Circular to elliptical orbit delta V requirements

this might be a dumb question, but i've been recently trying to calculate the delta V to deorbit a satellite, and I've run into a problem. Assuming a 400km circular starting orbit(and disregarding drag), how much delta V would be required to bring the perigee down to 0km altitude, or what equation could I use to find this out? I'm sure this should be a simple modification of a hohmann transfer equation or a keplerian law, but I just can't figure out how to do it. I'd be super grateful for some help since I'm stumped.

• You don't have to do a burn to get the perigee down to zero. You just burn a little to dip the orbit into the sensible atmosphere. space.stackexchange.com/questions/12011/… – Organic Marble Nov 2 '19 at 1:58
• Thanks, I'm trying to work out the delta v required for a precise deorbit i.e. for recoverable satellites, so I was trying to get a rough figure without involving the complexity of drag. – T.S Nov 2 '19 at 13:33

Assuming a 400km circular starting orbit(and disregarding drag), how much delta V would be required to bring the perigee down to 0km altitude, or what equation could I use to find this out?

The vis-viva equation is the go-to equation for a lot of things:

$$v^2(r)=GM\left(\frac{2}{r}-\frac{1}{a} \right)$$

$$a = \frac{r_{peri} + r_{apo}}{2}$$

You start in a circular orbit with $$r=a$$ of 6378+400 kilometers, don't forget to multiply by 1000 to change to meters! The standard gravitational parameter $$GM$$ of Earth is 3.986E+14 m^3/s^2. You should get an initial orbital velocity of about 7669 m/s.

Airless Earth

If your planet (which amazingly has exactly the same parameters as Earth) has no atmosphere and you want to change to an elliptical orbit with a periapsis 400 km lower so it is tangent to the Earth's surface, then when you do your delta-v maneuver your apoapsis will still be at 400 km altitude but the periapsis is zero altitude, or 6378 km. That makes your semimajor axis $$a$$ 6378 + 200 km.

You calculate your new velocity at apoapsis (where you do the burn) from the vis-viva equation using of course $$r=r_{apo}$$.

That will give you a target velocity of 7551 m/s, which is a change in velocity or delta-v of 118 m/s.

Earth

Answers to How hard do you have to throw something off the ISS to make it deorbit? which by coincidence is also in a 400 km circular orbit range from 93 m/s to reach 80km altitude when the atmosphere will do the rest for you right away, all the way down to 0 m/s because at such a low altitude of only 400 km, objects have (very roughly) a few months to a year or two before they re-enter the atmosphere due to drag.

• Thanks very much, I knew it was one of the fundamental orbital equations but i couldn't remember which one. The reason I didn't bother with drag is so i could get a rough ballpark figure without having to go into complicated atmospheric modelling, but the second part of your answer seems to have that covered. And yes, 400km was chosen because of the ISS, so it is slightly less of a coincidence :) – T.S Nov 2 '19 at 13:31
• @T.S I had a hunch about the 400 km. Thank you for the quick accept, but usually people wait a day or two to give a chance for other answers. The next one might be better or more useful than this, and quick accepting sometimes discourages other people from bothering to post another answer. – uhoh Nov 2 '19 at 13:45

To answer your question I entered 100 into cell F38 (Periapsis altitude) and 400 into cell F39 (Apoapsis altitude). I didn't enter 0 into F38 because getting the periapsis down into the upper atmosphere suffices to de-orbit a satellite.

Over in cell J40 is the apoapsis circularize burn. Since orbits are time reversible it takes the same burn to go from a 400 circular to an elliptical 100x400 orbit.

You can see it takes about .1 km/s to de-orbit from a 400 km circular orbit.

I use the vis-viva equation for much of this spreadsheet.

Just to complement uhoh's answer, note that your spacecraft would reach the surface of the Earth at a speed

$$v=\sqrt{2GM\left(\frac{1}{r_{Earth}}-\frac{1}{2a}\right)},$$

a little higher than 8 km/s = 28,889 km/h (not counting the rotation of the planet). To make the spacecraft land smoothly you'd need to brake to zero, and spend almost as much propellant as the required to put the satellite in orbit (the same, if the launch would be in atmosphereless planet as well). The atmosphere make us a great service by braking the spacecraft for us (with only the addition of an ablation shield and parachutes, for example). That is, if you want the spacecraft to survive the descent.