PeriapsisPrograde
Wannabe addon dev
I have recently taken an interest in doing the math for my orbital maneuvers in Orbiter and Kerbal Space Program by hand. I am mostly writing these down for my own reference (sorry for the sloppy disorganization), but perhaps they will be useful for someone else, so I am publishing them here.
There are many different things to consider when planning an interplanetary trajectory. You need to account for the delta-v and timings of everything for you heliocentric transfer orbit as well as the hyperbolic escape from the planet of origin and hyperbolic capture at the destination planet. To simplify the calculations, we make a ton of assumptions:
These assumptions allow us to simplify the problem enough to actually get anywhere in solving it. Perhaps later we can discard some of these assumptions, but for now we use this very simplified model of the universe. Let us begin by listing some key equations:
(1) [math] F = \frac{mMG}{r^2}[/math]
(2) [math] \mu = MG [/math]
(3) [math] \epsilon = \frac{v^2}{2} - \frac{\mu}{r} = \frac{-\mu}{2a}[/math]
(4) [math] v^2 = \mu \bigg( \frac{2}{r} - \frac{1}{a} \bigg) [/math]
(5) [math] T = 2\pi \sqrt{\frac{a^3}{\mu}} [/math]
Derivation of these equations is left as an exercise to the reader. One may note that the velocity of a circular orbit is constant by using (3) and substituting [math]r=a[/math]
(6) [math] v_{circular}= \sqrt{\frac{\mu}{r}}[/math]
Use (6) to find the orbital velocity of the origin and target planet. For the rest of this blog, I will use an Earth-Mars transfer as an example. The standard gravitational parameter of the Sun is 1.327e20 m^3/s^2 and the semi-major axes of Earth and Mars are 1.496e11 m and 2.273e11 m, respectively,
[math]v_{Earth} = \sqrt{\frac{1.327*10^{20} m^3s^{-2}}{1.496*10^{11} m}}[/math] = 29,783.1 m/s. The velocity of Mars is 24,162.2 m/s.
The semi-major axis of our transfer orbit is the average of the planets' axes:
(7) [math]a_{transfer} = \frac{a_{Earth}+a_{Mars}}{2}[/math] = 1.885e11 m.
Using (4) and the semi-major axis just found, we can find the necessary heliocentric velocity at Earth and the velocity at Mars:
[math]v_E = \sqrt{1.327*10^{20}m^3s^{-2} \bigg( \frac{2}{1.885*10^{11} m}-\frac{1}{1.496*10^{11} m} \bigg) } [/math] = 32,712.2 m/s. The velocity at Mars is 21,532.3 m/s.
Thus the velocity relative to Earth after ejection is 32,712.2 m/s - 29,783.1 m/s = 2929.1 m/s. This is our hyperbolic excess velocity from Earth. To determine our periapsis velocity from a parking orbit of radius r, use:
(8) [math] v_{periapsis} = \sqrt{v_{\infty}^2 + \frac{2\mu}{r}} [/math], which for r = 6.567e6 m and [math]\mu_{Earth}[/math] = 3.986e14 m^3s^-2 gives v = 11,400.6 m/s. A circular orbit at that altitude has velocity 7790.9 m/s, so the Delta-v of ejection is 3609.7 m/s.
The difference between our heliocentric velocity at apoapsis and the velocity of Mars is 2629.9 m/s. We can use (8) again to find our periapsis velocity at Mars, using r = 3.590e6 m and [math]mu_{Mars}[/math] = 4.280e13 m^2s^-2: 5546.2 m/s. The velocity of the circular orbit at that height is 3452.8 m/s, so the insertion burn around Mars will take 2093.4 m/s
Thus the sum of the burns will require 5703.1 m/s. More delta-v will be required for mid-course corrections, and some might be saved by using an aerocapture, but this is a minimum number a spacecraft should have trying to go to Mars. This concludes part one.
Part 2: Timing -- going at the right time, phase angles
Part 3: Dead reckoning -- no MFDs, no problem!
~~~~~~~~
Parts I will write once I lean how to do them:
-- How to account for eccentricities and inclinations
-- Non-Hohmann transfers
-- Non-instantaneous burns
--Scenic detours Gravity assists
-- Dealing with perturbations
There are many different things to consider when planning an interplanetary trajectory. You need to account for the delta-v and timings of everything for you heliocentric transfer orbit as well as the hyperbolic escape from the planet of origin and hyperbolic capture at the destination planet. To simplify the calculations, we make a ton of assumptions:
- We assume that only the gravity of the dominant body acts on the spacecraft at a given time (a patched-conics approximation)
- We assume the orbits of planets to be circular (zero eccentricity) and the orbits of the spacecraft are circular at the start and finish
- We assume that all motion occurs in the same plane (no relative inclinations)
- We assume that the planets have no axial tilt
- We assume a Hohmann transfer method
- We assume that burns occur instantaneously
These assumptions allow us to simplify the problem enough to actually get anywhere in solving it. Perhaps later we can discard some of these assumptions, but for now we use this very simplified model of the universe. Let us begin by listing some key equations:
(1) [math] F = \frac{mMG}{r^2}[/math]
(2) [math] \mu = MG [/math]
(3) [math] \epsilon = \frac{v^2}{2} - \frac{\mu}{r} = \frac{-\mu}{2a}[/math]
(4) [math] v^2 = \mu \bigg( \frac{2}{r} - \frac{1}{a} \bigg) [/math]
(5) [math] T = 2\pi \sqrt{\frac{a^3}{\mu}} [/math]
Derivation of these equations is left as an exercise to the reader. One may note that the velocity of a circular orbit is constant by using (3) and substituting [math]r=a[/math]
(6) [math] v_{circular}= \sqrt{\frac{\mu}{r}}[/math]
Use (6) to find the orbital velocity of the origin and target planet. For the rest of this blog, I will use an Earth-Mars transfer as an example. The standard gravitational parameter of the Sun is 1.327e20 m^3/s^2 and the semi-major axes of Earth and Mars are 1.496e11 m and 2.273e11 m, respectively,
[math]v_{Earth} = \sqrt{\frac{1.327*10^{20} m^3s^{-2}}{1.496*10^{11} m}}[/math] = 29,783.1 m/s. The velocity of Mars is 24,162.2 m/s.
The semi-major axis of our transfer orbit is the average of the planets' axes:
(7) [math]a_{transfer} = \frac{a_{Earth}+a_{Mars}}{2}[/math] = 1.885e11 m.
Using (4) and the semi-major axis just found, we can find the necessary heliocentric velocity at Earth and the velocity at Mars:
[math]v_E = \sqrt{1.327*10^{20}m^3s^{-2} \bigg( \frac{2}{1.885*10^{11} m}-\frac{1}{1.496*10^{11} m} \bigg) } [/math] = 32,712.2 m/s. The velocity at Mars is 21,532.3 m/s.
Thus the velocity relative to Earth after ejection is 32,712.2 m/s - 29,783.1 m/s = 2929.1 m/s. This is our hyperbolic excess velocity from Earth. To determine our periapsis velocity from a parking orbit of radius r, use:
(8) [math] v_{periapsis} = \sqrt{v_{\infty}^2 + \frac{2\mu}{r}} [/math], which for r = 6.567e6 m and [math]\mu_{Earth}[/math] = 3.986e14 m^3s^-2 gives v = 11,400.6 m/s. A circular orbit at that altitude has velocity 7790.9 m/s, so the Delta-v of ejection is 3609.7 m/s.
The difference between our heliocentric velocity at apoapsis and the velocity of Mars is 2629.9 m/s. We can use (8) again to find our periapsis velocity at Mars, using r = 3.590e6 m and [math]mu_{Mars}[/math] = 4.280e13 m^2s^-2: 5546.2 m/s. The velocity of the circular orbit at that height is 3452.8 m/s, so the insertion burn around Mars will take 2093.4 m/s
Thus the sum of the burns will require 5703.1 m/s. More delta-v will be required for mid-course corrections, and some might be saved by using an aerocapture, but this is a minimum number a spacecraft should have trying to go to Mars. This concludes part one.
Part 2: Timing -- going at the right time, phase angles
Part 3: Dead reckoning -- no MFDs, no problem!
~~~~~~~~
Parts I will write once I lean how to do them:
-- How to account for eccentricities and inclinations
-- Non-Hohmann transfers
-- Non-instantaneous burns
--
-- Dealing with perturbations
