MontBlanc2012
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In this note, I want to consider the single-burn manoeuvre needed to change the 'argument of periapsis' of an elliptical orbit by an angle [MATH]\Delta\omega[/MATH].
Whereas we know how to raise and lower the periapsis radius, [MATH]r_p[/MATH], and/or apoapsis radius, [MATH]r_a[/MATH], of an orbit to achieve a desired semi-major axis, [MATH]a[/MATH], and orbital eccentricity, [MATH]e[/MATH]; and whereas we know how to change the orbital inclination, [MATH]\iota[/MATH], and longitude of the ascending node, [MATH]\Omega[/MATH], at will, the procedure for changing the argument of periapsis of an orbit generally receives scant treatment. So, what is the general procedure for doing it? And, moreover, how can one do it is efficiently as possible?
I want to consider a standard 'single-burn' treatment of this topic. In a subsequent note, I want to present an alternative two-burn solution that uses roughly half the [MATH]\delta V[/MATH] of the single-burn solution.
The 'standard' treatment
To make the issue concrete, let's suppose that one is an elliptical orbit with am eccentricity of 0.5 and a semi-major axis of 1.0. And let's suppose that we want to transfer to a new elliptical orbit which is the same as our starting ellipse excepted that the line of apsides is rotated by 90 degrees in a counter-clockwise direction. A graph of this scenario is given below:
Here, the blue ellipse is our initial orbit which, in keeping with standard convention, we move around in our initial orbit in a counter-clockwise. The orange ellipse is the orbit that we wish to transfer to which, again, we move around a counter-clockwise direction. This is the same ellipse as the first - except the its argument of periapsis has been rotated counter-clockwise by 90 degrees.
Now, if one refers to standard texts on orbital mechanics (e.g., "Spacecraft Dynamics and Control: A Practical Engineering Approach" - Marcel J. Sidi - "Spacecraft dynamics and control") the recommended single-burn strategy for making the orbit transfer is to execute a burn at either of the points of intersection of the two ellipses (in the graph, the points 'A' and 'B'). The texts then go on to calculate the amount of [MATH]\Delta V[/MATH] required to execute this manoeuvre and come up with following formula:
[MATH]\Delta V = 2\,e\,\sqrt{\frac{\mu}{a\,(1-e^2)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
So where did this expression come from? The procedure for calculating this expression if first to calculate the points of intersection; then calculate the the velocity vector of the orbiting body on both the initial and final ellipses; and then take the magnitude of the difference between the initial and final velocity vectors. Now, the maths for finding the intersection of two co-planar ellipses has been set out in an earlier post "Calculating intersection of two orbits", so we ought to be able to go through the derivation ourselves.
Calculate the points of intersection
Basically, the procedure for calculating the points of intersection is to calculate three quantities:
[MATH]A = \beta _1 \, e_2+\beta _2 \, e_1-\left(\beta _1 \, e_1+\beta _2 \, e_2\right) \cos (\Delta \omega )[/MATH]
[MATH]B = e_1^2+e_2^2-2\, e_1 \,e_2\, \cos (\Delta \omega )-e_1^2 \,e_2^2 \,\sin ^2(\Delta \omega )[/MATH]
[MATH]\Delta = \sin ^2(\Delta\omega ) \left(\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\right)[/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 then calculate the radii of the two points of intersection as:
[MATH]r_+ = \frac{A+\sqrt{\Delta}}{B}[/MATH]
and
[MATH]r_- = \frac{A-\sqrt{\Delta}}{B}[/MATH]
In the case considered here, where there is no change in the shape of the ellipse - i.e., the semi-major axis and the orbital eccentricity of the initial and final orbits remains the same - we can simplify things by setting:
[MATH]a_1 = a_2 = a[/MATH]
and
[MATH]e_1 = e_2 = e[/MATH]
Then we can gird our loins and systematically work our way some fairly tedious algebra and trig manipulations. But, if we do this, we can show that:
[MATH]r_+ = \frac{a\,(1-e^2)}{1+e\,\cos\Delta\omega/2}[/MATH]
and
[MATH]r_- = \frac{a\,(1-e^2)}{1-e\,\cos\Delta\omega/2}[/MATH]
Calculate the velocity vectors at the intersection points
OK, so far so good. Now we need to calculate the velocity vectors at the intersection points. Fortunately, the velocity vector in radial and transverse component form at any point on an elliptical orbit is just a function of [MATH]r_a[/MATH] and [MATH]r_p[/MATH] (and, hence, [MATH]a[/MATH] and [MATH]e[/MATH]) and [MATH]r[/MATH] so we have just enough information to do this. Here, the radial component is the component of the velocity vector directed along the radial line between the orbiting vessel and the central gravitating body; and the transverse component is the rest of the velocity vector that is at right-angles to the radial component in the direction of travel. Specifically, we have:
[MATH]v_\theta(r) =\frac{1}{r}\,\sqrt{2\,\mu\,\frac{r_a\,r_p}{r_a+r_p}} [/MATH]
and
[MATH]v_r(r) = \pm\frac{1}{r}\,\sqrt{2\,\mu\,\frac{(r_a - r)\,(r-r_p)}{r_a+r_p}}[/MATH]
where [MATH]v_r(r)[/MATH] is the radial component of the velocity vector when the orbital radius is [MATH]r[/MATH]; and [MATH]v_\theta(r)[/MATH] is the transverse component of the velocity vector. The [MATH]\pm[/MATH] flags that the sign of the radial component varies according to whether or not we are approaching periapsis or moving away from it. If we are moving away from it, and heading towards apoapsis, then [MATH]v_r(r)[/MATH] is positive; and if we have passed apoapsis and heading back towards periapsis then [MATH]v_r(r)[/MATH] is negative.
So, now we are in a position to calculate the velocity vectors. If we go through some tedious algebra and trig again, we find that at the intersection point corresponding to [MATH]r=r_+[/MATH], then for both the initial blue and final orange orbits:
[MATH]v_\theta(r_+) = \sqrt{\frac{\mu }{a \left(1-e^2\right)}} \left(1-e \cos \frac{\Delta\omega }{2}\right)[/MATH]
in other words, the transverse component of the velocity vector does not change as we move from the blue ellipse to the orange ellipse.
Calculate the magnitude of he change in the velocity vector
What about the radial component. Here we find that for both the blue and orange ellipses, the magnitude of the radial component is the same:
[MATH]\left|v_r(r_+)\right| = e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
However, the sign of the radial component changes as we move between ellipses. For example, at the intersection point 'A', the radial component of the velocity vector on the blue ellipse is positive because we are moving away from periapsis whereas for the orange ellipse it is negative because we are heading towards periapsis. So, the only change in the velocity vector is a change in sign of the radial component which means that the magnitude of the change in the velocity vector is:
[MATH]\Delta V = 2\,\left|v_r(r_+)\right| = 2\,e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
We can go through exactly the same analysis for [MATH]r=r_-[/MATH] and if we do we will find that:
[MATH]v_\theta(r_-) = \sqrt{\frac{\mu }{a \left(1-e^2\right)}} \left(1 + e \cos \frac{\Delta\omega }{2}\right)[/MATH]
and
[MATH]\Delta V = 2\,\left|v_r(r_-)\right| = 2\,e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
In other words, the amount of [MATH]\Delta V[/MATH] needed to make the transfer from the blue to orange orbit is the same - irrespective of which intersection point that we choose.
An example
Let's move back to the example of "Tangent ellipses - a follow on from 'calculating the intersection of two orbits'". That thread used as an example an initial orbit with a perigee radius of 6,671 km and orbital eccentricity of 0.5. Equivalently, that corresponds to an orbit with a apogee of 20,013 km and semi-major axis of 13,342 km.
Using the expression that the [MATH]\Delta V[/MATH] required to make the transfer between the initial and final elliptic orbits with the argument of periapsis rotated by [MATH]\Delta\omega =[/MATH] 90 degrees:
[MATH]\Delta V = 2\,e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
we calculated that the [MATH]\Delta V = [/MATH] 4,462.8 m/s.
Ouch!
A connection with bi-tangent elliptic transfers
Now, the astute reader of "Tangent ellipses - a follow on from 'calculating the intersection of two orbits'" may have no noticed that the example tangent transfer started from the same initial elliptical orbit but rotated the argument of periapsis around by 45 degrees - exactly half of the target 90 degree transfer of this note - using just 983.1 m/s of fuel.
As it turns out, one can use another tangent transfer that is effectively the mirror of the first to rotate the argument of periapsis by a further 45 degrees (i.e., 90 degrees in total) and return the eccentricity and perigee to their original values. To do this, we need a further 983.1 m/s of fuel. In total then, using two tangent elliptical transfers, we can achieve the same rotation of 90 degrees of the argument of periapsis but use only 1,966.2 m/s of fuel - i.e., a saving of 2,496.6 m/s over the single-burn strategy!
In a subsequent note, I'll introduce and work through these bi-tangent elliptical transfers in more detail.
Whereas we know how to raise and lower the periapsis radius, [MATH]r_p[/MATH], and/or apoapsis radius, [MATH]r_a[/MATH], of an orbit to achieve a desired semi-major axis, [MATH]a[/MATH], and orbital eccentricity, [MATH]e[/MATH]; and whereas we know how to change the orbital inclination, [MATH]\iota[/MATH], and longitude of the ascending node, [MATH]\Omega[/MATH], at will, the procedure for changing the argument of periapsis of an orbit generally receives scant treatment. So, what is the general procedure for doing it? And, moreover, how can one do it is efficiently as possible?
I want to consider a standard 'single-burn' treatment of this topic. In a subsequent note, I want to present an alternative two-burn solution that uses roughly half the [MATH]\delta V[/MATH] of the single-burn solution.
The 'standard' treatment
To make the issue concrete, let's suppose that one is an elliptical orbit with am eccentricity of 0.5 and a semi-major axis of 1.0. And let's suppose that we want to transfer to a new elliptical orbit which is the same as our starting ellipse excepted that the line of apsides is rotated by 90 degrees in a counter-clockwise direction. A graph of this scenario is given below:
Here, the blue ellipse is our initial orbit which, in keeping with standard convention, we move around in our initial orbit in a counter-clockwise. The orange ellipse is the orbit that we wish to transfer to which, again, we move around a counter-clockwise direction. This is the same ellipse as the first - except the its argument of periapsis has been rotated counter-clockwise by 90 degrees.
Now, if one refers to standard texts on orbital mechanics (e.g., "Spacecraft Dynamics and Control: A Practical Engineering Approach" - Marcel J. Sidi - "Spacecraft dynamics and control") the recommended single-burn strategy for making the orbit transfer is to execute a burn at either of the points of intersection of the two ellipses (in the graph, the points 'A' and 'B'). The texts then go on to calculate the amount of [MATH]\Delta V[/MATH] required to execute this manoeuvre and come up with following formula:
[MATH]\Delta V = 2\,e\,\sqrt{\frac{\mu}{a\,(1-e^2)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
So where did this expression come from? The procedure for calculating this expression if first to calculate the points of intersection; then calculate the the velocity vector of the orbiting body on both the initial and final ellipses; and then take the magnitude of the difference between the initial and final velocity vectors. Now, the maths for finding the intersection of two co-planar ellipses has been set out in an earlier post "Calculating intersection of two orbits", so we ought to be able to go through the derivation ourselves.
Calculate the points of intersection
Basically, the procedure for calculating the points of intersection is to calculate three quantities:
[MATH]A = \beta _1 \, e_2+\beta _2 \, e_1-\left(\beta _1 \, e_1+\beta _2 \, e_2\right) \cos (\Delta \omega )[/MATH]
[MATH]B = e_1^2+e_2^2-2\, e_1 \,e_2\, \cos (\Delta \omega )-e_1^2 \,e_2^2 \,\sin ^2(\Delta \omega )[/MATH]
[MATH]\Delta = \sin ^2(\Delta\omega ) \left(\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\right)[/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 then calculate the radii of the two points of intersection as:
[MATH]r_+ = \frac{A+\sqrt{\Delta}}{B}[/MATH]
and
[MATH]r_- = \frac{A-\sqrt{\Delta}}{B}[/MATH]
In the case considered here, where there is no change in the shape of the ellipse - i.e., the semi-major axis and the orbital eccentricity of the initial and final orbits remains the same - we can simplify things by setting:
[MATH]a_1 = a_2 = a[/MATH]
and
[MATH]e_1 = e_2 = e[/MATH]
Then we can gird our loins and systematically work our way some fairly tedious algebra and trig manipulations. But, if we do this, we can show that:
[MATH]r_+ = \frac{a\,(1-e^2)}{1+e\,\cos\Delta\omega/2}[/MATH]
and
[MATH]r_- = \frac{a\,(1-e^2)}{1-e\,\cos\Delta\omega/2}[/MATH]
Calculate the velocity vectors at the intersection points
OK, so far so good. Now we need to calculate the velocity vectors at the intersection points. Fortunately, the velocity vector in radial and transverse component form at any point on an elliptical orbit is just a function of [MATH]r_a[/MATH] and [MATH]r_p[/MATH] (and, hence, [MATH]a[/MATH] and [MATH]e[/MATH]) and [MATH]r[/MATH] so we have just enough information to do this. Here, the radial component is the component of the velocity vector directed along the radial line between the orbiting vessel and the central gravitating body; and the transverse component is the rest of the velocity vector that is at right-angles to the radial component in the direction of travel. Specifically, we have:
[MATH]v_\theta(r) =\frac{1}{r}\,\sqrt{2\,\mu\,\frac{r_a\,r_p}{r_a+r_p}} [/MATH]
and
[MATH]v_r(r) = \pm\frac{1}{r}\,\sqrt{2\,\mu\,\frac{(r_a - r)\,(r-r_p)}{r_a+r_p}}[/MATH]
where [MATH]v_r(r)[/MATH] is the radial component of the velocity vector when the orbital radius is [MATH]r[/MATH]; and [MATH]v_\theta(r)[/MATH] is the transverse component of the velocity vector. The [MATH]\pm[/MATH] flags that the sign of the radial component varies according to whether or not we are approaching periapsis or moving away from it. If we are moving away from it, and heading towards apoapsis, then [MATH]v_r(r)[/MATH] is positive; and if we have passed apoapsis and heading back towards periapsis then [MATH]v_r(r)[/MATH] is negative.
So, now we are in a position to calculate the velocity vectors. If we go through some tedious algebra and trig again, we find that at the intersection point corresponding to [MATH]r=r_+[/MATH], then for both the initial blue and final orange orbits:
[MATH]v_\theta(r_+) = \sqrt{\frac{\mu }{a \left(1-e^2\right)}} \left(1-e \cos \frac{\Delta\omega }{2}\right)[/MATH]
in other words, the transverse component of the velocity vector does not change as we move from the blue ellipse to the orange ellipse.
Calculate the magnitude of he change in the velocity vector
What about the radial component. Here we find that for both the blue and orange ellipses, the magnitude of the radial component is the same:
[MATH]\left|v_r(r_+)\right| = e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
However, the sign of the radial component changes as we move between ellipses. For example, at the intersection point 'A', the radial component of the velocity vector on the blue ellipse is positive because we are moving away from periapsis whereas for the orange ellipse it is negative because we are heading towards periapsis. So, the only change in the velocity vector is a change in sign of the radial component which means that the magnitude of the change in the velocity vector is:
[MATH]\Delta V = 2\,\left|v_r(r_+)\right| = 2\,e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
We can go through exactly the same analysis for [MATH]r=r_-[/MATH] and if we do we will find that:
[MATH]v_\theta(r_-) = \sqrt{\frac{\mu }{a \left(1-e^2\right)}} \left(1 + e \cos \frac{\Delta\omega }{2}\right)[/MATH]
and
[MATH]\Delta V = 2\,\left|v_r(r_-)\right| = 2\,e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
In other words, the amount of [MATH]\Delta V[/MATH] needed to make the transfer from the blue to orange orbit is the same - irrespective of which intersection point that we choose.
An example
Let's move back to the example of "Tangent ellipses - a follow on from 'calculating the intersection of two orbits'". That thread used as an example an initial orbit with a perigee radius of 6,671 km and orbital eccentricity of 0.5. Equivalently, that corresponds to an orbit with a apogee of 20,013 km and semi-major axis of 13,342 km.
Using the expression that the [MATH]\Delta V[/MATH] required to make the transfer between the initial and final elliptic orbits with the argument of periapsis rotated by [MATH]\Delta\omega =[/MATH] 90 degrees:
[MATH]\Delta V = 2\,e\,\sqrt{\frac{\mu }{a \left(1-e^2\right)}}\,\sin\frac{\Delta\omega}{2}[/MATH]
we calculated that the [MATH]\Delta V = [/MATH] 4,462.8 m/s.
Ouch!
A connection with bi-tangent elliptic transfers
Now, the astute reader of "Tangent ellipses - a follow on from 'calculating the intersection of two orbits'" may have no noticed that the example tangent transfer started from the same initial elliptical orbit but rotated the argument of periapsis around by 45 degrees - exactly half of the target 90 degree transfer of this note - using just 983.1 m/s of fuel.
As it turns out, one can use another tangent transfer that is effectively the mirror of the first to rotate the argument of periapsis by a further 45 degrees (i.e., 90 degrees in total) and return the eccentricity and perigee to their original values. To do this, we need a further 983.1 m/s of fuel. In total then, using two tangent elliptical transfers, we can achieve the same rotation of 90 degrees of the argument of periapsis but use only 1,966.2 m/s of fuel - i.e., a saving of 2,496.6 m/s over the single-burn strategy!
In a subsequent note, I'll introduce and work through these bi-tangent elliptical transfers in more detail.
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