How to eject at the right angle on interplanetary transfers

[PeriapsisPrograde]

I have recently been interested in doing math by hand for my flights in Orbiter. However, there is one thing that I haven't been able to find: How do I align my ejection burn so that my final velocity is tangent to my path around the Sun?

For example, consider an ejection burn from Earth to Mars. The burn requires a velocity of 11307 m/s (at 6.68e6 m periapsis) to eject from Earth with 2926 m/s excess, which changes the orbit around the Sun to intersect with Mars at aphelion. Since circular orbital velocity at 6.68e6 m around Earth is 7723 m/s, the ejection requires 3584 m/s of dV. I understand the math to do those calculations.

My problem is determining beforehand what direction my velocity at infinity will be. How can I make it tangent to Earth's velocity around the Sun?

The equations relating angles in hyperbolic orbits seem to all use eccentricity, which I can't find a useful equation for in the case of hyperbolic orbits. Many sources say that e>1 for hyperbolic orbits, but don't give a formula.

Is there a way to predict the angle necessary to eject with all excess velocity tangent to a certain vector? Logically, the orbit is very nearly curved 90 degrees from periapsis to infinity of a barely hyperbolic orbit, but it curves less and less as velocity increases. I feel like this should have a relatively simple formula.

Thanks for helping.

 

[Keithth G]

The angle between the velocity vector at periapsis and the asymptotic value of the velocity vector on the outgoing hyperbolic arm is given by the expression

[MATH]\theta = \frac{\pi }{2}-\cos ^{-1}\left(\frac{\mu }{\mu +r_p \,v_{\infty }^2}\right) [/MATH]
where

[MATH]\mu[/MATH] is the gravitational constant for the gravitating body (km^3 s^-2)
[MATH]r_p[/MATH] is the periapsis radius (km)
[MATH]v_\infty[/MATH] is the hyperbolic excess velocity (km/s)

and where [MATH]\theta[/MATH] is measured in radians.

For the Earth, [MATH]\mu = 398600.4418\,km^3/s^{2}[/MATH]; you have a periapsis radius of [MATH]6680\,km[/MATH]; and you have a hyperbolic excess velocity of [MATH]2.926\,km/s[/MATH].

Consequently, for your specific case, I calculate a change in angle from periapsis of 1.0067 radians or 57.677 degrees.

It is easy to check that if the hyperbolic excess velocity is zero (i.e., a parabolic orbit), the change in angle is 90 degrees. And if the hyperbolic excess velocity is very high, then the change in angle tends 0 degrees.

I hope that helps.

 

[sorindafabico]

The angle between the velocity vector at periapsis and the asymptotic value of the velocity vector on the outgoing hyperbolic arm is given by the expression

[MATH]\theta = \frac{\pi }{2}-\cos ^{-1}\left(\frac{\mu }{\mu +r_p \,v_{\infty }^2}\right) [/MATH]
where

[MATH]\mu[/MATH] is the gravitational constant for the gravitating body (km^3 s^-2)
[MATH]r_p[/MATH] is the periapsis radius (km)
[MATH]v_\infty[/MATH] is the hyperbolic excess velocity (km/s)

and where [MATH]\theta[/MATH] is measured in radians.

For the Earth, [MATH]\mu = 398600.4418\,km^3/s^{2}[/MATH]; you have a periapsis radius of [MATH]6680\,km[/MATH]; and you have a hyperbolic excess velocity of [MATH]2.926\,km/s[/MATH].

Consequently, for your specific case, I calculate a change in angle from periapsis of 1.0067 radians or 57.677 degrees.

It is easy to check that if the hyperbolic excess velocity is zero (i.e., a parabolic orbit), the change in angle is 90 degrees. And if the hyperbolic excess velocity is very high, then the change in angle tends to 0 degrees.

I hope that helps.

If I understood well, if Earth's longitude around the Sun is, for instance, 90 degrees, and theta is 30 degrees, the burn should occur when spaceship's longitude around Earth is 60 degrees?

 

[PeriapsisPrograde]

Thanks for the quick response, Keithth G. Would you mind showing how to derive the equation?

 

[Keithth G]

If I understood well, if Earth's longitude around the Sun is, for instance, 90 degrees, and theta is 30 degrees, the burn should occur when spaceship's longitude around Earth is 60 degrees?

Yes, that's the right idea.

To be a little more specific, though, imagine that you have a ship in a circular orbit around Earth with radius [MATH]r_p[/MATH]. Imagine also that the orbital plane of the ship happens to coincide with the plane of the ecliptic - i.e., the plane of the Earth's orbit around the Sun. Finally, imagine that you know at what point in this circular orbit, the ship's velocity vector is exactly parallel to the velocity vector of the Earth in its orbit around the Sun - and that the reference angle for this point is, say, [MATH]\theta_0[/MATH]. Then, to eject from Earth's orbit with a given hyperbolic excess velocity [MATH]v_{\infty}[/MATH], one calculates the angle [MATH]\theta[/MATH] using the formula given earlier. Then one subtracts this from [MATH]\theta_0[/MATH] to identify the point in the ship's orbit where the TMI burn should take place.

After the burn has been completed, and the ship has moved a reasonable distance away from the Earth - out to the edge of the SOI, say, the ship will then have a velocity vector parallel to the Earth's - and the direction of motion will be entirely prograde relative to Earth's orbit around the Sun. Using the TransX vocabulary, there will be no 'outward' velocity component nor will there be a 'plane change' velocity component. Consequently, one will have ejected the ship from Earth in a Hohman-like transfer with a trajectory that will (barring perturbations) continue to lie in the Earth's ecliptic plane. To arrive at Mars, then, one needs to time the ejection from Earth, and to gauge [MATH]v_{\infty}[/MATH] so that the arrives at the intersection of the Earth and Mars orbital planes at the same time that Mars passes through the intersection (line of nodes).

If, on the other hand, one wishes to eject the ship with an 'outward' velocity component, one will have to time the TMI burn so that it occurs earlier (or for an 'inward' velocity component later). So, for example, if one wishes to eject the ship at an angle of 5 degrees from the pure prograde direction, one would need to execute the TMI burn 5 degrees earlier.

Of course, all of this takes no account of perturbations - and so high accuracy is not likely to do things this way - but it is a good way of thinking about the problem.