Astrodynamics made simple (no, really!)

[GoSlash27]

I will try to explain the math and physics behind orbital mechanics as simply as I can in text and pictures. If I do it right, you should fully understand how to solve these problems in order to plan your missions.

I will start with gravity and uniform circular motion, and progress from there to elliptical orbits/ vis-viva, Orbital DV, escape/ interplanetary transfers, gate orbits, etc.

This is liable to take some time to complete, so please bear with me. If you have any questions/ corrections/ comments, please post them here:

Best,
-Slashy

Edited April 10, 2016 by GoSlash27

[GoSlash27]

1a) Gravity

 Gravity is the force of attraction between two bodies due to their mass and proximity. You can think of it as a bowling ball on a trampoline; the closer a marble is to the bowling ball, the more attraction will be produced.

"G" is the universal gravitational constant, and can be thought of as the "looseness" of the trampoline surface. This value is constant both in our universe and KSP itself;

G=6.67408x10^-11

M is the mass of our bowling ball (or moon/ planet/ sun) in kilograms.

r is our distance from the center of the bowling ball in meters.

a marble's attraction to the bowling ball is directly proportional to the mass of the bowling ball and inversely proportional to the square of the distance.

Our marble's attraction to the bowling ball is therefore:

a=GM/r² 

Some examples:

Jeb in his skivvies in the VAB.

Kerbin's mass is 5.292E22. G is always 6.674E-11. Kerbin's sea level radius is 600 kilometers.

a=GM/r²

His acceleration towards the planet's core is therefore (5.292E22)*(6.674E-11)/(600E3)² = 9.811 m/sec²

Standing on the Mun, it's 9.76E20*G/(200E3)²=1.628 m/sec²

2 million meters away from Duna... you get the idea.

Now... G and M are best buddies and we never see the one without the other in these problems. To simplify things, we therefore multiply them together and refer to them as "mu".

u=GM

In other discussions, you may see it referred to as mu or μ. In this tutorial, it will be represented as u.

 

So to summarize,

a=GM/r²
u=GM
therefore..
a=u/r²
 

_________________________________________________________________________________________________

1b) Uniform circular motion

To understand a uniform circular orbit, we must first start with centrifugal acceleration.

Swinging an object around in a circle will produce an acceleration felt by the body away from the center. This acceleration is directly proportional to the square of the rotational velocity and inversely proportional to the radius.

a_c=V²/r

A circular orbit results when the acceleration of gravity (a_g) is counterbalanced by the centrifugal acceleration of orbiting the planet (a_c)

Spoiler

Note: This is a convenient fib. It's not *actually* what's going on. The truth is that our satellite has just enough velocity to fall around the body and miss it instead of falling into it or getting flung further away. There's no such thing as "centrifugal force", it's just a mirage.
 Having said that... it's "correct" enough for our purposes and the math works, so let's just roll with it.

a_g=a_c

To find the velocity this happens at, we set them as equal,

u/r²=V²/r

and reorganize to isolate V.

u*r/r²=V²*r/r

u/r=V²

sqrt(u/r)= sqrt(V²)

sqrt(u/r)= Vorb

 

Edited April 11, 2016 by GoSlash27

[GoSlash27]

2) Orbital periods

We can derive a simple formula for orbital periods based on radius. This is extremely handy when we start playing with it later.

Since we have our radius, our circumference is 2πr.

We have already defined our orbital velocity as sqrt(u/r)

Our orbital period is the time it takes to complete an orbit, so

period (in seconds) = circumference (meters) divided by orbital velocity (m/sec)

p=C/Vorb.

C= 2πr and Vorb= sqrt(u/r) , so

p=2πr/[sqrt(u/r)]

Throwing some algebra at it to tidy it up...

p=2πr[sqrt(r/u)]

p=2π[sqrt(r³/u)]

Perhaps we want to know the reverse of this; we need an orbital radius with a specific period. This is just a matter of reorganizing the equation.

p=2π[sqrt(r³/u)]

p/2π=sqrt(r³/u)

(p/2π)²=r³/u

up²/4π²=r³

r= cuberoot(up²/4π²)

 

[GoSlash27]

3a) Elliptical orbits

We often have a need to make orbits that are elliptical instead of perfectly circular. Let's take a circular orbit and "squish" it a bit to see what happens...

As you can see, our central body will lie at a focal point of the ellipse. This means that our orbit will have a high point and a low point. They are referred to as "Apoapsis" (high like apex) and "Periapsis" (low like a periscope... or Paris Hilton ). They are abbreviated as Ap and Pe, respectively.
 The orbit will behave like it's average radius as far as period and average velocity are concerned. This "average radius" is called Semi-major axis, or SMA.

 The first thing to notice about it is that the periapsis is just as far below the semi-major axis as the apoapsis is above it. This holds true for all elliptical orbits and gives us our first set of equations:

SMA= (Ap+Pe)/2

Likewise...

Ap= 2SMA-Pe
and
Pe= 2SMA-Ap

 

Now as for the motion...

That's a bit more complicated. As you can see, an object in an elliptical orbit moves much faster at Pe than it does at Ap.

Because our object sweeps out the same area of the ellipse in the same period of time, our velocity at periapsis divided by average velocity will always equal average velocity divided by velocity at Apoapsis.

V_pe/V_avg= V_avg/V_ap

This gives us some new handy rules:

V_avg=sqrt(V_apV_pe)

V_pe=V_avg²/V_ap

V_ap=V_avg²/V_pe

Another observation worth noting:

V_pe*Pe=V_ap*Ap, so

V_pe=V_ap*Ap/Pe

V_ap=V_pe*Pe/Ap

Pe=Ap*V_ap/V_pe

Ap=Pe*V_pe/V_ap

3b) The Vis-viva equation

Putting all of this together and throwing some math at it yields a supremely valuable equation:

If we have the u for our system, the SMA for our orbit, and our current radius, this will yield our velocity at the moment.

 

So now to start putting all of this to use...

 

 

 

 

 

Edited April 13, 2016 by GoSlash27

[GoSlash27]

4) Applied single- body mechanics

4a) Jeb launches from the Mun!

Launching from the surface of an airless body requires 4 distinct phases:

1: Burn horizontally (without losing altitude) to establish orbit at sea level

2: Burn horizontally to get enough velocity to establish your desired apoapsis

3: Go eat a snack until you reach apoapsis,

and

4: Burn horizontally at apoapsis to circularize your orbit.

For step 1, we must figure out what velocity would put us into a circular orbit at sea level.

The Mun's mass is 9.760x10²⁰ kg and G is always 6.674x10^-11, so u = MG = 6.514x10¹⁰.

Our sea level orbital velocity would be sqrt(u/r) and the Mun's sea level radius is 200km.

6.514x10¹⁰/2x10⁵=3.257x10⁵

sqrt(3.257x10⁵)=570.7 m/sec.

For step 2, we must define the SMA of our transfer altitude.
 Let's say our desired orbital altitude will be 14km. This means our Ap will be 14km plus the Mun's sea level radius of 200km =214 km.
Our periapsis will be sea level, or 200 km. Our SMA will therefore be (214+200)km/2=207km

Our velocity at periapsis in this elliptical transfer orbit is defined by the vis-viva equation.

Vpe=sqrt[u(2/Pe-1/SMA)]

Vpe=sqrt[6.514x10¹⁰(2/2x10⁵-1/2.07x10⁵)]

=sqrt[6.514x10¹⁰(5.169x10^-6)]

=sqrt(3.367x105)

Vpe=580.3 m/sec

We will lose speed as we sit there eating snacks, so we need to know how fast we will be going when we reach Apoapsis. This is figured the same way as we did Vpe.

Vap=sqrt[u(2/Ap-1/SMA)]

I'll skip the figuring this time.

Vap=542.3m/sec

Finally in step 4, we need to increase this speed to a proper speed for orbit at Ap; 214km. This is figured the same as step 1.

Vorb=sqrt(u/r)=551.7 m/sec.

So we started at zero. We increased our speed to 570.7, then increased it further to 580.3. Our speed dropped to 542.3 as we coasted, and finally we boosted it to 551.7 m/sec.

Our total change in velocity was 580.3+(551.7-542.3) m/sec, or 589.7 m/sec.

Since the Mun rotates clockwise at 9 m/sec, We could subtract that if we launch East, yielding 580.7 m/sec DV.

This is the DV required to reach orbit from the Mun.

 

 

 

 

 

Edited April 13, 2016 by GoSlash27

[GoSlash27]

4b) BadS="True", period!

 I love to use orbits that appear overhead KSP on a regular schedule whenever possible. My favorite is the 36 minute schedule, which passes overhead KSC 10 times a day.

Since Kerbin rotates 1 time per day itself, this means our orbit will need to occur 11 times per day in the sidereal frame.

Kerbin's sidereal period is 5 hours, 59 minutes, 9.4 seconds. That's 21,549.4 seconds. Our desired orbital period would therefore be 21,549.4/11= 1,959.04 seconds.

r= cuberoot(up²/4π²)

r= cuberoot(3.433x10¹⁷)

r= 700,212 m

Subtracting Kerbin's radius leaves an altitude of 100,212m

If we put something into this orbit, we will know precisely when it will appear overhead KSC simply by looking at the clock. It will pass overhead once every 36 minutes. Not only that, but we will know precisely where it is in it's orbit around Kerbin at any given time. It's always moving East at 10° per minute, 1° every 6 seconds. If we mark it's passage overhead KSC, we know where it is simply by the clock on the wall.

This allows us to use launch windows to intercept stations in this orbit from KSC, pinpoint the longitude of interesting landmarks on the ground, use windows to set up resonant transfer orbits and intercepts for the Mun and beyond.

 

 

 

 

 

Edited April 13, 2016 by GoSlash27