Planning asteroid capture with orbital physics

[reaction_wheel]

I'm planning my first asteroid capture mission, and (without having played the tutorial mission) I got curious to learn how much fuel and thrust I would need to put on the capture vessel in order to bring the asteroid into a useful orbit around Kerbin. I spent a lot of time trying to learn something about it, and thought I would share. I would really love to hear your thoughts and feedback!

 

The lovely little space rock I have in mind to make my own is one SDD-569, a class A asteroid. SDD-569 is approaching Kerbin for a leisurely flyby at a periapsis of ~2,079 km, well inside the orbit of the Mun. This puts SDD-569 into a sharply looping orbit around the planet, before it flies back out into parts unknown. I'd like to drag the asteroid into a circular orbit, and then bring it down to about 500 km for future research and exploitation.

 

Can we calculate how fast is SDD-569 going with respect to Kerbin, based on what we already know? Since energy is always conserved, the total kinetic energy for a given object (from orbital speed) plus its potential energy (from gravity) never changes. The relationship of kinetic energy to velocity is a consequence of Newton's third law:

 

 

In other words, for a constant mass, v² is a measure of kinetic energy.

 

The vis-viva equation describes the conservation of energy for a small body orbiting a much larger one:

 

 

• GM, also known sometimes as μ, as is Kerbin's gravitational parameter, which the KSP wiki reports is 3.5316 x 10¹² m³/s². This parameter is the product of the gravitational constant of the universe with Kerbin's mass, which is effectively constant.
• a is the semi-major axis of the orbit as measured from the center of the celestial body. Kerbin's radius is 600km, so we add that to the altitude of SDD-569 at periapsis to give a = 2,679km.
• r is the distance between the two objects at a given time.

 

An object moving fast enough to escape Kerbin's gravity is in an orbit with a semi-major axis that is effectively infinite. At periapsis, this simplifies the vis-viva equation to describe the kinetic energy that an object must have in order to overcome Kerbin's gravity from a given distance r:

 

 

In other words, escape speed from Kerbin orbit at the moment of periapsis (r = 2,679km) is v_e = 1,623 m/s.

But since SDD-569 is tracing a hyperbolic (i.e. open) trajectory through Kerbin's SoI, it must be traveling faster than this, or else it would be captured. How much faster?

 

Consider the other extreme case of the vis-viva relation, where the asteroid has shot past Kerbin and the distance r between them trends towards an infinite apoapsis. Setting r =∞ in the vis-viva equation tells us how fast the object is still going at that point, which is called its hyperbolic excess velocity:

 

 (where μ = GM)

 

So for the flyby of SDD-569, the hyperbolic excess velocity is v_∞ = 1,148 m/s. This characteristic energy is over and above the energy needed to escape Kerbin's SoI from that distance, so the total energy possessed by SDD-569 relative to Kerbin at periapsis is:
 

This gives a total velocity for SDD-569 relative to Kerbin at periapsis of 1,988 m/s! By how much do we need to reduce this so that it drops into a nice 2Mm circular orbit from periapsis?

In a circular orbit, the distance between the two objects r and the orbital radius a are always the same. Thus the orbital velocity is:
 

 

Not coincidentally, this is the same as its hyperbolic excess velocity, because r = a. So at r = 2,679km, an object in a circular orbit around Kerbin travels at 1,148 m/s.

 

So, to get SDD-569 into a circular orbit from its flyby periapsis, we need to bleed off Δv = 840 m/s.

 

To then bring SDD-569 down to a more convenient altitude of 500km, we would do a Hohmann transfer, which can be calculated with the standard formula, and works out to another 614m/s Δv to descend to a 500km circular orbit, for a total of 1,454 m/s.

 

What’s more, the spacecraft sent to capture SDD-569 needs to match orbits with the asteroid in order rendezvous. That means that  if the spacecraft starts from, say, 500km above Kerbin, it will need to expend that much to get to the asteroid in the first place. So, starting from a 500km orbit around Kerbin, the total Δv budget for this mission is 2,909 m/s.

 

Next up:

 

• Asteroid capture planning, part 2: How much fuel do we need to bring?
• Mission to SDD-569: Where the rubber meets the regolith!

Did I get this right? If you have feedback or ideas, I would love to hear them!  

 

Edited September 16, 2016 by reaction_wheel
(edit: tweaked the numbers because they were slightly off)

[Curveball Anders]

8 hours ago, reaction_wheel said:

[math]

I can't stop to wonder over this forum.

It's one click on "next unread" to go from a 100% silly discussion on Kerbal biology to dissertion on orbital mechanics and math.

For the latter I need more coffee ...

[klesh]

I have never mathed it out like the OP, but have had 100% succes in capturing asteroids by simply overbuilding my vessel.   You need alot less that you might imagine just by looking at them.  

[mrclucks]

You say 2909 dv, but you have to remember to account for the mass of the astroid once you dock with it.  Now I don't know the mass of the actual craft that will capture the astroid nor do I know the masses for each class of astroid, but since its only a class A then (and I'm just spit balling) I'd say you need about 3500 dv unmodified.  

I think you can see the actual mass of the astroid in the contract but I wouldn't know since I never do astroid contracts and only load up a craft with tons of dv if I'm going astroid hunting.  

If your craft has a mining system then that's alot of dv you don't have to think much about since you'll probably get tons of it drilling the astroid

[neil470]

It seems you've mixed up semi-major axis (a in  the equation) with periapsis (r) - they are not interchangeable except in the case of a circular orbit. The semi-major axis for a hyperbolic orbit is negative, not infinite. We would need to know some other information about the asteroid's initial orbit to obtain the semi-major axis - eccentricity for example. Though, you did correctly set the lower bound for the asteroid's velocity using the escape velocity equation, calculated at the asteroid's periapsis. I'm still trying to figure this problem out - it's got me stumped.

And, just a nitpick - the equation for kinetic energy comes from Newton's first law (F = m*dv/dt) and some calculus, not the third law.