac{2GM}{r}} \tag{3}$$

So as you say, the escape velocity is $\sqrt{2}$ times the orbital velocity.

The geostationary orbit is the point where the tangential velocity is the same as the orbital velocity, so equating equations (1) and (2) gives:

$$r\omega = \sqrt{\frac{GM}{r}}$$

or rearranging to give the altitude of the geostationary orbit, $r_g$:

$$r_g = \sqrt[3]{\frac{GM}{\omega^2}}$$

We get the position on the pole where the velocity equals the escape velocity by equating equations (1) and (3) to get:

$$r_e = \sqrt[3]{\frac{2GM}{\omega^2}} = r_g\sqrt[3]{2}$$

So for $r > r_e$ the orbit is hyperbolic and for $r < r_e$ the orbit is ellitical. At $r = r_e$ the orbit is parabolic.

For $r_g < r < r_e$ the semi-major axis of the orbit is roughly along your direction of travel, while for $r < r_g$ the semi-major axis is along the line connecting you to the Earth.

The last step is to work out when we'll just graze the Earth, and we do this using the vis-viva equation. From this we can derive the relationship between the perigee $r_p$ and the apogee $r_a$ of the orbit:

$$r_p = \frac{r_a}{\frac{2GM}{v^2r_a} - 1}$$

The apogee is your distance $r$, and the velocity $v$ is equal to $r\omega$. You just graze the Earth when the perigee is equal to the radius of the Earth $R$ (I'm running out of useful subscripts!) so the distance along the pole where you just hit the Earth is given by:

$$R = \frac{r}{\frac{2GM}{r^3\omega^2} - 1}$$

And rearranging this to solve for $r$ we get the quartic equation:

$$r^4 + Rr^3 - \frac{2GMR}{\omega^2} = 0$$

There is a formula for solving quartics, but it's hideously complicated so I solved the equation numerically to get:

$$r \approx 29,779,000$$

So to summarise:

• $r > r_e$ - hyperbolic
• $r = r_e$ - parabolic
• $r_g < r < r_e$ - elliptic
• $r = r_g$ - circular
• $r < r_g$ - elliptic
• $r <= 29,820,000$ metres - splat!
• Our results for the "splat" altitude differ by 1288 km. Any idea? – mic_e Nov 26 '14 at 12:03
• @mic_e: can you convert your equations to MathJax? I tried comparing my working to yours (we've obviously both used the vis-viva equation) but reading your equations is making my head hurt. – John Rennie Nov 26 '14 at 12:07
• @mic_e: if you're not sure about the MathJax notation click "Edit" on my answer to see the raw text. – John Rennie Nov 26 '14 at 12:08
• @mic_e: aha! I used a 24 hour day not a sidereal day. Let me see what difference that makes ... no, that makes very little difference. – John Rennie Nov 26 '14 at 12:09
• I also used equatorial radius, which might also make a difference. Will convert. BTW fun trivia: The landing spot for one of the Mercury missions was off by a few tens of kilometers because they forgot about sidereal days. – mic_e Nov 26 '14 at 12:10

The "pole" you are referring to is also known as a space elevator. Creating a cable of sufficient tensile strength is currently unfeasible, but carbon nano- and macrotubes are promising.

Your question can be generalized, ignoring Earth's non-spherical shape, to apply to any position in space around earth by replacing "height above the equator" with "distance from Earth's rotation axis" (henceforth r).

All questions except for the one for the minimum r to achieve orbit can be answered using elemental math and energy conservation.

The velocity at any r can be easily calculated via

$$\omega = \frac{2 \cdot \pi}{\textit{1 sidereal day}}$$

Note the difference: One year has 366(.2425) sidereal days, but only 365(.2425) days

$$v(r) = \omega r \approx 7.29212 \cdot 10^{-5} \frac{r}{s}$$

At ground level ($r=6378137 m$), that is $465 \frac{m}{s}$. At geostationary altitude, it's about $3\frac{km}{s}$. The speed of light is reached at a distance of about 27 AU (read: far, far away).

The specific (per unit of mass) energy of your orbit, $E_{spec}$, is the sum of the specific kinetic and potential energies, and negative for a bound (circular, elliptical and/or sub-orbital) orbit.

$$E_{spec,kin}(r) = \frac{1}{2} v(r)^2 = \frac{1}{2} \omega^2 r^2$$ $$E_{spec,pot}(r) = \int_{r}^{\inf}{\frac{GM}{r^2}dr} = -\frac{GM}{r}$$

The parabolic escape orbit has a specific energy of zero:

$$E_{spec}(r) = 0$$ $$\frac{GM}{r} = \frac{1}{2} \omega^2 r^2$$ $$2 GM = \omega^2 r^3$$ $$r = \sqrt[3]{\frac{2 GM}{\omega^2}}$$ $$r = 53123487m$$

That is 46,745 km above the surface of earth, or 10,959 km above geostationary altitude. Note that slightly lower altitudes will also ultimately lead to an escape due to the three-body nature of the Earth/Sun/Object system (further reading).

All trajectories above this altitude are hyperbolic, with increasing eccentricities. Between the geostationary and escape altitude, you have various elliptical orbits with eccentricities ranging from zero (at geostationary altitude) to one (at escape altitude). The Periapsis (closest-to-earth point) is at release altitude. The calculation of the Apoapsis (furthest-from-earth) altitude and further orbital parameters requires Kepler equations; this site provides an excellent listing of all relevant equations.

We already know the Periapsis altitude, plus the orbit's specific energy $E_{spec}$. From that we can calculate the semi-major axis of the ellipse:

$$a = -\frac{GM}{2E_{spec}}$$

The Apoapsis altitude can then be calculated via

$$r_{ap} + r_{pe} = 2a = -\frac{GM}{2E_{spec}}$$ $$r_{ap} = -\frac{GM}{E_{spec}} - r_{pe} = -\frac{GM}{E_{spec}} - r_{geostat}$$

All orbits from below geostationary altitude are elliptical, with their Apoapsis fixed at their release altitude, and Periapsis below; the Periapsis can be calculated as above:

$$r_{pe} = -\frac{GM}{E_{spec}} - r_{ap} = -\frac{GM}{E_{spec}} - r$$

The point at which the orbit touches Earth's surface (and thus becomes suborbital) can be calculated by setting $r_{pe} = r_{earth}$:

$$r_{earth} = -\frac{GM}{E_{spec}} - r$$ $$E_{spec} = -\frac{GM}{r_{earth} + r}$$ $$\frac{1}{2} \omega^2 r^2 - \frac{GM}{r} = -\frac{GM}{r_{earth} + r}$$ $$\frac{\omega^2 r^2}{2GM} + \frac{1}{r_{earth} + r} = \frac{1}{r}$$

The solution to this equation is not nice, and left as an exercise. The numerical solution is 29,790 km, i.e. 23,412 km above earth's surface, or 12,373 km below geostationary altitude. To achieve a stable orbit (periapsis = 200 km), an altitude of 23,610 km (198 km higher) is required.

• "All orbits from below geostationary altitude are elliptical, with their Apoapsis fixed at geostationary altitude" - I don't get this. If I let go of the pole at the Earth's surface the apogee of my orbit is not the geostationary altitude. Surely the apogee is just the distance you were along the pole when you let go? – John Rennie Nov 26 '14 at 12:54
• I just noticed that fundamental error in my reasoning myself. Stand by for edit. – mic_e Nov 26 '14 at 12:57
• My fix has reduced the error between our predictions to 30 kilometers. – mic_e Nov 26 '14 at 13:15
• Aha. That will be down to me not using a sidereal day. – John Rennie Nov 26 '14 at 15:41