# Low(est) earth orbit

What’ the lowest a space ship can be and still stay in orbit? I realize that there is orbital decay at virtually any height. For my question, let’s ask what’s the lowest a circular orbit can be and still be up there after 24 hours? Or, at what altitude would the decay be 100m per orbit?

Secondary question: Why was Branson’s flight lower than 100km? His vehicle looked a lot like Spaceshipone which reached 100km (twice) many years ago. Was it because the passengers weight reduced the fuel capacity?

Here’s the back-of-the-napkin explanation: If you hop into a rocket and go really high like Branson or Bezos did, the Earth’s gravity is still pulling you back, and you’ll eventually fall back to earth.

But if you go fast enough around the earth, you’ll get to a speed (~7.8 m/s or 17000 MPH) where even though you’ve gone high and the Earth’s gravity is still pulling you back, you’re now going fast enough that rather than fall back to Earth, you basically fall around the Earth. Put differently, you’re still falling back to Earth, but you’re going fast enough that you fall “past” the Earth.

So with that in mind, the main thing stopping low Earth orbits is the presence of the atmosphere which slows you down, and causes you to stop falling past the Earth (orbiting), and you fall back to Earth. Not incidentally, spacecraft do retrograde burns to re-enter the atmosphere- they fire their engines in the opposite direction that they’re going, and slow down to the point where they fall back to Earth.

To answer your question, according to the ESA, it’s as low as 160 km/99 miles @ about 7.8 m/s/17000 MPH.

I don’t know about Branson’s flight altitude only being about 55 miles- probably because they were just trying to get into “space” on a relatively safe first commercial flight.

That is pretty low… what kind of satellite is that, and how long is it supposed to stay up?

I don’t know- I got it from here, and meant to paste it into my last post.

This paper seems to suggest that if you really want it to stay up for 24 hours, you need 200 km.

It’ll also depend on the size of your satellite. Air resistance will be less significant for a larger (or denser) satellite than for a smaller (or less dense) one. Mitigating this, of course, is the fact that large, dense satellites are exactly the ones that you don’t want falling out of the sky after a couple of days.

A microscopic black hole (if they exist) could actually continue orbiting even after its orbit decays all the way to the surface of the Earth, or below.

Science fiction thought experiments involving tiny black holes is that they would continue to orbit within a planet, slowly consuming the mass of that planet.

I doubt it’s ever been observed, though.

The Apollo spacecraft after it Trans Earth Injection burn was actually no longer in orbit, even though it was 400,000 km up.

For what it’s worth, I asked a very similar question a while back.

It depends on the mass of the hole: A sufficiently small one would evaporate quicker than it would eat.

Let me remove your doubt, though: It’s absolutely definite that no black hole less massive than a star has ever been observed.

The Japanese SLATS satellite investigated very low orbits:

It was able to maintain an orbit at 180 km using an ion thruster. I don’t have the details on its performance, but it’s likely in the ballpark of a few meters/second per day. It would quickly deorbit at that rate, but not after 24 hrs.

It was also able to maintain orbit at 167 km, but only with RCS thruster support. Again I don’t have the details, but it was likely losing tens of meters/s of orbital velocity at that altitude.

Very low orbits are interesting for imaging companies since the resolution increases for a given diameter of optics. 200 km vs. 400 km altitude means you only need 1/2 the telescope diameter, and 1/8 the satellite volume.

One problem that makes the question hard to answer directly is that there’s extreme variation in atmospheric density based on where you are in the 11-year solar cycle, or the year, or the day, and whether there is a geomagnetic storm going on. These factors can add to a 100x difference between min and max drag.

This link shows some of the factors involved. It has a graph of the Starshine satellite during a particular date range, which of course is just one example, but we can use it to give one answer to your question.

100 m/orbit is ~1600 m/day, or 48 km/mo. On page 6 they have a graph where the boxes are in units of 40 km/mo, so we’re looking for the line where the slope is about -1.2. That’s at roughly the 260 km level. On the other hand, it takes only a few days to drop from 200 km to 160 km.

This is clearly much worse than the SLATS satellite I mentioned above, but it was probably not meant to be aerodynamic, and it was happening close to a solar maximum.

FWIW, Shuttle flights used 400,000 feet as the altitude for “nominal reentry”.

This discussion reminds me of the SF story The Holes Around Mars,

in which a third moon of Mars is discovered, which is about 4 inches in diameter and orbits at high speed a few feet above the surface.

A good SF book on the subject: Thomas Wren’s “The Doomsday Effect”.

Larger means less air resistance?

For a given density (and a given shape), an object twice as large will have 8 times the mass but only 4 times the surface area, so less air resistance per mass.

More air resistance, but less significant. What matters is how big the air resistance is compared to all the other forces. Air resistance is proportional to the area, but all of the other relevant forces are proportional to the mass, and hence (assuming constant density) to the volume. It’s yet another case of the square-cube problem.