Nuclear Waste Disposal into Sun or outside Solar System?by osxraider Tags: delta v requirements, nuclear waste, orbit determination 

#1
Dec312, 01:56 AM

P: 26

Hello fellow aerospace engineers and physicists, I am currently working on a project and trying to figure this out. I was wondering if you guys could help me out.
Here is the problem statement: A nuclear waste disposal spacecraft is to be sent from Earth in order to carry radioactive waste either out of the solar system or into the Sun. Which alternative is cheaper in terms of delta v requirements ? The lesser delta v, the cheaper! Remember that planetary flybys are allowed which means Venus and Mercury for this specific problem going to Sun. We will assume that whenever the spacecraft reaches Venus's sun orbit, venus will be there and the same for mercury. In case we are dumping outside the Solar System, we need to consider all planets beyond Earth. These should be used to minimize the required delta v. Further, both trajectories lie in the elliptic. Can people get me started off on how to go about designing the orbits and then figuring out which one is ultimately cheaper? I will also need to use STK to design the orbits! Thanks! PS: I thought about whether it was more appropriate to post this in the coursework section but given the specific nature of this problem and the poor reply rate for aerospace/highly specific problems, I decided on posting here. I feel I will get more responses on here. 



#2
Dec312, 04:59 AM

Mentor
P: 14,481





#3
Dec312, 01:38 PM

P: 26

Yes D H, that is what I meant. Sorry about that!




#4
Dec312, 01:41 PM

P: 1,784

Nuclear Waste Disposal into Sun or outside Solar System?
Don't forget that Venus and Mercury can also help you get out of the Solar System.




#5
Dec312, 02:25 PM

P: 26

skeptic, I never considered that. So I use planetary flybys to first send the spacecraft towards sun and then at Mercury, a decision needs to be made whether to turn around or head into the Sun?
that seems a little counter intuitive as both options would have the same delta v till Mercury but then you would need a small delta v to head into the Sun versus turning back and in spite of using all planets for fly bys, I feel the delta v will be more? For the purposes of this problem, we assume our garbage (truck) spacecraft is already in Earth orbit or in other words, the launching into Earth orbit for both choices is the same and hence no cost assessment can be made for that part of the problem. 



#6
Dec312, 04:19 PM

Mentor
P: 14,481

Needless to say, so is using the Earth out of scope. Using the Earth as a gravity assist for a payload of nuclear waste perhaps isn't that good an idea. (Aside: The Earth has been used for multiple gravity assists, including satellites with RTGs. The problem statement however doesn't appear to allow using the Earth for gravity assists.) 



#7
Dec312, 05:14 PM

P: 26

Actually, DH, That wasn't part of the original problem statement. It was simply my assumption that when we fly towards the Sun, we will only encounter the inner planets and when we fly out, we will encounter all outer planets. The problem would allow for using the Earth itself as a Gravity Assist.
Evidently, I am finding out that my assumption was a bad one and so i apologize for that. Also, we will assume that there will be no safety hazard of ay form. That is irrelevant to this problem. We assume that both missions will have a 100% success rate/ no failures. Also, it is irrelevant whether either methods are realistically feasible. The purpose of the problem is simply to investigate which trajectory will have lesser delta v requirements and therefore turn out to be cheaper. All planetary fly by's are allowed. The orbits of all planets are in the same plane (assumed) and the circular earth orbit in which the spacecraft is currently residing is this same plane. Therefore, the problem is essentially reduced to 2D. Later this evening, I will post my basic build up for solving this problem and you guys can comment on whether I got everything right or wrong. My basic plan is to use an initial delta v to lodge the spacecraft out of earth orbit and then towards the Sun. One its way there, it will encounter Venus and Mercury. Mercury and Venus must simply accelerate the craft faster (as Voyager was by Jupiter and Saturn). As for leaving the Solar System, the method should be using a same initial delta v but when the spacecraft is moving in the opposite direction and then head outwards. My professor has clarified that to keep it simple, I must assume that the craft will encounter every planet when it reaches the planet's Sun orbiting radius/distance. So essentially, there will 5 planetary fly by's (Mars, Jupiter, Saturn, Uranus and Neptune.) Once the craft has crossed Neptune, it is considered to have left the Solar System. I guess what I am asking essentially to map out Voyager's journey as it left the System but only in this case, it will encounter 5 planets that it can use for gravity assist instead of just 2. 



#8
Dec312, 05:27 PM

P: 26

Also, as for using Earth itself as a gravity assist. I don't see how that might be possible since the spacecraft is initially already in Earth orbit. It needs to leave this orbit to go anywhere and that is only possible by apply delta v, not using gravity assist.
As for using inner planets (Mercury and Venus) to leave the Solar System, I am not sure I would need to do that. I just wondering, they didn't send Voyager inward first? did they? I think since this is a simple undergrad problem and we have already made simplifying assumptions, it is fair to assume that only planets between the desired destination should be considered for gravity assist. That being said, would Mercury and Venus actually make enough of a difference to tip the scale in favor of sending out to Solar System? In this case, I suppose you could also use Earth 2 and essentially get all 8 planets doing gravity assist. How should I begin? Should I start with a simple Hohmann transfer from circular Earth orbit to orbit around the Sun? that would mean 2 delta v's. One to eject and one to insert. Finally, one last delta v to send into Sun from Sun orbit. In between, we use to gravity assists to speed it up? The thing I am a little confused about right now is I don't even understand clearly where these gravity assists would be helpful? delta v's are use to make trajectory changes. there will always be a minimum number or delta v's required. The only thing that gravity assists to do is speed up the craft and this problem doesn't ask to find which one is faster. I am reading the problem and only seeing what is required which is minimum delta v scenario. I think my professor simply wants me to include gravity assists/ planetary flybys since this will make the report more detailed and he will be grading the reports based on use of these planetary flybys. 



#9
Dec312, 05:51 PM

Mentor
P: 14,481

Or you can swing by Earth after a gravity assist by another planet. That's what Cassini did, first encountering Venus twice, then Earth, then Jupiter, before finally arriving at Saturn and entering Saturn orbit. 



#10
Dec512, 03:38 AM

P: 166

It would seem to me that deltaV into the sun is always going to require less energy than escaping the solar system. After all, all you need to do is get it into the same orbit as earth and then slow it down enough for its orbit to naturally decay. If you don't care that it might take hundreds or thousands of years that is, but you didn't specify a time constraint. ;)
I suppose a little extra calculation to ensure it doesn't gravity assist from Venus and come back to smack into us would be smart. 



#11
Dec512, 04:37 AM

Mentor
P: 14,481

Scientists worry about near Earth objects impacting the Earth. Your solution has just created an extremely hazardous NEO. 



#12
Dec512, 10:49 AM

P: 1,784





#13
Dec512, 11:52 AM

Emeritus
Sci Advisor
PF Gold
P: 2,352

When the rocket reaches Venus' orbit it will have picked up speed from falling in towards the Sun and will be moving faster than Venus. We let the probe catch to Venus from behind. Let's assume that Venus' orbital velocity is V, and the difference between the Rocket's velocity and Venus' is Y. This means that before the rocket starts being pulled in by Venus' gravity its orbital speed is V+Y. The rocket is pulled in towards Venus, does a parabolic whip around and heads back out again. Once it reaches its original starting distance from Venus its relative speed to Venus will be again Y, However, where before this velocity was towards Venus in the same and in the same direction as Venus' orbital direction, coming out it will be away from Venus and counter to Venus orbit. Thus the rocket's new orbital velocity with respect to the Sun is VY and is less than it was before. 



#14
Dec512, 12:04 PM

Emeritus
Sci Advisor
PF Gold
P: 2,352

In order to escape the Solar system, it would take ~ 41% of that or ~ 12.4 km/sec. 



#15
Dec512, 12:06 PM

Mentor
P: 14,481

A planet can be used to change both the magnitude and the direction of the velocity as observed in a suncentered frame of reference. These gravity assist maneuvers are used for almost every interplanetary mission that isn't going to the Moon, Venus, Mars, or Jupiter. To envision these gravity assist maneuvers, first look at things from the perspective of a heliocentric frame or a barycentric frame. The vehicle is approaching some planet at a high speed. The planet will do something to the vehicle; the issue is to determine what that something is. The best way to see what that planet does is to switch the perspective to a planetcentered frame of reference. From this perspective, the approaching vehicle is moving toward the planet with a velocity that is significantly greater than escape velocity. The vehicle is on a hyperbolic trajectory. Because the velocity is so high, the velocity won't change all that much until the vehicle is fairly close to the planet. This point where the velocity does start changing is the point at which one switch perspective from the heliocentric frame to the planetcentered frame. The velocity of the vehicle at this point is called [itex]v_{\infty}[/itex]. (Better: [itex]v_{\infty}[/itex] would be the vehicle's velocity at an infinite distance if only the vehicle wasn't also orbiting the Sun.) The vehicle would follow a hyperbolic trajectory if the Sun wasn't present. The vehicle will continue on a mirror image of the inbound trajectory after reaching periapsis (closest approach). The velocity will drop until its magnitude reaches this same [itex]v_{\infty}[/itex] on the outbound leg of the trajectory. From the perspective of this planetcentered frame, all that this flyby accomplished was to rotate the velocity vector. The encounter does not change the magnitude. From this planetcentered perspective, the Sun's gravitational influence on the vehicle is fairly small during this encounter. To first order, one can model the encounter with the planet as comprising three phases:
We can simplify step #2, the hyperbolic encounter, even further. Because the encounter is fairly short in duration, the distance traversed during this encounter is rather small compared to the hugeness of the initial orbit about the Sun. To first order, all that this encounter does is to rotate the velocity vector rom the perspective of the planetcentered frame. What happens from the perspective of the heliocentric frame? From the perspective of the planetcentered frame, this rotation of the velocity vector is described by some rotation matrix T: [tex]\vec{v}_{\text{out},\text{planet}} = T\,\vec{v}_{\text{in},\text{planet}}[/tex] From the perspective of the Suncentered frame, this becomes [tex]\begin{aligned} \vec{v}_{\text{out},\text{sun}} &= \vec{v}_{\text{out},\text{planet}} + \vec v_{\text{planet}} \\ &= T\,(\vec{v}_{\text{in},\text{sun}}  \vec v_{\text{planet}}) + \vec v_{\text{planet}} \\ &= T\, \vec{v}_{\text{in},\text{sun}} + (\pmb{1}T)\,\vec v_{\text{planet}} \end{aligned}[/tex] That first term on the right hand side, [itex]T\, \vec{v}_{\text{in},\text{sun}}[/itex], is just a rotation of the suncentered velocity vector. The second term, [itex](\pmb{1}T)\,\vec v_{\text{planet}}[/itex], changes both the direction and the magnitude of the suncentered velocity vector. 



#16
Dec712, 01:03 PM

P: 26

D H and Janus, thank you so much for providing some perspectives!
Janus, you provided some direct numbers. I'm curious as to how you came up with them. Could you give me a brief walk through? D H, you hit it right on. This project was handed out right after we covered the method of patched conics. For now, if I might, I would like to reduce the problem to just figuring out dropping it into the Sun. For that, I am thinking that I will first assume a circular parking orbit (probably the geostationary orbit) as it would allow us to choose a a suitable region of Earth's atmosphere over which the the waste will always orbit. Ok, so we have the geostationary orbit as our circular earth parking orbit. Here at periapsis, the spacecraft execute's a single delta v to leave the circular parking orbit and go into a Hohmann transfer ellipse. Normally, this is used for interplanetary trajectories but here is my idea: The Sun is really huge so we can imagine that we are sending the craft to an imaginary planet that is orbiting the Sun's center at a distance/radius that is equal to the radius of the Sun. This means that the apoapsis simple lies on the surface of the Sun on the other side. Basically, imagine an ellipse whose periapsis touches the RIGHT side of earth and the left side of the Sun. The major axis of this ellipse is simply Diameter of Sun+ Distance between Sun's equatorial surface and Earth's equatorial surface + Diameter of Earth. As the spacecraft reaches the apoapsis of its heliocentric transfer ellipse, it begins to get very close to the Sun's surface and can touches it at apoapsis. Though it will not encounter any planet here and normally would have just continued on the second half of the transfer ellipse, this is the surface and so the spacecraft will simply get destroyed. In reality, the spacecraft will begin melting long before it reaches this point. My professor has said that this is acceptable. Now my question [for part one of this problem] is how do we put gravity assists in and will they help? In gravity assist, after the fly by, the spacecraft leaves with hyperbolic trajectory and although I have a choice as to where I'd like to encounter the planet along the transfer ellipse, I can necessarily choose the outbound trajectory and ensure that it then heads towards the Sun without burning again and remember, we want to burn as little as possible. Hohmann transfer are supposed to be the most efficient? correct? so the way I am seeing this problem right now is that there will definitely be at least one burn during departure from earth parking orbit, let's call it delta v1. So if and when we burn again, we will be adding to total delta v {delta v1+....} Therefore, I think this is the most efficient trajectory in terms of minimum delta v requirements. [ON A SIDE NOTE, BACK TO PART 2 LEAVING STAR SYSTEM] Here I am employing similar reasoning that I employed for the problem with hyperbolic trajectory problem and minimum delta v requirements. Again, a Hohmann transfer to Mars (very simple), then a fly by/gravity assist from Mars and then Mars shoots it off on a hyperbolic trajectory. The final trajectory for leaving the Solar System must by hyperbolic so that the junk never returns. OK, I'm done. PS: I did not get alerts for new posts. I didn't even know there had been more replies. 



#17
Dec712, 01:12 PM

P: 26

Also, the post about just letting the craft decay is incorrect. The craft will still remain in orbit around the Sun trailing Earth. To get it to decay, you would have to slow it down to 0. Since it is following Earth in it's orbit, its speed must be the same as that of the Earth for that orbital radius around the Sun which means that your delta v would have to be the negative of Earth's velocity which is HUGE! That would definitely not be the minimum delta v. One thing to remember about why delta v is so important is that the craft will have to carry 79 times the amount old compared to cargo depending on specific impulse.
Also, I am not worrying too much about which one is lesser right now. Either way, I have to solve both and then simply find out so which one is lesser is actually of little relevance in the end as the motivation for the project is simply to design the orbits and perform the calculations. In the end, the conclusion would simply compare the 2 different delta v's and make a recommendation. I am still very interested in Janus's numbers! Was it really that easy to calculate? Also, I apologize for not quoting. I'm a newbie here. 



#18
Dec712, 08:46 PM

Mentor
P: 14,481

The problem is that the delta V needed to go from a circular orbit at 1AU to an elliptical orbit with a 1.39 million km perihelion is extremely expensive. 


Register to reply 
Related Discussions  
nuclear waste disposal by breakdown  Nuclear Engineering  13  
Solar Radiation and Nuclear Waste  General Physics  4  
Why don't we drop medical waste and nuclear waste into active volcanoes?  Nuclear Engineering  23  
A stupid question about nuclear disposal ..  Earth  9  
Offshore Nuclearwaste disposal.  Earth  13 