MontBlanc2012
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This note presents the conditions that need to be satisfied for two elliptical orbits to 'touch' at a point. It follows on from an earlier thread by 'maki' entitled "calculating intersection of two orbits" (https://www.orbiter-forum.com/showthread.php?t=39048.
Knowledge of these conditions is useful for understanding how one can transition from one elliptical orbit to another elliptical orbit by executing either a prograde or retrograde burn at some point in the orbit other than at the apoapsis or periapsis.
In turn, this understanding is useful for designing strategies for changing the argument of periapsis of an orbit while keeping the orbital periapsis and apoapsis constant. These orbital transfer strategies will be the subject of a subsequent post.
An illustration
As an example of the kind of orbit transfer consider the following diagram below:
Here, the black circle represents the Earth with a radius of around 6371 km. The blue ellipse is an initial elliptical orbit with periapsis radius of 6671 km (~300 km altitude) and orbital eccentricity of 0.5. The orange ellipse is tangent the the blue orbit at one point (upper-left quadrant) and at this point, it is possible for the vessel on the blue trajectory to transfer to the orange orbit by executing a prograde burn. In so doing, the argument of periapsis of the orbit is rotates around by 45 degrees (in this example) and both the semi-major axis and orbital eccentricity are altered. There is a relationship between the initial and final apoapsis and periapsis radii and the change in the argument of periapsis. That relationship is the subject of this note.
A quick recap and some new stuff
In response to 'maki's' post ("calculating intersection of two orbits"), I stated that one can calculate a quantity, [imath]\Delta[/imath], such that if [imath]\Delta<0[/imath], there are no points of intersection of two co-planar elliptical orbits. Equally, if [imath]\Delta>0[/imath], then one can calculate two points of intersection of the ellipses and a way of calculating those points was set out in that post.
But what happens if [imath]\Delta=0[/imath]?
In this case, the two ellipses are tangent to each other - i.e., they 'kiss'. Following on from that post [imath]\Delta=0[/imath] if:
[math]\left(\beta _1^2-2 \,\cos (\Delta\omega ) \,\beta _1 \,\beta _2+\beta _2^2\right)\, e_1^2 \,e_2^2 = \left(\beta _1\, e_1-\beta _2 \,e_2\right){}^2[/math]
where:
[math]\beta _1 = a_1 \, e_2 \, \left(1-e_1^2\right)[/math]
[math]\beta _2 = a_2 \, e_1 \, \left(1-e_2^2\right)[/math]
and where [imath]a_1[/imath] and [imath]e_1[/imath] are the semi-major axis and orbital eccentricity of the first ellipse; [imath]a_2[/imath] and [imath]e_2[/imath] are the semi-major axis and orbital eccentricity of the second ellipse; and [imath]\Delta\omega[/imath] is the difference in argument of periapsis of the two ellipses.
After a little algebra, one can show that this reduces to the expression:
[math]\cos (\Delta\omega ) = 1-2\,\frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)}[/math]
or, equivalently:
[math]\sin (\frac{\Delta\omega}{2} ) = \pm\sqrt{\frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)}}[/math]
where [imath]r_{a,1}[/imath] and [imath]r_{p,1}[/imath] are the apoapsis and periapsis of the initial elliptical orbit; and [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] are the apoapsis and periapsis of the final elliptical orbit. The [imath]\pm[/imath] sign flags that there are two possible values for [imath]\Delta\omega[/imath] that differ by a sign change.
Using either formula, these expressions yield sensible values for [imath]\Delta\omega[/imath] only if either:
[math]r_{a,2}>r_{a,1}>r_{p,2}>r_{p,1}>0[/math]
or
[math]r_{a,1}>r_{a,2}>r_{p,1}>r_{p,2}>0[/math]
If these conditions cannot be met, then there is no value of [imath]\Delta\omega[/imath] that permits the two ellipses to 'touch'.
Assuming that these conditions are satisfied, then we can calculate not only [imath]\Delta\omega[/imath] but also the orbital radius, [imath]r^*[/imath] at which the two ellipses touch. This is given by the following expressions:
[math]r ^* = \frac{r_{a,1} \,r_{p,1} \,\left(r_{a,2}+r_{p,2}\right)-r_{a,2} \,r_{p,2} \,\left(r_{a,1}+r_{p,1}\right)}{r_{a,1}\, r_{p,1}-r_{a,2} \,r_{p,2}}[/math]
Now, we know the extent to which our argument of periapsis will be rotated, [imath]\Delta\omega[/imath], and also the orbital radius [imath]r^*[/imath] when we need to execute our manoeuvre. But how much [imath]\Delta V[/imath] do we require? The answer to this is given by the expression:
[math]\begin{aligned} \Delta V &= v_2 - v_1 \\ \\ v_2 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,2}+r_{p,2}}\right)} \\ v_1 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,1}+r_{p,1}}\right)} \end{aligned}[/math]
and where [imath]\mu[/imath] is the gravitational parameter for the central gravitating body (e.g., the Earth).
A quick example
As an example of this, let's return to the initial illustration. The initial orbit has [imath]r_{p,1} = 6671.0[/imath] km; and [imath]r_{a,1} = 20013.0[/imath] km.
For the sake of argument, let's choose [imath]r_{p,2} = 12096.5[/imath] km; and [imath]r_{a,2} = 24468.5[/imath] km. (Yes, these values have been carefully chosen!) Using the above expressions, we calculate:
[math]\cos(\Delta\omega) = 0.707107[/math]
or, in other words, [imath]\Delta\omega = 45^{\circ}[/imath] or [imath]\Delta\omega = -45^{\circ}[/imath] and:
[math]r^*=18564.8[/math] km
and, finally,
[math]\Delta V = 983.06[/math] m/s
So, if starting in our initial elliptical orbit and after having passes through periapsis (but before having arrived at apoapsis), we wait until our orbital radius is 18564.8 km and then execute a 983.06 m/s prograde burn, we will transfer to the orange elliptical orbit and, in so doing, rotated our argument of periapsis around by +45 degrees.
An Orbiter scenario
To test that this maths actually works, here are the contents of an Orbiter 2016 .scn file that sets a standard DeltaGlider in the initial orbit. By executing the above burn when the orbital radius reaches the target value of 18564.8 km (or a little before 'to balance' the burn), you will find the orbit transform to the target orange elliptical orbit.
Knowledge of these conditions is useful for understanding how one can transition from one elliptical orbit to another elliptical orbit by executing either a prograde or retrograde burn at some point in the orbit other than at the apoapsis or periapsis.
In turn, this understanding is useful for designing strategies for changing the argument of periapsis of an orbit while keeping the orbital periapsis and apoapsis constant. These orbital transfer strategies will be the subject of a subsequent post.
An illustration
As an example of the kind of orbit transfer consider the following diagram below:
Here, the black circle represents the Earth with a radius of around 6371 km. The blue ellipse is an initial elliptical orbit with periapsis radius of 6671 km (~300 km altitude) and orbital eccentricity of 0.5. The orange ellipse is tangent the the blue orbit at one point (upper-left quadrant) and at this point, it is possible for the vessel on the blue trajectory to transfer to the orange orbit by executing a prograde burn. In so doing, the argument of periapsis of the orbit is rotates around by 45 degrees (in this example) and both the semi-major axis and orbital eccentricity are altered. There is a relationship between the initial and final apoapsis and periapsis radii and the change in the argument of periapsis. That relationship is the subject of this note.
A quick recap and some new stuff
In response to 'maki's' post ("calculating intersection of two orbits"), I stated that one can calculate a quantity, [imath]\Delta[/imath], such that if [imath]\Delta<0[/imath], there are no points of intersection of two co-planar elliptical orbits. Equally, if [imath]\Delta>0[/imath], then one can calculate two points of intersection of the ellipses and a way of calculating those points was set out in that post.
But what happens if [imath]\Delta=0[/imath]?
In this case, the two ellipses are tangent to each other - i.e., they 'kiss'. Following on from that post [imath]\Delta=0[/imath] if:
[math]\left(\beta _1^2-2 \,\cos (\Delta\omega ) \,\beta _1 \,\beta _2+\beta _2^2\right)\, e_1^2 \,e_2^2 = \left(\beta _1\, e_1-\beta _2 \,e_2\right){}^2[/math]
where:
[math]\beta _1 = a_1 \, e_2 \, \left(1-e_1^2\right)[/math]
[math]\beta _2 = a_2 \, e_1 \, \left(1-e_2^2\right)[/math]
and where [imath]a_1[/imath] and [imath]e_1[/imath] are the semi-major axis and orbital eccentricity of the first ellipse; [imath]a_2[/imath] and [imath]e_2[/imath] are the semi-major axis and orbital eccentricity of the second ellipse; and [imath]\Delta\omega[/imath] is the difference in argument of periapsis of the two ellipses.
After a little algebra, one can show that this reduces to the expression:
[math]\cos (\Delta\omega ) = 1-2\,\frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)}[/math]
or, equivalently:
[math]\sin (\frac{\Delta\omega}{2} ) = \pm\sqrt{\frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)}}[/math]
where [imath]r_{a,1}[/imath] and [imath]r_{p,1}[/imath] are the apoapsis and periapsis of the initial elliptical orbit; and [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] are the apoapsis and periapsis of the final elliptical orbit. The [imath]\pm[/imath] sign flags that there are two possible values for [imath]\Delta\omega[/imath] that differ by a sign change.
Using either formula, these expressions yield sensible values for [imath]\Delta\omega[/imath] only if either:
[math]r_{a,2}>r_{a,1}>r_{p,2}>r_{p,1}>0[/math]
or
[math]r_{a,1}>r_{a,2}>r_{p,1}>r_{p,2}>0[/math]
If these conditions cannot be met, then there is no value of [imath]\Delta\omega[/imath] that permits the two ellipses to 'touch'.
Assuming that these conditions are satisfied, then we can calculate not only [imath]\Delta\omega[/imath] but also the orbital radius, [imath]r^*[/imath] at which the two ellipses touch. This is given by the following expressions:
[math]r ^* = \frac{r_{a,1} \,r_{p,1} \,\left(r_{a,2}+r_{p,2}\right)-r_{a,2} \,r_{p,2} \,\left(r_{a,1}+r_{p,1}\right)}{r_{a,1}\, r_{p,1}-r_{a,2} \,r_{p,2}}[/math]
Now, we know the extent to which our argument of periapsis will be rotated, [imath]\Delta\omega[/imath], and also the orbital radius [imath]r^*[/imath] when we need to execute our manoeuvre. But how much [imath]\Delta V[/imath] do we require? The answer to this is given by the expression:
[math]\begin{aligned} \Delta V &= v_2 - v_1 \\ \\ v_2 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,2}+r_{p,2}}\right)} \\ v_1 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,1}+r_{p,1}}\right)} \end{aligned}[/math]
and where [imath]\mu[/imath] is the gravitational parameter for the central gravitating body (e.g., the Earth).
A quick example
As an example of this, let's return to the initial illustration. The initial orbit has [imath]r_{p,1} = 6671.0[/imath] km; and [imath]r_{a,1} = 20013.0[/imath] km.
For the sake of argument, let's choose [imath]r_{p,2} = 12096.5[/imath] km; and [imath]r_{a,2} = 24468.5[/imath] km. (Yes, these values have been carefully chosen!) Using the above expressions, we calculate:
[math]\cos(\Delta\omega) = 0.707107[/math]
or, in other words, [imath]\Delta\omega = 45^{\circ}[/imath] or [imath]\Delta\omega = -45^{\circ}[/imath] and:
[math]r^*=18564.8[/math] km
and, finally,
[math]\Delta V = 983.06[/math] m/s
So, if starting in our initial elliptical orbit and after having passes through periapsis (but before having arrived at apoapsis), we wait until our orbital radius is 18564.8 km and then execute a 983.06 m/s prograde burn, we will transfer to the orange elliptical orbit and, in so doing, rotated our argument of periapsis around by +45 degrees.
An Orbiter scenario
To test that this maths actually works, here are the contents of an Orbiter 2016 .scn file that sets a standard DeltaGlider in the initial orbit. By executing the above burn when the orbital radius reaches the target value of 18564.8 km (or a little before 'to balance' the burn), you will find the orbit transform to the target orange elliptical orbit.
Code:
BEGIN_DESC
END_DESC
BEGIN_ENVIRONMENT
System Sol
Date MJD 51982.0290677529
Help CurrentState_img
END_ENVIRONMENT
BEGIN_FOCUS
Ship GL-02
END_FOCUS
BEGIN_CAMERA
TARGET GL-02
MODE Cockpit
FOV 50.00
END_CAMERA
BEGIN_SHIPS
GL-02:DeltaGlider
STATUS Orbiting Earth
RPOS 6670999.831 -3.208 -1838.070
RVEL 1.7390 16.5233 9467.1307
AROT -91.263 74.202 108.198
AFCMODE 7
PRPLEVEL 0:1.000000 1:0.999985
NAVFREQ 586 466 0 0
XPDR 0
HOVERHOLD 0 1 0.0000e+000 0.0000e+000
NOSECONE 1.0000 0.0000
AAP 0:0 0:0 0:0
SKIN BLUE
END
END_SHIPS
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