guest blogger john hare

In my last couple of posts involving gravity turns, there seems to be some legitimate doubt whether useful maneuvers can be done with Lunar gravity turns. My numbers are quite different from those obtained by others and my methods are probably unacceptable for serious further debate. I believe there are some clear relationships in gravity turn mechanics just as there are in escape velocity relative to orbital velocity.

If a particular altitude has an orbital velocity of x, then escape velocity will be 1.41x for any gravity field or velocity. A 1,500 m/s circular orbit has an escape velocity of 2,115 m/s whether it is low Lunar orbit, high Earth orbit, or extremely distant Solar orbit. That is a constant relationship.

Somewhere, somebody has derived constant relationships for gravity turn angles relative to local orbital velocity. A useful table for this could fit on an index card. Something like this.

Orbital velocity 1.0x

escape velocity 1.41x

170 degree turn velocity 1.4-x

160 degree turn angle 1.5-x

150 degree turn angle 1.5-x

The real tables are out there somewhere because they are too obvious and useful to not exist. I will put up $100.00 to the first person that finds a reliable, citable source for this information. I think my reasons are obvious.

I said it was a cheap contest.

#### johnhare

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While other people are searching calculate your own table. You can then publish the table in an academic paper or on your website.

I think you might want to spell things out a little more clearly.

I assume we’re talking about an impulsive (i.e. nearly instantaneous) turn and the turn angles are measured between the tangent directions before and after the turn. Otherwise, one would have to account for that part of the turn which is due to the unmodified orbital trajectory.

For these turn angles, are you referring to gravity assisted turns, or free space turns (e.g. plane change turns)? Are we talking about turns in elliptic orbits, or hyperbolic orbits, or both?

A_M_,

The idea is to have tables available that have been profesionally done and used so that any discussion can be about the uses of the gravity manuevers instead of the methods of calculation. This not being my field, I do not expect anyone to take my word for it that I got it right. Doing your own for your own uses is very different than working with other people. They need either a reputable source, or a very well done and documented paper.

Eric,

The total turn angle during the time the object is close enough to feel any effects from the body in question. I’m talking about a gravity turn with no propulsive assistance. Hyperbolic orbits that escape. The total change during the hours of a Lunar flyby or the days of an Earth flyby.

@John Hare

I got that. I just suspect that there is no official source.

You may be right that there is no official source. I would be quite surprised if there isn’t at least an AIAA paper on the subject. My hope is that someone will remember where it is. The tables I am interested in would represent man-years of work, just as the references most of us use represent man-millenia of work.

My method of graphing many short time segments on an x-y axis is time consuming and inaccurate compared to what I would like to use. And using that would lead to a long discussion on calculation methods rather than means of improving future missions.

John, I found this paper:

“A Gravity Assist Primer”, R. J. Cesarone, AIAA Student Journal, 27(1), Spring 1989, pp. 16-22

An online copy can be found at:

http://www.gravityassist.com/IAF1/Ref.%201-85.pdf

On page 18 (third page of the paper) it has this formula for total turn angle of a hyperbolic flyby:

turn_angle = 2 * arcsin{1 / [1 + (r_p * v_inf^2 / GM)]}

r_p is the radius of periapsis of the hyperbola, v_inf is the hyperbolic excess velocity of the satellite, G is the universal gravitational constant, and M is the mass of the planet.

Orbital velocity for a circular orbit at radius r_p is:

v_orbit = sqrt(GM/r_p)

so:

r_p/GM = 1/v_obrit^2

substitute that into the previous equation:

turn_angle = 2 * arcsin{1 / [1 + (v_inf / v_orbit)^2]}

This gives you the turn angle from hyperbolic excess velocity and the velocity of a circular orbit at the same altitude as the periapsis of the hyperbola. It sounded like you were asking for an equation based on velocity of the hyperbolic orbit at periapsis, but this formulation is so darn convenient. In table form:

v_inf / turn_angle

v_orbit

0 180

0.5 106

1 60

1.5 36

2 23

I hope this is good enough to get the $100. You’ll be happy to know it will go towards the SpeedUp rocket development budget.

Kudos on guessing it would be an AIAA paper.

Congrats on getting there first Bob. I was going to suggest something similar. The only part I was missing was relating the turn angle to the v_inf/v_orb.

These numbers look similar to what I was getting. Unfortunately for John’s retro-station idea, the hyperbolic excess velocity when coming in from a Mars transfer orbit towards the moon is on the order of a couple of thousand times greater than the orbital velocity – hence the very small turn angles.

Those turn angles may not be sufficient for the purposes intended, but they are far less than a couple of thousand times off. Coming in at double orbital velocity at infinity still gives 23 degrees of turn if I’m reading his chart correctly. This gives me a base to work from with some honest turn angles available. I’ll run some numbers a to see what I can get. Coming in with with one orbital velocity of excess gives 60 degrees which just might work for some missions.

I am going to take a look at an accelerating tanker that doesn’t leave the Earths’ vicinity to see if there is a realistic way of substantially reducing return mission mass numbers and cost. A fairly small burn at Earth perigee ‘might’ be profitable with some propellant that doesn’t have to be carried there and back as payload. I think a hard look at the trades for that one might help.

The check’s in the mail.