# How to calculate the lowest possible altitude a satellite can orbit at due to aerodynamic heating if provided with a sufficient propulsion system?

If a satellite is equipped with a propulsion system which is enough for compensating the local drag and maintaining the orbit, then aerodynamic heating would be the limiting factor for attaining the lowest possible altitude.

How can one calculate or at least estimate the limiting altitude for a given satellite? What major parameters or aspects of the satellite are required?

• +1 I've adjusted your question so that it's not closed for "needs detail or clarity". People will comment "It depends on the specific satellite" etc. I also adjusted your title to match the body of your question. While people can't give you an exact altitude until you give them an exact satellite (and then they still won't do it) written this way an answer can explain how this might be calculated and what factors you'll need to know. You can then ask a follow-up question. You are welcome to edit further or roll back. Welcome to Space! – uhoh Mar 10 at 15:37
• 1. Are you sure atmospheric heating would be the limiting factor? Why? – Dragongeek Mar 10 at 16:05
• 2. What you're describing sounds less like a satellite and more like an aircraft or cruise missile. What exactly do you mean by satellite? – Dragongeek Mar 10 at 16:06
• The limiting factor is likely to be the propulsion system (and that's before you consider one can often dump heat into your propellant). ESA'a GOCE might be of interest – JCRM Mar 10 at 16:39
• It sounds like the answer should be simple. Just find instantaneous drag acceleration and compute for required thrust. Altitude should be in that equation somewhere, so just evaluate for it. But, it's probably not that simple... – BMF Mar 10 at 19:15

Following @JCRM's lead: rocket horsepower questions to the rescue! See this answer and this answer for derivations and explanations.

## Power

If we assume that most of the kinetic energy of the air molecules striking the spacecraft is converted to heat (perhaps it's more like half or 2/3) then we can use the concept of "rocket power" which is really just the kinetic energy of the gas leaving a spacecraft calculated in the frame of the spacecraft.

$$\frac{dE}{dt} = P = \frac{v^2}{2}\frac{dm}{dt}$$

$$\frac{dm}{dt}$$ would be the mass of air encountered per unit time and is the density times the velocity time the area $$\rho v A$$.

If our drag shield were a plate of metal held "into the wind" thermally shielded and on insulating posts maintained at a temperature $$T$$ of 1000 Kelvin (about 730C) it could dissipate about $$\sigma A T^4$$ by thermal radiation assuming a shock wave hasn't formed in front that is so dense it starts radiating back and blocking radiation out. If that were the case then you'll need to absorb heat in front and re-radiate it out the back using a circulating liquid to transfer the heat, which sounds hard and also sounds like someone may have thought of this in the past.

$$P = \sigma A T^4 = \frac{v^2}{2}\frac{dm}{dt} = \frac{v^2}{2} \rho v A$$

$$P = \sigma A T^4 = \frac{1}{2} \rho v^3 A.$$

I'm leaving the drag coefficient equal to one, otherwise that is what Wikipedia gets as well. Solving for density;

$$\rho = \frac{2 \sigma T^4}{v^3}.$$

The Stefan Boltzmann constant $$\sigma$$ is about 5.67E-08 W m-2 K-4.

Put in 1000 K and 7800 m/s for example and we get roughly 2E-07 kg/m^3 or (also roughly) 2E-07 bar which puts it roughly at (found here) the Karman line at 100 km which makes @JCRM's comment about this being another "Karman plane question" either eerily prescient or profoundly insightful!

## What thrust is needed?

Since force is just power divided by velocity we remove one power of $$v$$ to get

$$F = \frac{1}{2} \rho v^2 A.$$

At 2E-07 kg/m^2 that's 12 Newtons which is much bigger than you could easily do with solar-electric on a spacecraft with a 1 square meter cross section orbiting at 100 km. You'll need a conventional thruster and so you'll run out of propellant quickly.

I'll leave it as an exercise for the reader to calculate the thruster's horsepower ;-)

• If thrust $$T = \frac{dm}{dt} v$$ then shouldn't the $$\frac{1}{2}$$ in the last formula not be omitted ? – Cornelisinspace Jul 1 at 10:15
• @Cornelisinspace I don't say it is a derivative, I just divide by velocity (power divided by velcity is force). Of course to remain in orbit thrust has to equal drag, no? It's not a coincidence that that is the result! – uhoh Jul 1 at 11:45
• I meant one had to take the derivative of the velocity but i now see that 's not right because you assume that v is a constant. And because you compared "gas leaving a spacecraft" with "air molecules striking the spacecraft" and used the concept of "rocket power" i wrongly thought the thrust equation should be used. Just an explanation ! – Cornelisinspace Jul 1 at 12:35
• @Cornelisinspace okay got it. I got the assumption that $v$ is a constant from the question: "...a propulsion system which is enough for compensating the local drag and maintaining the orbit..." – uhoh Jul 1 at 12:55
• This is a nice answer! – tfb Jul 1 at 18:08