ohmannExpress:

When it Absolutely, Positively Has To Be There Overnight

The Hohmann Transfer Orbit is one of the most delta-V efficient transfer orbits

there is but the efficiency comes at the price of having to use very precise

departure and arrival times as well as very long travel times. (There are more

efficient orbits but they have even longer travel times and often require

gravitational slingshot maneuvers.)

Efficiency is important because even in the 21^{st} century, we are stuck using

wimpy chemical rockets that make you watch every milligram of payload. Even

atomic rockets try to use Hohmann transfers orbits or suffer large payload mass

penalties.

Calculating Hohmann orbits is also useful as a measuring rod to illustrate the tradeoffs

you have to make between different types of rockets, mass ratios and flight

times.

It’s not like the good old days where Flash, Buck, Tom or Dan just jumped into their

rockets, lit the candle and a short while later crashed into Venus, Mars or

Mongo. The average Hohmann transfer time going one way between Earth and Mars

is about 9 months. Being packed like spam in a can for months on end is, let’s

face it, boring. Unless your crewmate is a homicidal maniac, flying between the

planets isn’t as exciting as most eager young space cadets think.

That’s the reality.

Basically, reality stinks

Quick philosophical question. What is reality?

Answer. Reality is what you can get away with.

If you can’t do something one way, try another. If you have to, change the

conditions so you can do what you want. (I learned that in a movie) In other

words, cheat.

Most science fiction stories cheat by changed the way gravity works or just by

ignoring the huge amount of energy required to move stuff around quickly in

space. It’s an understandable cheat, but everyone does that.

I want to do something different.

I want to make a really, really small solar system that is easier to get around in.

Something that doesn’t take month or years to move around in. I’m calling it the Ognom System.

Ognom is a large Gas Giant in the habitable zone of a star. It has several very large

planet sized moons orbiting it. Some of which are life bearing. The basic data about the Ognom System is in the

following tables.

Ognom Mass KG |
2.079E+28 |

Ognom Radius KM |
70000 |

Ognom Volume KM^3 |
1.436755E+15 |

Ognom Density KG/M |
1.447E+05 |

Here is the basic data for five of the major moons in orbit around Ognom.

Name |
Kalmor |
Palenmor |
Malmor |
Balanmor |
Xalamor |

SemiMajor Axis |
450257 |
658709 |
1045635 |
1370169 |
2175008 |

SemiMajor Axis |
3.01E-03 |
4.40E-03 |
6.99E-03 |
9.16E-03 |
1.45E-02 |

Eccentricity |
0.00200 |
0.00100 |
0.00150 |
0.00200 |
0.03000 |

Inclination |
0.03 |
0.2 |
2 |
3 |
5 |

Orbital Period |
0.5898 |
1.0437 |
2.0874 |
3.1311 |
6.2622 |

Orbital Velocity |
55.51349 |
45.896772 |
36.428292 |
31.82304 |
25.25797 |

Radius KM |
1821.6 |
5285 |
3430 |
4025 |
3750 |

Radius |
0.285607 |
0.8286297 |
0.5377861 |
0.631076 |
0.587959 |

Diameter KM |
3643.2 |
10570 |
6860 |
8050 |
7500 |

Circumference |
11445.45 |
33206.634 |
21551.326 |
25289.82 |
23561.94 |

Surface Area |
4.17E+07 |
3.51E+08 |
1.48E+08 |
2.04E+08 |
1.77E+08 |

Surface Area |
0.08 |
0.69 |
0.29 |
0.40 |
0.35 |

Mass KG |
1.32E+23 |
2.50E+24 |
6.57E+23 |
1.17E+24 |
1.10E+24 |

Mass (Earth=1) |
0.02 |
0.51 |
0.11 |
0.20 |
0.18 |

Volume |
2.53E+10 |
6.18E+11 |
1.69E+11 |
2.73E+11 |
2.21E+11 |

Density Kg/M^3 |
5202 |
4050 |
3887 |
4273 |
4971 |

Surface Gravity |
0.27 |
0.61 |
0.38 |
0.49 |
0.53 |

Surface Gravity |
2.649 |
5.983 |
3.727 |
4.808 |
5.211 |

Escape Velocity |
3.11 |
7.95 |
5.06 |
6.22 |
6.25 |

DeltaV Liftoff |
2196.48 |
5622.83 |
3575.17 |
4398.59 |
4420.24 |

DeltaV Liftoff |
2.20 |
5.62 |
3.58 |
4.40 |
4.42 |

Since I am interested in quickly traveling between worlds, the first question I want to

ask is how long does it take to get from, lets say Malmor to Palenmor?

The equation for the transfer times for Hohmann transfer orbits is:

Where:

t_{H}= Hohmann transfer time

π = pi = 3.14159265359

a = Semi-Major Axis of the Hohman Orbit (Has to be calculated. We don’t have

this yet)

r_{1} = Radius of the Interior (Or Inferior) Planet’s Orbit

r_{2} = Radius of the Exterior (Or Superior) Planet’s Orbit

µ = Standard Gravitation Parameter which is found by multiplying the universal gravitation constant times the mass of the primary body. In this case Ognom is the primary body and its mass is 2.079E+28 kg. So the standard gravitational parameter is: 6.67348E-11 m^{3} kg^{-1} s^{-2 } * 2.079E+28 kg = 1.388E^{+18 } m3 s^{-2}

We’re going to use the second equation that just uses the orbital radius of the two planets. To get the transfer time, insert the orbital radius (or semi-major axis) of the planet you are departing from and the one you are arriving at to find the length of time it takes to travel between the two worlds.

Worked example: Traveling from Malmor to Palenmor

π * sqrt(((658709KM + 1045635KM)^{3})/(8*1.388E^{+18} m3 s^{-2})

= 66344.832 seconds

= 1105.7472 minutes

= 18.42912 hours

= .76788 days

18 and a half hours travel time from Malmor to Palenmor. Not bad. It’s a heck of a lot better than the 9 month average to get from Earth to Mars.

Finding the one way travel times between the other worlds is left as an exercise for the reader.

Traveling using a Hohmann transfer orbit requires that some specific conditions be met before you can start. You can only do this when the two planets are aligned in the correct position. If they are out of alignment you can go ahead and rocket off but the planet you are wanting to visit won’t be there to meet you.

Since we now know how long it takes to get from our departure and destination planet we now need to find what position your destination planet has to be at when you depart..

To do this we try and find the phase angle of departure between the two planets. The phase angle is made by forming a triangle between the current position of planet A, the central mass, or orbital focus, and the current position of planet B.

The easy way to calculate this is to use the following equation.

r_{1} = Radius of the departure Planet’s Orbit

r_{2} = Radius of the destination Planet Orbit

So for our worked example we can plug in the numbers to find that the phase angle needs to be

180 * ((1- (658709KM + 1045635KM)/(2*658709KM))^{1.5}) = -84.864 degrees

What this means is that the destination planet needs to be 84.864 degrees behind the departure planet when viewing their orbits from above. When the calculation gives you a negative angle then the destination planet needs to be trailing behind

the departure planet as they travel counter clockwise. If the calculation gives you a positive angle then the destination planet needs to be ahead of the departure planet.

Basically finding the phase angle is just a fancy, but exact way to try to hit a moving target. Basically you are just leading your target. When shooting at a moving target, you don’t aim at where the target currently is, you aim at where the target will be when your shot arrives. The phase angle gives you the proper amount of lead for the shot so you can hit the target.

Take a look at the diagram below.

At a start time, lets call it t1, Malmor and Palenmor are in the correct position so you can use a Hohmann transfer orbit to travel from Malmor to Palenmor. At time t1, Palenmor is on the left side of the diagram and Malmor is at the bottom. The phase angle

between Palenmor and Mallmor is -84.864 degrees. Since Palenmor is in the interior orbit, it’s orbital velocity is faster then Malmor’s. As times passes, Palenmor will move ahead of Malmor until it reaches it’s new position toward the top of the diagram at our arrive time t2. The difference between times t1 and t2 is the same as the Hohmann transfer time of 18.42912 hours. In 18.5 hours Palenmor moves around almost ¾’s of it’s orbit. In the same amount of time Malmor moves through almost 1/3^{rd} of it’s orbit. You, traveling in your rocketship, move half way through your Hohmann orbit until you arrive at your destination.

The next thing we need to know is how often do the planets to line up in this position. The amount of time this takes is called the Synodic Period. To find this we use this equation.

p_{1} = Period of the Departure Planet’s Orbit

p_{2} = Period of the Destination Planet’s Orbit

For our worked example the synodic period is

1/((1/1.0437) – (1/2.0874)) = 2.0874 days

That means we can use the Hohmann orbit in a little over every two days to get from Malmor to Palenmor.

To summarize, if you miss your first change to go from Palenmor to Malmor you have

to wait a little over 2 days before the planets to align again so you can start

on you flight to Palenmor.

So compared to getting from earth to mars, arriving in less then 3 days ain’t bad.

Below is a table of Hohmann transfer information for moons of Ognom.

To Kalmor | |||

From | Travel Time Days | Phase Angle | Synodic Days |

Palenmor | 0.40 | -65.99 | 1.36 |

Malmor | 0.63 | -205.38 | 0.82 |

Balanmor | 0.85 | -337.36 | 0.73 |

Xalamor | 1.47 | -715.98 | 0.65 |

To Palenmor | |||

From | Travel Time Days | Phase Angle | Synodic Days |

Kalmor | 0.40 | 40.98 | 1.36 |

Malmor | 0.77 | -84.86 | 2.09 |

Balanmor | 1.00 | -164.01 | 1.57 |

Xalamor | 1.65 | -387.84 | 1.25 |

To Malmor | |||

From | Travel Time Days | Phase Angle | Synodic Days |

Kalmor | 0.63 | 71.11 | 0.82 |

Palenmor | 0.77 | 47.57 | 2.09 |

Balanmor | 1.30 | -43.49 | 6.26 |

Xalamor | 1.99 | -164.01 | 3.13 |

To Balanmor | |||

From | Travel Time Days | Phase Angle | Synodic Days |

Kalmor | 0.85 | 82.54 | 0.73 |

Palenmor | 1.00 | 65.33 | 1.57 |

Malmor | 1.30 | 31.01 | 6.26 |

Xalamor | 2.30 | -84.86 | 6.26 |

To Xalamor | |||

From | Travel Time Days | Phase Angle | Synodic Days |

Kalmor | 1.47 | 95.61 | 0.65 |

Palenmor | 1.65 | 85.36 | 1.25 |

Malmor | 1.99 | 65.33 | 3.13 |

Balanmor | 2.30 | 47.57 | 6.26 |

By looking at the table, the shortest travel time is less than half a day, the longest is less than two and a half days. Synodic Period range from .65 days to a little over six and a quarter days.

Finally we need to know that amount of delta v, or change in velocity, we need to make to move from one orbit to another. For this I am assuming we need to make two rocket burns break orbit on departure and then brake at our destination. The equations for the total delta-v using two instantaneous rocket impulses for departure and arrival is:

Δv1 = Change in velocity when we depart.

Δv2 = Change in velocity when we arrive.

Δvtotal = Total change in velocity

r_{1} = Radius of the Interior (Or Inferior) Planet’s Orbit

r_{2} = Radius of the Exterior (Or Superior) Planet’s Orbit

µ = Standard Gravitation Parameter = 1.388E^{+18} m3 s^{-2}

Δv1 for our example is found by

sqrt(1.388E^{+18} m3 s^{-2} / 658709KM) * (sqrt((2*1045635KM)/( 658709KM + 1045635KM)) – 1)

Δv1 = 4.40087 KM/S

Δv2 for our example is found by

sqrt(1.388E^{+18} m3 s^{-2} / 1045635KM) * (1 – sqrt((2*658709KM)/( 658709KM + 1045635KM)))

Δv2 = 4.94358556 KM/S

Δvtotal is therefore:

Δvtotal = 4.40087 + 4.94358556 = 9.344459 KM/S

So any spaceship we use has to be capable of changing it velocity by a total of 9.34 KM/S. This is actually more then the amount of Δv you need to get from Earth to Mars. So that is another trade off we have to deal with by making this miniature planetary system.

For more details on orbital mechanics and rocketing around space I highly recommend the Atomic Rockets and the Basics of Space Flight websites.