Hyperbolic orbits

[reachshijo]

The hyperbolic excess speed of a spacecraft is the excess velocity of the spacecraft at the exit of the sphere of influence of the planet. Which when we are on the ecliptic of the planet will be difference between the planets velocity on the ecliptic plane minus the speed of the spacecraft.

How can we make a hyperbolic exit such that the end velocity will be tilted by an inclination.?

For example I have an incoming s/c to Earth on the ecliptic with a known velocity. What should I do to make the orbit a polar orbit?

 

[ADSWNJ]

Short answer: get familiar with either of these two add-ons: TransX or IMFD.

To do this, read their documentation (especially if Ripley has a version in Italian for you!), and search YouTube for "Orbiter 2010 David Courtney" (David = blixel on this forum). David has tons of tutorials explaining how to play around with maneuvers on TransX to get your target orbital insertion to be exactly what you need.

Oh ... and fly in the XR-5 or the XR-2, rather than the default Delta Glider, as this will give you much richer autopilots and realism (e.g. stress & heat effects), if you are into that.

Hope this helps.

 

[Urwumpe]

For example I have an incoming s/c to Earth on the ecliptic with a known velocity. What should I do to make the orbit a polar orbit?

Remember that Earth moves as well and then choose the proper conditions at which you meet Earth. If your solar orbit plane is slightly inclined to the ecliptic, you can arrive north or south of Earth.

 

[Keithth G]

How can we make a hyperbolic exit such that the end velocity will be tilted by an inclination.?

Let's suppose, for sake of discussion, that you want a hyperbolic escape velocity of 5,000 m/s. You want this to be largely in the direction of Earth's orbital path, but once you have escaped Earth's gravity well, you want in inclination of, say, 5 degrees relative to Earth's orbital plane. Although we normally just refer to the magnitude of the hyperbolic escape velocity - i.e., here, 5000 m.s - it really is a vector quantity with three components - a 'forward' component, an 'outward' component' and a 'plane' component - which one can write as [MATH](v_f, v_o, v_p)[/MATH].

If one wants the hyperbolic escape velocity to lie in the direction of the Earth's velocity vector, one just uses the 'forward' component so that the hyperbolic escape velocity vector is written as [MATH](5000,0,0)[/MATH].

If, on the other hand, one wants the orbit to be tilted at a 5 degree inclination relative to Earth's orbit plane, then one has to use both the 'forward' and 'plane' components of the velocity vector. The 'forward' component takes the value [MATH]5000\,\cos\, 5 {}^{\circ}[/MATH] and the 'plane' component takes the value [MATH]5000\,\sin\, 5 {}^{\circ}[/MATH].

At this point, one would normally use TransX or IMFD and insert these velocity components into an 'escape plane' and then let the MFD work out the details of getting the spacecraft into the right position to execute the escape burn. But one can go through these calculations manually, though, if one really wants to. The key to doing this is to define a plane using a) the position vector of the Earth (relative to the Sun) at the time of departure, b) the hyperbolic escape velocity vector of the spacecraft relative to the Earth and then c) centring this plane on the Earth. Essentially, the hyperbolic escape trajectory will lie in this plane and, moreover, the (instantaneous) escape burn will need to be executed at the perigee of that trajectory.

 

[Keithth G]

Actually, a few more quick thoughts on this:

The equation given above:
[math]
\cos 5{}^{\circ}=\frac{5000 \cos (\iota )+v_E}{\sqrt{5000^2+2\times5000 \cos (\iota )\, v_E+v_E^2}}
[/math]

has four solutions for the angle [MATH]\iota[/MATH]. Above, only one was given - 36.53 degrees. The four solutions are:

[MATH]\iota = +36.53{}^{\circ}[/MATH]

[MATH]\iota = -36.53{}^{\circ}[/MATH]

[MATH]\iota = -153.47{}^{\circ}[/MATH]

[MATH]\iota = +153.47{}^{\circ}[/MATH]

The first corresponds to an escape plane for which the 'forward' velocity component is +4017.76 m/s and the 'plane' velocity component is +2976.17 m/s. The resulting orbital inclination of the spacecraft with respect to the Sun is +5 degrees.

The second corresponds to an escape plane for which the 'forward' velocity component is +4017.76 m/s and the 'plane' velocity component is now -2976.17 m/s. The resulting orbital inclination of the spacecraft with respect to the Sun is -5 degrees.

The third corresponds to an escape plane for which the 'forward' velocity component is -4473.53 m/s and the 'plane' velocity component is -2233.128 m/s. The resulting orbital inclination of the spacecraft with respect to the Sun is +5 degrees.

The fourth corresponds to an escape plane for which the 'forward' velocity component is -4473.53 m/s and the 'plane' velocity component is +2233.128 m/s. The resulting orbital inclination of the spacecraft with respect to the Sun is -5 degrees.

The first two cases correspond to 'throwing' the spacecraft in front of the moving Earth with the sign of the 'plane' component determining the sign of the final heliocentric orbital inclination. The remaining two cases correspond to 'throwing' the spacecraft backwards behind the moving Earth.


Maximum final orbital inclination
With a hyperbolic escape velocity of 5000 m/s, there is a maximum orbital inclination (relative to the Sun) that can be achieved. If we take Earth's orbital speed around the Sun to be 30,000 m/s, the value of this maximum angle is given by:

[MATH]\sin^{-1}\left(\frac{5000}{30000}\right) = 9.59{}^{\circ}[/MATH]

To get this maximum final orbital inclination of 9.59 degrees, the value of [MATH]\iota[/MATH] is given by the expression:

[MATH]\iota = \cos^{-1}\left(\frac{-5000}{30000}\right) = 99.59{}^{\circ}[/MATH]

To achieve this maximum final orbital inclination, one needs an escape plan with -833 m/s of 'forward' and 4930 m/s of 'plane' (to achieve a +9.59 degrees final orbital inclination); and -4930 m/s of 'plane (to achieve a -9.59 degrees final orbital inclination)

 

[Keithth G]

Just a few more quick thoughts on this.

There is a special case of these 'out of plane' Earth escape plans that seems quite interesting. If one gets the hyperbolic escape velocity and escape angle just right, one can set the spacecraft on a 1:1 resonance with Earth so that, after one year )or even 6 months), the spacecraft arrives back at Earth.

Consider the escape plane for which the forward velocity is [MATH]v_E\,(\cos \iota{}^\circ - 1)[/MATH]; and the plane velocity is [MATH]v_E\, \sin \iota{}^\circ[/MATH] where [MATH]v_E[/MATH] is the Earth's orbital speed at the date of departure; and [MATH]\iota{}^\circ[/MATH] is the target inclination of the spacecraft's orbit relative to the ecliptic once it escapes Earth. Then, relative to the Sun, and after Earth escape, the spacecraft is moving at the same speed as the Earth - and is essentially on the same orbit as the Earth - but just one that happens to be inclined at [MATH]\iota{}^\circ[/MATH] to the ecliptic.

For example, if the target inclination is, say, 2 degrees and the velocity of the Earth is 30,000 m/s on the date of departure, then the escape plane would consist of a 'forward' component of -18.3 m/s; and a 'plane' component of 1046.98 (i.e., a total hyperbolic escape velocity of 1047.1 m/s. Moreover, in an Earth-centric reference frame, the spacecraft escapes Earth via a trajectory that passes over either the North or the South pole. Once it has escaped Earth, its Sun-centric speed will be 30,000 m/s (the same as Earth's); the period of its orbit will also be 365.25 days; the aphelion and perihelion of the orbit will also be the same as earth's. But the orbit will be inclined by 2 degrees. And the spacecraft will next encounter Earth after 6 months - when it next crosses the plane of the ecliptic.

And if anyone thinks that 1047.1 m/s hyperbolic escape velocity is high (fuel) price to pay for a 2 degree inclination, it is worth considering that from a 300 x 300 km orbit the escape burn needs to be 3,251 m/s. In comparison, the escape plan needed to achieve a 0 m/s hyperbolic escape is 3,201 m/s. So, the incremental cost of the 1047.1 m/s is a modest 50 m/s. Such are the wonders of the Oberth Effect.

 

[reachshijo]

I also have a very brief understanding of the effects of the rotation of the planet on the hyperbolic speeds. If we are on the ecliptic plane in the same direction as that of the rotation of the planet, what is the effect of the rotation of the planet on the speeds of the incoming spacecraft which is in the same direction.

For example, if the spacecraft reaches the planet in the same direction as the rotation of the planet(no inclination), considering the same hyperbolic parameters, what is difference between the velocities of this trajectory compared to when the spacecraft is approaching from a 90 degree tilt (ie on a polar orbit) with the same approach trajectory?

 

[boogabooga]

No effect.

The rotation does not matter, except if you start on that planet's surface.

 

[Keithth G]

'Boogabooga' is quite right: the standard 'inverse square law' for gravity doesn't care if the gravitating body (e.g., the Earth, the Moon, the Sun, etc) is rotating - the gravitational force on the spacecraft is exactly the same.

When people talk about 'escape burns', speed at periapsis and hyperbolic excess velocity, they unwritten assumption is that the reference frame is centred on the gravitating body and doesn't rotate. Because the force doesn't care about the speed of rotation of, say, the Earth, then it becomes convenient to use a 'non-rotating' reference frame: it makes the maths simpler.

If course, if you are trying to get off the ground into orbit, then Earth's rotational velocity is significant because in the non-rotating reference frame you are already moving at, say, 400 m/s. So, it makes sense to take off in the easterly direction if you can. And the effect is more profound the closer to the equator that you are. But once in orbit and clear of the atmosphere, you can go back to forgetting about the rotation of the planet - that is, until you want to land again.

 

[Keithth G]

Consider the escape plane for which the forward velocity is v_E\,(\cos \iota{}^\circ - 1); and the plane velocity is v_E\, \sin \iota{}^\circ where v_E is the Earth's orbital speed at the date of departure; and \iota{}^\circ is the target inclination of the spacecraft's orbit relative to the ecliptic once it escapes Earth. Then, relative to the Sun, and after Earth escape, the spacecraft is moving at the same speed as the Earth - and is essentially on the same orbit as the Earth - but just one that happens to be inclined at \iota{}^\circ to the ecliptic.

Earlier, I suggested this as an 'escape plan' for achieving a 1:1 Earth resonant orbit with a prescribed target inclination. Indeed, this is correct - provided that:

a. the Earth is assumed to move in a circular orbit around the Sun
b. the Earth and Moon do not co-rotate around a common barycentre
c. we base our trajectory planning on a simple 'linked-conics' gravity model

Of course assumptions a and b are false and one can easily improve upon those simple assumptions by taking into accounting Earth's elliptical motion around the Sun, and the effect that the Moon has on Earth's motion. To do that, though, one has to write a little script (rather than a simple equation). And here is a script that takes into account these particular effects:

--------------------------------------------------------
-- A routine to calculate the escape burn to achieve a 'out of plane' 1:1 resonance using a simple linked-conics model
-- 
-- Inputs:
--   1. The required inclination of the final heliocentric orbit relative to Earth's orbital plane
--   2. The altitude of the LEO prior to Earth escape
--
-- Outputs:
-- 	 1.  The elements of the Escape Burn in TransX coordinates (prograde, outward, plane-change)
--	 2.  The magnitude of the escape burn needed to effect this escape burn from LEO
--
------------------------------------


-- define relevant constants for this script
muE			= 398600441800000.0     -- Gravitational constant for the Earth

-- get the handles for the Sun, the Earth and the Moon
sun         = oapi.get_objhandle("Sun")
earth		= oapi.get_objhandle("Earth")
moon		= oapi.get_objhandle("Moon")

-- pause the simulation and set the MJD to the date of interest

str 		= proc.wait_input("Input the tilt angle (degrees):")
theta 		= tonumber(str) * math.pi /180

str 		= proc.wait_input("Altitude of LOE departure orbit (km):")
r			= (tonumber(str) + 6371) * 1000

-- get positions and masses of the gravitating bodies
q			={}
q[0]		= oapi.get_globalpos(sun)
q[1]		= oapi.get_globalpos(earth)
q[2]		= oapi.get_globalpos(moon)

p			= {}
p[0]		= oapi.get_globalvel(sun)
p[1]		= oapi.get_globalvel(earth)
p[2]		= oapi.get_globalvel(moon)

m			= {}
m[1]		= oapi.get_mass(earth)
m[2]		= oapi.get_mass(moon)

-- swap over to a conventional co-ordinate system
for i = 0, 2, 1 do
	tempq 	= q[i].y
	tempp	= p[i].y
	q[i].y	= q[i].z
	p[i].y	= p[i].z
	q[i].z	= tempq
	p[i].z	= tempp
end

-- calculate the state vectors of the Centre of Mass of the Earth-Moon system in heliocentric coordinates.
q[3]		= vec.sub( vec.div( vec.add( vec.mul(q[1],m[1]), vec.mul(q[2],m[2]) ), m[1]+m[2] ), q[0] )
p[3]		= vec.sub( vec.div( vec.add( vec.mul(p[1],m[1]), vec.mul(p[2],m[2]) ), m[1]+m[2] ), p[0] )

-- construct the rotation matrix for the velocity vector
w			= vec.div( q[3], vec.length(q[3] ))

R 			= {}

R.m11		= (1 - math.cos(theta)) * w.x * w.x + math.cos(theta)
R.m22		= (1 - math.cos(theta)) * w.y * w.y + math.cos(theta)
R.m33		= (1 - math.cos(theta)) * w.z * w.z + math.cos(theta)

R.m12		= (1 - math.cos(theta)) * w.x * w.y - math.sin(theta) * w.z
R.m13       = (1 - math.cos(theta)) * w.x * w.z + math.sin(theta) * w.y
R.m23		= (1 - math.cos(theta)) * w.y * w.z - math.sin(theta) * w.x

R.m21       = (1 - math.cos(theta)) * w.x * w.y + math.sin(theta) * w.z
R.m31		= (1 - math.cos(theta)) * w.x * w.z - math.sin(theta) * w.y
R.m32		= (1 - math.cos(theta)) * w.y * w.z + math.sin(theta) * w.x

-- calculate the change in heliocentric velocity needed to effect the tilt in the coordinates.
p[4]		= mat.mul( R   , p[3])
pE			= vec.sub( p[1], p[0])
qE			= vec.sub( q[1], q[0])
deltaV		= vec.sub( p[4], pE  )

-- work out the TransX reference frame
fward		= vec.div( pE, vec.length(pE))
plane		= vec.crossp( pE, qE)
plane		= vec.div( plane, vec.length(plane))
oward		= vec.crossp( plane, fward)

-- calculate the escape plan needed - using TransX coordinates
vf			= vec.dotp( fward, deltaV)
vo			= vec.dotp( oward, deltaV)
vp			= vec.dotp( plane, deltaV)

velp		= math.sqrt( vec.dotp( deltaV, deltaV) + 2 * muE / r) - math.sqrt(muE / r)

-- print out the results
term.out(' ')
term.out('TransX escape plan - current time & date')
term.out('------------------------------')
term.out('Prograde:     ' .. vf .. ' m/s')
term.out('Outward:      ' .. vo .. ' m/s')
term.out('Plane change: ' .. vp .. ' m/s')
term.out(' ')
term.out('Escape burn altitude:  ' .. (r - 6371000) / 1000 .. ' km' )
term.out('Escape burn magnitude: ' .. velp .. ' m/s')
term.out(' ')
term.out(' ')

Just cut & paste this into a text file and place in Orbiter's 'script' directory with a '.lua' extension. Then, in Orbiter, run the script in a lua console window. The script will calculate the parameters of the required escape plane (TransX coordinates) once you have entered the target inclination of your heliocentric orbit; and your departure orbital altitude.