# Modelling a circularizing burn at Apogee

I am relatively new to rocket physics and am looking into modelling burns.

Is there a good place to start modelling an orbital burn? I know about the rocket equation but perhaps that is not what is needed here. I could go to fundamentals and model the ODE from a FBD?

So my question is, where would you begin/what papers would you read if you wanted to model an orbital burn?

Cheers!

Although in my question I was just really asking about some pointers on where to start, I thought I would clarify the final problem that I am trying to solve. I have a spacecraft at the apogee of an elliptical orbit and I want to model a burn to circularize the orbit. I know the altitude of periapsis and apoapsis, $$h$$ and $$H$$ respectively and all the constants such as current velocity in $$x$$ and $$y$$ directions, mass, radius of planet etc...

• These aren't direct answers to your question but you may find some background information helpful (both the answers and links therein) How to solve the two-body problem in the ECI frame through numerical integration? (coupled first order ODE solving) Simulating engine burn with scipy ode solve But in this case your "m dot" (dm/dt) is central to the problem, since fuel is roughly 90% of the mass of a rocket, so your ODEs will need to reflect this. – uhoh Nov 28 '20 at 14:58
• See en.wikipedia.org/wiki/Gravity_turn#Mathematical_description Since there are to many free parameters (you can change thrust magnitude and direction at any time in any way) usually start with some simple polynomial steering law. See answers to For an Apollo Lunar Module Ascent Stage launch, what is the optimal profile of β (or γ) vs time? to get started, where there is a link to Derivation of Linear Tangent Steering Laws – uhoh Nov 28 '20 at 15:06
• – uhoh Nov 28 '20 at 15:07
• @nv0id Welcome to Space stackexchange! By the sound of it you want to start somewhere simpler though I was wondering what you meant by "apoapsis point to get the spacecraft into orbit. " - if you mean that you imagine a newly launched vehicle to be on a sub-orbital trajectory and then needs to circularise at apogee then my answer will help. If you meant something else then could you explain more of your problem in the question? – Puffin Nov 28 '20 at 21:49
• @Puffin, thanks for your answer, I was working on the problem last night. You actually pointed me in the right direction so thank you. The Hohmann transfer is exactly the wiki page I needed so I am trying to put it into python today. I will post an answer with exactly what I did in case others are wanting to do a similar thing. – nv0id Nov 29 '20 at 10:49

From the comments and the OP reply, it sounds as if this is "a good place to start":

Hohman transfer

1. Learn the equation for orbital velocity as a function of the apogee and perigee of the orbit. Determine those velocities for the start orbit and the finish orbit (step back from your homework problem here and just put any circular orbits in, just to get used to it).
2. For the situation where you want to manoeuvre from the low circular orbit to the high circular orbit imagine an ellipse between them acting as a transfer orbit.
3. Manoeuvre 1 is performed where the lower circular orbit meets the elipse. The deltaV required is the difference between the two orbital velocities at that point of intersection. Assuming the manoeuvre is impulsive, the satellite has changed from the first orbit to the ellipse.
4. Manoeuvre 2 happens where the ellipse meets the higher circular orbit and its deltaV is again the difference between the velocities at that point of intersection. The satellite has now transitioned to the higher circular orbit. Minimum transfer time is half the orbital period of the ellipse.
5. Try this for different types of orbit just to get used to the numbers. If you want the start and finish orbits to be non-circular then be prepared to experiment to find the most efficient manoeuvre. If you want to do manoeuvres at points other than the apogee and perigee of the ellipse, then learn about the Vis-Viva Equation.

Wikipedia: Hohmann_transfer_orbit

Wikipedia: Vis-viva_equation

So I have spend a few hours couple days going down this rabbit hole and I thought I would give my findings of going from knowing little about orbital mechanics to someone who knows a little more... Many things could be wrong so it would be great if someone who actually knows what they are talking about could correct and explain to me why I am wrong.

Ok, end of pre-amble...

## Hohmann Transfer

As I will clarify in my original post, my end goal is getting the spacecraft from path 2 to path 3 (circularized orbit):

Conveniently, the equation for the change in speed was already there:

$$\Delta v_2 = \sqrt\frac{\mu}{r_2} \bigg( 1- \sqrt \frac{2r_1}{r_1+r_2} \bigg)$$

to leave the elliptical orbit at $$r = r_2$$ to the $$r_2$$ circular orbit, where $$r_1$$ and $$r_2$$ are respectively the radii of the departure and arrival circular orbits; the smaller (greater) of $$r_1$$ and $$r_2$$ corresponds to the periapsis distance (apoapsis distance) of the Hohmann elliptical transfer orbit.

So I just sub in the variables I know about my spacecraft, $$h$$, the periapsis altitude, $$H$$, the apoapsis altitude and $$R$$ the radius of the planet:

$$\Delta v_2 = \sqrt\frac{GM}{H+R} \bigg( 1- \sqrt \frac{2(h+R)}{h+H+2R} \bigg)$$

## Apogee Kick

For my problem I want to do a kick burn to circularize my orbit. Considering I know know $$\Delta v$$, I thought the rocket equation would work in my case:

$$\Delta v = v_e ln \frac{m_0}{m_f}$$

This is as far as I have got, I will edit this if/when, I have done more or realized I am being stupid.

## Edit: Guess what... I was being stupid

After a light banging of the head on the desk, I realized how to actually solve this problem. Whats really cool and encouraging is that my theoretical value was the same as the model's value!

Here is how I did it:

### 1. The vis-viva equation

As user:Puffin kindly mentioned in his answer above, you can use the vis-viva equation to work out the required speed for an orbit.

$$v^2 = \mu \bigg(\frac 2 r - \frac 1 a \bigg) \quad \text{vis-viva equation}$$

where $$r$$ is the distance between the two bodies and $$a$$ is the semi major axis.

So this allows me to work out the final speed I want to achieve $$v_f$$ (path 3 from the diagram:

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

Then I can work out the theoretical speed of the elliptical orbit (path 2 from the diagram above) and make an equation for the change in speed:

$$\Delta v = v_f-v_i = \sqrt{GM}\Bigg( \sqrt{\frac {1} {H+R}} - \sqrt{ \frac 2 {H + R} - \frac 1 {\frac{H+h}2 + R}}\Bigg)$$

(NOTE: $$H$$ and $$h$$ are the apoapsis and periapsis altitudes, its problem specific)

The theoretical speed was 0.0055 km/s faster than the actual speed! This deviation is probably due to the drag or something... That's how I know I was on the right track.

### 2. The Rocket Equation

Now all I had a value for $$\Delta v$$ I could simply sub it into the rocket equation assuming the Apogee kick motor has a specific impulse of 320 seconds (typical value). Keeping it general, the equation for mass of propellant required was:

$$m_{\text{propel}} = m_i - m_f = m_i - \frac {m_i}{e^{\big( \frac{\Delta v}{I_{\text{sp}}\cdot g_0}\big)}}$$

Et voila, I now have the mass of propellant, everything I wanted to achieve! Now I know you could go into a lot more detail and worry about thrust vectoring and go through all the links that uhoh posted but I am happy with this level for now.

Maybe this will help someone, maybe it wont but it might help me if I need to do this again one day...

• That's awesome! I really liked your enthusiastic tone in this answer - it shows how well you figured out these things by yourself given a slight nudge into the right direction. So happy for you mate! And yes, it definitely helps me as it shows the practical application of the equations I already knew about :) – OrangeDurito Nov 30 '20 at 18:52
• Yeah, who knew it feels good not being just given the answers ahah. Thanks for the kind words @OrangeDurito – nv0id Nov 30 '20 at 20:27