Flight Question DG fail to escape from Earth SOI using both TransX and Transfer MFD

Nicholas Kang

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Hello,

Before telling the question, I would like to tell you about my plan. I plan to demonstrate Orbiter in my school's annual Science and Math exhibition. I already mastered the technique to fly, rendezvous and dock to the ISS and also fly to moon and back with the former using Transfer MFD and the later using TransX.

However, I encountered problems while flying to Mars. I started up with the default DG Mk4 in orbit scenario. First, I align my orbital plane with respect to Mars by burning normal/anti-normal using the Align Plane MFD, that is to reduce relative inclination to zero. Next, I circularized my orbit to approximately 300 km i.e. an LEO. Then, I used the Transfer MFD to set up my trip to Mars, using Sun as reference, Earth as source and finally Mars as target.

I warped time until some 2000 seconds DTe, i.e. time before ejection point and burned normal/anti-normal to reduce relative inclination once again to zero. Finally, I orientated my DG prograde to begin for TME (Trans-Martian-Ejection). Just 145 seconds before ejection point, that is 145 seconds DTe. I applied full main engine burn until DTe counted zero. Yet, I hadn't escape earth gravity. My orbital eccentricity did not exceed 1 but was only approximately 0.96.

Here is how I calculated the DTe to be 145 seconds.

I used the kinematics formula v=u+at. V is final burn-out velocity, u is initial velocity, a is acceleration and t is the value in interest. I started with u-value of 7.746 km/s. Delta-V is approx. 3.110km/s. So, v-value is 10.856km/s. To calculate a-value, use Newton 2nd Law of Motion, F=ma. Rearranging,

a=F/m

Full main engine thrust provides 400 kN of force, thus F=400kN
Mass of DG before burn=18673.7 kg (Mass of DG remains constant throughout the simulation period as I unchecked the limited fuel option in the launchpad)

Substituting the F and m values, I get an a-value of 21.42 m/s^2.

Substituting all values into the equation v=u+at,
10856=7746+21.42t
t=3111/21.42=145.2 seconds.

I was using ASUS A53TA with AMD A6 APU 1.4GhZ, 6 GB RAM and 1GB HD6720G2 Radeon graphic card (Dedicated).

Hopefully the information provided above is sufficient enough for kind experts to help out.

Thank you.

Regards,

Nicholas
 

boogabooga

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You started your burn too early. You need to pass your theoretical burn point half-way through your burn, not at the end of your burn. Start your burn 72.5 seconds before the burn point and see what happens.

Having said that, you seem sufficiently serious that you should learn to use TransX and IMFD. Both are much more sophisticated navigation tools than Transfer MFD.
 

Nicholas Kang

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Hi boogabooga,

I will now try out your plan. I had succeeded in using IMFD to fly to Mars yesterday (Malaysian Local Time). Thanks for your advice. Now, I will burn to Mars. Stay tune for my reply.

Regards,

Nicholas
 

Keithth G

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Hello,
I started with u-value of 7.746 km/s. Delta-V is approx. 3.110km/s. So, v-value is 10.856km/s.

Supposing that you are in a near circular orbit with an orbital speed of 7.746 km/s, you need 3.209 km/s of dV to escape Earth's gravity - i.e, escape velocity is 10.955 km/s. So, with a target dV of 3.110 km/s, you are not even going to make escape velocity.

What I think is going on here, I think, is that the 3.110 km/s is the 'hyperbolic escape velocity' - i.e., the speed the spacecraft needs to have in order to reach Mars once it has escaped Earth's gravity well. I comparison, escape velocity is where you 'only just' escape Earth with a hyperbolic escape velocity of 0.0 km/s.

To achieve a hyperbolic excess velocity of 3.110 km/s from a circular orbit with orbital speed 7.746 km/s, you need to increase your speed to 11.387 km/s (not the 10.856 km/s that you originally targeted).

N.B. The general formula for calculating escape velocity from a circular orbit with speed [MATH]v_c[/MATH] is:

[MATH] v_{esc}=\sqrt{2}\,v_c [/MATH]
And, the general formula for calculating the speed you need to reach in order to attain a hyperbolic excess velocity of [MATH]v_\infty[/MATH] is:

[MATH] v_{target}=\sqrt{v_\infty^2 + 2\,v_c^2} [/MATH]
This second equation is the same as the first when [MATH]v_\infty = 0[/MATH]
 
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Nicholas Kang

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Hi Keithth G,

I understand the words in bold. But I don't understand how you get the value 11.387 km/s.

Also, you stated the value 0.0 km/s in one of the lines above. Is it due to typing mistake/error?

Regards,

Nicholas
 

Keithth G

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Also, you stated the value 0.0 km/s in one of the lines above. Is it due to typing mistake/error?

No, it isn't an error. Basically, the hyperbolic escape velocity is the speed the spacecraft has when it has finished climbing up out of the Earth's gravity well. The lowest speed that a spacecraft can have when it climbs out is 0.0 km/s. (N.B., this 0.0 km/s is with respect to Earth's motion. Relative to the Sun, the spacecraft will still be moving at Earth's orbital speed around the Sun - i.e., 29.8 km/s).

Or, to put another way: The energy to climb out of the gravity well has to come from somewhere. In this case, the energy comes from the spacecraft's kinetic energy - i.e., in a reduction of its speed (relative to Earth).
 
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Urwumpe

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No, it isn't an error. Basically, the hyperbolic escape velocity is the speed the spacecraft has when it has finished climbing up out of the Earth's gravity well. The lowest speed that a spacecraft can have when it climbs out is 0.0 km/s. (N.B., this 0.0 km/s is with respect to Earth's motion. Relative to the Sun, the spacecraft will still be moving at Earth's orbital speed around the Sun - i.e., 29.8 km/s).

It helps remembering that 0.0 is a theoretical value for a two-body problem, not a real world scenario. You leave the gravity well of planet after an infinite distance there and all this funky math works out easily in a theoretical world with frictionless vacuum, infinitetly long conductors and Siemens air anchors.

In reality, it is a lot more complicated at such low excess velocities, but the big excess velocities will work with only small differences to the theory. Just without the need to travel an infinite distance.

You have a three body problem then (Earth-Sun-Spacecraft) and you simply need to imagine the excess velocity as the velocity relative to Earth that you will have left when the gravity of the Earth has only a minor influence on you - or the other way around.
 

Nicholas Kang

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I would like to say a big thank you to both Keithth G and Urvumpe. I can successfully calculated the value required. Please verify below:

v-target=square root (3110 squared)+2(7746 squared)
=11387.4 m/s
=11.387 km/s

* I apologize because I don't know how to type equations like what Keithth G had done in previous reply. Maybe both experts can teach me?

Also, I would like to confirm if I need to use the equations provided by Keithth G whenever I calculate hyperbolic escape speeds from any planetary body.

Finally, may I know how to derive the equations provided by Keithth G. Any source(s) available?

Thanks for both of your help. It proves invaluable to me.

Your mentee,

Nicholas
 

Urwumpe

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Finally, may I know how to derive the equations provided by Keithth G. Any source(s) available?

Practically you already start out there with Newton.

But a cheap textbook is for example "Fundamentals of Astrodynamics".
 

Keithth G

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Nicholas

Yes, that's the right way to calculate it.

I apologize because I don't know how to type equations like what Keithth G had done in previous reply. Maybe both experts can teach me?

If you double-click on the equations in the earlier post, a little box should 'pop-up'. This box shows how to write the equation using ASCII characters. If you cut & paste all of the text in that box and insert it into your post, others will see an equation when they read your posts. Moe broadly, the equations are written in a type-setting language called LaTeX. I suggest you google to find an introduction to mathematical type-setting using LaTeX.


Also, I would like to confirm if I need to use the equations provided by Keithth G whenever I calculate hyperbolic escape speeds from any planetary body.

Yes, these equations are quite general. They apply to any gravitating body.


Finally, may I know how to derive the equations provided by Keithth G. Any source(s) available?

The second of these expressions is just the law of energy conservation in disguise. You probably know that the sum of kinetic and potential energy for systems of this kind is conserved. The general expression for the total energy of an object in the gravitational field of, say, a planet is (in LaTeX again!):

[MATH] E = \frac{1}{2}m\,v^2 - m\,\mu / r [/MATH]
where [MATH]E[/MATH] is the total energy of the object; [MATH]v[/MATH] is the speed of the object (relative to the planet); [MATH]m[/MATH] is its mass; [MATH]\mu[/MATH] is the gravitational constant for the planet; and [MATH]r[/MATH] is the distance from the centre of the planet.

Now, this equation holds true no matter where the object is - or how far it is travelling. It is true, then, in two specific circumstances: a) when the object is very far away from the planet (Udwumpe's 'at infinity'); and b) when the object is at periapsis - the point of closest approach to the planet. Let's calculate E in these two cases:

At infinity:

[MATH] E = \frac{1}{2}m\,v_\infty^2 [/MATH]
where [MATH]v_\infty[/MATH] is the speed of the object (relative to the planet) when it is very far away. This is also known as the 'hyperbolic excess velocity'.


At periapsis:

[MATH] E = \frac{1}{2}m\,v_p^2 - m\,\mu / r_p[/MATH]
where [MATH]v_p[/MATH] is the speed of the object at periapsis; and [MATH]r_p[/MATH] is the distance of the object from the centre of the planet at periapsis.

Since energy is conserved, these two quantities are equal:

[MATH] \frac{1}{2}m\,v_\infty^2 = \frac{1}{2}m\,v_p^2 - m\,\mu / r_p[/MATH]
We can immediately get rid of [MATH]m[/MATH] in this equation:

[MATH] \frac{1}{2}\,v_\infty^2 = \frac{1}{2}v_p^2 - \mu / r_p[/MATH]
And we can multiply through by '2':

[MATH] v_\infty^2 = v_p^2 - 2 \mu / r_p[/MATH]
Now, it turns out if you are in a circular orbit around a planet with radius [MATH]r_p[/MATH] then your (circular) orbital speed is:

[MATH]v_c = \sqrt{\mu / r_p}[/MATH]
So, putting this into the above, we get:

[MATH] v_\infty^2 = v_p^2 - 2 v_c^2[/MATH]
or

[MATH] v_p = \sqrt{v_\infty^2 + 2 v_c^2}[/MATH]
which is the result as required.

N.B. Note that this derivation didn't make any assumptions about 'which' planet we were talking about - or 'which' object we had. So, the above result is quite general. It always applies - so long as we don't have to worry about perturbations from smoother gravitating body.
 
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Nicholas Kang

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What a great explanation, Keithth G and special thanks to Urwumpe for your kind sharing of cheap textbooks.

However, for Urwumpe, you should know that weakened Malaysian Ringgit is an extra burden for those live in Malaysia. Any free online academic books that that can be downloaded free. I mean textbooks, not JPL's book provided in the Orbiter link. I had downloaded that book, but I think it is not technical enough. When I learn things in orbital mechanics, I want to be serious, not through colorful pictures and simple analogies. You know, I am preparing for my university courses in the near future.

For Keithth G, I would like to ask where can I learn LaTeX?

Thanks for both of your help and assistance. What a warm welcome. My first question is "actively answered."

Thanks a lot.

Regards,

Nicholas
 

Urwumpe

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However, for Urwumpe, you should know that weakened Malaysian Ringgit is an extra burden for those live in Malaysia. Any free online academic books that that can be downloaded free. I mean textbooks, not JPL's book provided in the Orbiter link. I had downloaded that book, but I think it is not technical enough. When I learn things in orbital mechanics, I want to be serious, not through colorful pictures and simple analogies. You know, I am preparing for my university courses in the near future.

Well, it did cost me about 10€ and was worth much more of it. :lol: And more technical than even using generic units for the math is hard to find.

https://books.google.de/books/about/Fundamentals_of_Astrodynamics.html?id=UtJK8cetqGkC&redir_esc=y


If you google for "Bate Mueller White Fundamentals of Astrodynamics" you can also find a link to a uni-lj.si URL with a (illegal) PDF version of the book, which looks correct and complete. I will report them to the men in black later. I hope you can understand that I can't post links to copyright violations here. :tiphat:

---------- Post added at 05:14 PM ---------- Previous post was at 05:07 PM ----------

For Keithth G, I would like to ask where can I learn LaTeX?

Most universities you can find Latex tutorials in PDF form, especially the German ones have many tutorials and cheatsheets for mathematical functions.

Maybe this is already helpful for you:

https://www.tug.org/twg/mactex/tutorials/ltxprimer-1.0.pdf (Tutorial VIII is about math)


I have still decided to buy a good German book on the topic because the tutorials don't help you much about doing bigger projects in LaTeX or which module out of five similar alternatives existing in LaTex is the best solution to a problem (There are thousands of modules in LaTex, but Orbiter-Forum supports only the basic mathematical functions)
 
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Keithth G

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For Keithth G, I would like to ask where can I learn LaTeX?

As Urwumpe notes, this forum is only capable of dealing with LaTeX 'snippets' - it can't process full LaTeX. But the following is a 'cheat sheet' for type-setting equations in LaTeX. It is all you really need to know to get going writing equations in posts in this forum.

http://users.dickinson.edu/~richesod/latex/latexcheatsheet.pdf
 
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Nicholas Kang

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Again I would like to thank Urwumpe for your kind assistance. That is a great PDF book, albeit illegal. I still enjoy the contents, especially the formulae that seem natural to me. Sadly, it is illegal. Hopefully ebooks in astrodynamics can be freely distributed one day. I know, not many would be like Dr. Martin spending his free time creating Orbiter. But for ebooks, I think the community should help out. I suggest the forum members unite to create a technical ebook like the one Urwumpe mentioned that is freely available.

Anyway, thank you Urwumpe.

For both Keithth G and Urwumpe, thank you for your links to the LaTeX language "manual" PDF. Both of you have been a great help in my learning journey. I learnt a lot from both of you. Glad to receiver help from both Asia and Europe at the same time.

Lastly, to both Urwumpe and Keithth G, can I ask both of you is what boogabooga said in the first reply true? I mean is it true that my eject burn should be done at half of my DTe value calculated?

Thanks.

Regards,

Nicholas.
 

Urwumpe

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Lastly, to both Urwumpe and Keithth G, can I ask both of you is what boogabooga said in the first reply true? I mean is it true that my eject burn should be done at half of my DTe value calculated?

Depending on how TIG (Time of ignition) is calculated, yes.

In reality you calculate that differently depending on the strategy your guidance system follows. About half of the DV before the node and half of it after the node is a common scheme, but for eject burns other solutions may be more effective. Also, you don't burn in one direction in reality, but slowly change your attitude to minimize gravity losses during longer burns.

IMFD does that optimization pretty well. A pretty common software algorithm in the Orbiter world is Powered Explicit Guidance, you can find it in many add-ons which have an autopilot included.
 
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...Lastly, to both Urwumpe and Keithth G, can I ask both of you is what boogabooga said in the first reply true? I mean is it true that my eject burn should be done at half of my DTe value calculated?...
The "right" point in space where your MFDs tell you to ignite your engines in order to go to your selected destination is one.
IF you had MEGA-futuristic magic engines that could deliver ALL your needed dV in just one click, you shouldn't worry about splitting your burn half before and half after that point.

Bur since your engines deliver their power through a burn extended in time (several seconds at least), you need to resort to that technique.

After a critical burn duration (mainly depending on your Isp), your needed dV will gradually be reached along the course of several orbits, where each time you are at "burn point" you burn your engines and increase your speed (see Mangalyaan).
 

Keithth G

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Lastly, to both Urwumpe and Keithth G, can I ask both of you is what boogabooga said in the first reply true? I mean is it true that my eject burn should be done at half of my DTe value calculated?

If you work through the maths, if you are in uniform gravitational field with and apply a constant acceleration, doing half the burn before the manoeuvre date and half the burn after the manoeuvre date is equivalent to doing all of the burn, instantaneously, at the manoeuvre date. This is known as 'balancing the burn'.

For short burns, it is a useful (and accurate) technique. But, as Urwumpe points out, for longer burns the underlying assumptions - constant gravitational field and constant acceleration - break down. For escape burns, in particular, given their long duration, you need to use more sophisticated approaches if you want an accurate result. Not only does the curvature of the Earth's gravitational field become significant, but the spacecraft's acceleration will increase near exponentially through the burn because of its decreasing mass.

Working out optimal burn vectors and timings is not straightforward. For the most part, its easiest to use a solution that has already been developed - e.g. TransX, IMFD.
 
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Nicholas Kang

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Hello Ripley.

So, basically I can only only rely on external MFD to fly to Mars? Also, Keithth G and others, when do you consider a burn as short burn and when is it a long burn?

Do you consider flying to Moon a short burn?

Without any external MFD, my burn using Transfer MFD to Mars can NEVER be accurate?
 
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