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Disposal of radioactive materials

Posted by: slycker - Thu Dec 02, 2004 10:33 pm
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Post    Posted on: Mon Dec 20, 2004 1:45 pm
Ekkehard Augustin wrote:
Gravity isn't sufficient to calculate an orbit - another vector pointing to a different direction does exist in reality and is required. This vector is the velocity of the object.

There are two vectors in the diagram. The black vector is velocity. It is not a force. The red vector is the force of gravity (actually the acceleration due to the force of gravity) at the point X,Y due to the sun at point 0,0. These 3 quantities, Position, Velocity and Acceleration are all that is needed to calculate an orbit. Only one of the 3 is a force. Or more properly the gravity acceleration vector is the RESULT of a force, the force of gravity.

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In reality too there is a force working slycker mentioned recently - the centrifugal force which isn't included in orbit.xls and isn't included too in the mathematics I myself described. The centrifugal force is working to the opposite direction of gravity.

:oops: I knew I was making a mistake mentioning centrifugal force in the funnel example because there is no such thing as centrifugal force. You need to exert a centripetal force on a moving object to make it curve around a point in space. The centripetal force you exert to keep the object curving can be misinterpreted as opposing an imaginary force directed away from the center, but centrifugal force is not real in physics.


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Post    Posted on: Mon Dec 20, 2004 2:13 pm
Alright - we agree on the vectors again.

But I still feel unsatisfied by orbit.xls. As I said earlier in this thread the mathematics seem to be rigth to me except the problems I had. Leave these problems aside now - then I continue to be missing something. orbit.xls simply seems to do recalculations after entering a DeltaV - it simply is calculating a new orbit. In other word there is starting orbit and a resulting new orbit - but orbit.xls doesn't show the transition course for an object that moves from the starting orbit to the resulting orbit.

To move from a starting orbit to a certain resulting orbit requires a special course always that has to be calculated - what NASA always is doing while planning a mission. Given a certain launch date and time there is only one course to enter another orbit at a special date and time at a special point of that orbit.

Additionaly there are very much orbits between the surface of sun and the earthian orbit. To keep an orbit at an altitude above sun's surface of let's say 10,000 km a tangential velocity of very much more than at Mercury's distance is required. If a package gets there and has a tangential velocity significantly less than required then it will fall into the sun instead of going to an orbit somewhere between a few meters above the surface and 10,000 km. And the reason or cause will be gravity.

You may disagree but I want to calculate courses to see what really will happen - a simple result calculated by orbit.xls is too insufficient. For this reason I want to use something like I described earlier - I need alternatives to changing DeltaV in an Excel spreadsheet.

It will take time to do all the steps that are required but I will start to do them in the coming days.



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Post    Posted on: Mon Dec 20, 2004 4:04 pm
Ekkehard Augustin wrote:
Alright - we agree on the vectors again.

Now let me go through in detail how orbit.xls calculates the vectors.

At the start the sun is assumed to be at X,Y = 0,0 and the user (me) has placed the space craft at X,Y = 0,10. The black velocity vector is specified also, by both you and me, but in a slightly indirect way. I have asked the user to input DeltaVx and DeltaVy. I multiply these by a scale factor (to avoid using the huge numbers associated with the real size and speed of an orbit). Then I add these to other values I have chosen and the result is used as the first value of the V vector. The values I chose are such that if both DeltaVs you enter are 0 then the orbit will be a circle.

Now let’s consider how the black and red vectors are used in orbit.xls.

The black velocity vector has magnitude V at angle Theta. It can be decomposed into the two blue components parallel the X and Y axes. The black and blue vectors are related this way:
Vx = V * COS(Theta)
Vy = V * SIN(Theta)
V = SQRT(Vx^2 + Vy^2)
The black vector V never appears in orbit.xls but Vx and Vy together carry all the direction and magnitude information of V.

The red acceleration vector has magnitude A at some other angle Phi. It can be decomposed into the two green components parallel the X and Y axes. The red and green vectors are related this way:
Ax = A * COS(Phi)
Ay = A * SIN(Phi)
A = SQRT(Ax^2 + Ay^2)
The red vector A never appears in orbit.xls but Ax and Ay together carry all the direction and magnitude information of A.

Although A never appears in orbit.xls, I calculate Ax and Ay directly from information that IS in orbit.xls this way.
Ax = -X / R^3
Ay = -Y / R^3
(For now please just accept this calculation. We can go over it in detail later if you agree with the rest of this explanation)

To add the V and A vectors all I need to do is add their components. The angle between V and A never needs to be used directly.
New_Vx = Vx + Ax * Time
New_Vy = Vy + Ay * Time
Where Time is assumed to be 1.

I could calculate the yellow velocity vector magnitude this way
New_V = SQRT(New_Vx^2 + New_Vy^2)
but I don’t need to. All I want to use the yellow vector for is to find the new position of the space craft in X,Y coordinates. It is done this way:
New_X = Old_X + Vx * Time
New_Y = Old_Y + Vy * Time
Where Time is again assumed to be 1.

But I will use the NEW velocity to calculate this, not the original initial value entered by the user. So the math is really like this:
New_X = Old_X + New_Vx
New_Y = Old_Y + New_Vy

Now we are at the second line of orbit.xls. (ignoring the header line of course) The values of X,Y and Vx,Vy are different than what was entered by the user on the 1st line. Excel calculates new values for R and Ax,Ay to complete the information on the 2nd line.

These new values are then used in place of the original values to repeat the calculations for the next line in orbit.xls. This is repeated as often as needed to calculate any length orbital path. It is a brute force numerical calculation, but it works.


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Post    Posted on: Tue Dec 21, 2004 7:50 am
Hello, Peter,

your explanations seem to prove that I am right in mising something in orbit.xls:

1. DeltaVx and DeltaVy do specify the black velocity vector only - but this is not the only vector that can be controlled. I will say something about this below.

2. The user of orbit.xls has indirect control over the black velocity vector only. This means that if he wants a special black velocity vector he has to calculate the blue vectors based on the wanted black velocity vector first and then to enter the results into orbit.xls to get the black velocity vector again. This is a preventable source of errors and causes unproductive extended mathematical requirements some people don't like - but this other people do fell different.

Now the announced words why the black velocity vector doesn't be the only vector that can be controlled:

You are right - the red acceleration vector is caused by gravity. Acceleration is measured by km/s^2 and this you can use as red vector and it is to be reduced by the cube of distance from the sun as the center of gravity. But consider an object moving fast through a true empty space. Nothing is effecting its course then and it will be moving straight forward (Newton's theoretical model).

For simplicity let's say that this straight foward direction of motion is the X-axis. The object than has no second vector causing an Y<>0. Now suddenly a strong gravitational center like the sun appears aside the direction of motion. This causes acceleration to the side of the straight-forward-direction and suddenly there really is a vector causing an Y<>0... which means that there is not only acceleration but velocity too. And this velocity can be used as vector instead of the acceleration.

The velocity to the side was 0 km/s before the gravitational center appeared and if the acceleration caused by the center is 0.0001 km/s^2 then the vector is 0.0001 km/s after one second and this vector can be used to calculate angle and distance of motion for that second. After that second the body is closer to the center because of that vector which means that the acceleration by gravity has increased - for this reason the vector to the side of the former direction has to be increased by adding x km/s. "x" is given by the acceleration and will be higher than the first 0.0001. And so on.

Now the former straight-forward-velocity had a certain amount independant of any gravity and the suddenly appeared gravitational center causes an acceleration depending only on the mass concentrated within the center. there is a certain relation possible between the former velocity and the acceleration by the center which can cause a motion so that at the end of each second the distance to the center is identical and the acceleration by gravity is identical after each that second and at each point of the course. Then the object is going a circular orbit.

This way the gravitational vector is be a velocity vector like calculated by my own mathematical way I described.

Here now something is possible that doesn't seem to be possible by orbit.xls. Acceleration to the direction of the gravitational velocity vector is possible - by rockets causing thrust to that direction (to the inner of the orbit) with their exhaust going to the opposite direction (outer of the orbit). The acceleration by this thrust then has to be added to the gravitational velocity vector each second. This I cannot see to be possible in orbit.xls - DeltaVx and DeltaVy don't provide acceleration to the center of gravity.

It reminds me to integrals a little bit.

The formulars you listed I saw display pointing to the graphics shown in orbit.xls but I am interested in the way you are interpreting them and do something by them.

Your mathematics are right but they don't fit fully into the requirements here.



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Post    Posted on: Tue Dec 21, 2004 2:25 pm
The key to understanding this is in the definition of acceleration. Acceleration is velocity change per unit time. In other words acceleration can be used as velocity if done correctly. The red vector is the acceleration caused by gravity, but I am really using it as another velocity.

Consider a velocity of 4 meters per second to the right along the X axis. This is the black vector and it has 0 Y component. It will continue forever to the right at 4 meters per second if no force appears.
Now a gravity force suddenly appears far below on the Y axis, so the red vector appears. Let us assume it causes an acceleration of 3 meters per second squared directly down. This will cause the black velocity to curve down in the –Y direction without changing the 4 meters per second in the X direction. How do we calculate the curving path? The definition of acceleration is change in velocity per unit time. An acceleration of 3 meters per second squared means the speed changes by 3 meters per second every second. So if the 3 meter per second squared gravity suddenly appears, a stationary object will start moving. After one second it will be moving at 3 meters per second. After 2 seconds it will be moving 6 meters per second, and so on. What I am doing in orbit.xls is simply considering the amount of velocity change caused by the red vector in one unit of time as a velocity instead of an acceleration. I assume that the 3 meter per second squared red vector results in a VELOCITY in the direction of the red vector of 3 meters per second after it has been acting for one second. Then it becomes a 3 meter per second Y velocity to be added to the 4 meter per second X velocity of the black vector. So just before gravity appears the object is moving to the right at 4 meters per second. One second after gravity appears the object is moving to the right at 4 meters per second AND down at 3 meters per second. This is a 3-4-5 right triangle so the new velocity (one second after gravity starts) is 5 meters per second angled down at 37 degrees. After another second the Y velocity has increased another 3 meters per second. Add this additional 3 meters per second to the existing 3 and after 2 seconds the Y velocity is 6 meters per second down. The X velocity is still unchanged at 4 meters per second so the total velocity is now SQRT(6^2 + 4^2) = 7.2 meters per second angled down at 56 degrees. If this continued forever then the object would end up going down at infinite velocity and to the right at 4 meters per second.

But wait, there is more!

The gravity field is not always directly down. If it comes from a point on the Y axis, then as you move to the right and descend below the X axis the direction of the gravity force changes and so the related velocity change starts to have an X component. Now the 4 meters per second X velocity starts to change too. And if you started the object with EXACTLY the right location, speed and direction it will move in a circle. If you change any of the starting conditions the shape will not be a circle. What I am trying to show is that if it isn’t a circle then it has to be a straight line, parabola, hyperbola or ellipse. It can’t be a spiral.

Now a word about thrust. I have provided only one place in orbit.xls to calculate thrust, just at the start. Thrust is a force. The force results in an acceleration. And I am considering this acceleration as another velocity to be added to the black vector, just like I do with gravity. The only difference is that the total velocity change due to thrust is assumed to take place all at once, just before orbit.xls starts computing gravity. So I just add the DeltaV caused by thrust to the starting velocity. I have written versions of this program where the user has the ability to command thrust in any direction at any time, but I didn’t do it in Excel, it complicates the code and it isn’t needed for what I am trying to show here.

What I AM trying to show is that if you apply any amount of thrust for A SHORT TIME and then coast according to the laws of gravity that you cannot spiral into the gravity source.


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Post    Posted on: Wed Dec 22, 2004 10:11 am
Hello, Peter,

your last explanation is convincing me that I'm right in that by orbit.xls something isn't possible what's possible using my method. And I always agree that under these restricted possibilities within orbit.xls do allow orbits only and no spirals. So orbit.xls don't include any argument for spirals being impossible. To find such arguments other methods have to be applied.

What orbit.xls is proving using correct mathematics and physics are relations between gravity, velocity and angle that make sure that an object is going along an orbit - orbit.xls is providing answers to the question what orbits are surely possible.



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Post    Posted on: Wed Dec 22, 2004 2:14 pm
:? OK…..
So you agree that both my math and physics are correct but STILL believe an object can spiral into the Sun without a continuous force being applied.

This would violate conservation of energy.

The funnel example is a purely physical tool that also shows spirals to be impossible. Remember that the ball spirals into the center of the funnel due to the continuous force of friction. If the funnel and ball could be made with absoultely no friction, the ball would never spiral in. Do you agree?


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Post    Posted on: Wed Dec 22, 2004 2:38 pm
The Energy Conservation Theorem isn't violated anyway by what I'm thinking - I don't state that the spacecraft is loosing energy.

In opposite - it's getting energy: By the thrust of the rocket or engine. This thrust is working to the direction of sun and gets the spacecraft this way kinetic energy for motion to the direction of sun.

This energy is moved over from the exhaust etc. to the spacecraft - like it happens whenever a spacecraft is launched out of an orbit around earth into the interplanetary space. The energy connected to the tangential component isn't destroyed - the energy connected to the vertical movement is increased. Still the only loss of energy is caused by gravity of sun - but this loss of energy doesn't hurt the Energy Conservation Theorem too.

I'm considering the spacecraft - not its energy. No change of the spacecraft's energy does mean any change of the total amount of energy - a change of the spacecraft's energy does mean only a move of a portion of the total amount of energy, a cahnge in the distribution of energy among objects, gases, dust etc.



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Post    Posted on: Wed Dec 22, 2004 5:22 pm
Ekkehard Augustin wrote:
Still the only loss of energy is caused by gravity of sun - but this loss of energy doesn't hurt the Energy Conservation Theorem too.


The gravity of the Sun does NOT cause a loss of energy. It causes a GAIN in KINETIC energy, just like a falling rock gains kinetic energy as it falls. What you ARE losing is potential energy. The two forms of energy always add up to the same total energy to obey conservation of energy.

Kinetic energy is ½MV^2. This does not depend of the direction of V, only the magnitude. (Gravitational potential energy is mgh, more on that later) A moving space craft has kinetic energy and a stationary space craft does not. When you fire your rocket the potential energy of the fuel is converted into kinetic energy. On Earth, a rock has kinetic energy if you throw it. If you just drop it, it will fall. As it falls it speeds up. As it speeds up it’s kinetic energy increases. An object gaining energy like that would be violating conservation of energy so the kinetic energy must be coming from someplace. You didn’t throw it so that is not the source of it’s kinetic energy. The source is the conversion of gravitational potential energy. A space craft gains kinetic energy from the chemical or nuclear energy of the rocket fuel if the engine is operating. If it is just moving in space without power, all increased kinetic energy comes from gravitational potential energy.

http://en.wikipedia.org/wiki/Potential_ ... ial_energy

Gravitational potential energy is mass * gravitational acceleration * height or:
E = mgh

The source of gravitational potential energy is only due to your POSITION with respect to a gravity source. It does not depend on speed or direction. A rock sitting still on a table has gravitational potential energy. A rock sitting at the bottom of a hole on Earth has gravitational potential energy because it COULD fall into a deeper hole. Only a rock sitting at the center of the Earth has no gravitational potential energy (assuming the Earth is the only object in the universe).

When a space craft is 100 million kilometers from the Sun the energy calculation has to also take into account the changing value of “g” over that 100 million kilometer distance, but it is really the same type of calculation. A moving or even STATIONARY spacecraft has energy due purely to the presence of the distant Sun. If the space craft starts stationary and is allowed to fall into the sun along a straight line then potential energy gets converted to kinetic energy. It hits the surface of the sun going very fast. The energy associated with the impact all came from the potential energy the stationary space craft had just before you let it fall. And if the space craft starts with some kinetic energy with a velocity perpendicular to the falling direction, then you will get some kind of orbit.

So when I say energy has to be conserved I am saying the sum of the potential and kinetic energy has to remain constant. Converting chemical or nuclear potential energy with a rocket is only one way to change kinetic energy. This will not apply to an unpowered package of waste falling toward the Sun. In that case the kinetic energy increase is exactly balanced by the potential energy decrease. The Orbital Energy Equation

http://en.wikipedia.org/wiki/Specific_orbital_energy

can be used to calculate the exact amount of speed (but not direction) you will get just by approaching nearer to the Sun. If you do this you will see that you need to continuously apply a force against the direction of motion to keep the space craft from gaining a speed that is much faster than the speed of the various circular orbits it is passing as it approaches the Sun.


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Post    Posted on: Thu Dec 23, 2004 7:49 am
Hello, Peter,

you wrote "The gravity of the Sun does NOT cause a loss of energy. It causes a GAIN in KINETIC energy, just like a falling rock gains kinetic energy as it falls. What you ARE losing is potential energy. The two forms of energy always add up to the same total energy to obey conservation of energy. "

Alright - I thought you were arguing that the gravity of sun were causing a loss of energy responsible for the tangential velocity. This seems to be a misunderstanding as I see now.

But then I really never have been using arguments that would hurt the Energy Conservation Theorem.

I still have to read the remainder of your post but I fear it would again lead us away from the topic of the disposal of radioactive materials into the sun.



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Post    Posted on: Thu Dec 23, 2004 1:30 pm
Math and physics are not convincing you. Let me try authority and common sense. Please refer to this web page

http://en.wikipedia.org/wiki/Orbit

Or in German:
http://de.wikipedia.org/wiki/Orbit_%28H ... echanik%29

This is one of many documents stating that objects in space near a point source of gravity must move in conic sections. A spiral is not a conic section.

Have you noticed that all orbital texts speak of conic sections but never mention a spiral? Don’t you think that at least SOME of them would mention a spiral explicitly, and maybe show math examples calculating spirals, if spirals were possible?

Doesn’t it seem strange that planets and asteroids manage to maintain just exactly the velocity they need to keep from spiraling into the Sun? If it were so easy to drop a package of waste into the sun just by giving it a small push in the right direction don’t you think that a large meteoroid or asteroid impact would have done that to the Earth long ago?

Doesn’t it seem just a little odd that ESA and NASA never use a spiral path to go anywhere? Or maybe they do.
http://www.esa.int/SPECIALS/SMART-1/SEMQXBXO4HD_0.html
Oops, that is under constant thrust from it’s ion engine. And it is spiraling up, not down.

I challenge you to find and post an authoritative reference describing a spiral into a point source of gravity that does not rely on some mechanism such as friction to remove energy from the orbiting body. If you cannot find such a reference you must consider the possibility that your idea of spiraling into the Sun is a misconception simply based on every day experience where friction is always present and gravity seems like an irresistible force pulling everything down.


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Post    Posted on: Thu Dec 23, 2004 1:50 pm
Mathematics can convince me because I do understand mathematics very good - so that's really no problem and has never been.

I still didn't find time to read the remainder of your previous post. But whatever you wrote and I read doesn't include any information where my mathematics don't represent physics.

Regarding the Energy Conservation Theorem:

Is it wrong that the exhaust of the rocket moves kinetic energy to the vehicle?
Is it wrong that this will increase the velocity to that direction the thrust was going to?
Is it wrong that this velocity will be kept?
Is it wrong that the ECT doesn't mean transformations from one kind of enrgy to another kind of energy only but is including the transmission or move of energy from one object to another?
Where is the point requiring the ECT to be hurt in my posts or in my mathematics?

Please point directly to the error or mistake I made seen from your point of view - that would be helpful for this discussion.

There is a real and certain difference between an orbiting planet and a vehicle launched by intention and conscious to a goal because in the last case velocity and angle may and often will be different to velocities and angles caused by apparently "random" events of nature. I don't insist on spirals being possible - I suppose that it is possible to throw something into the sun without providing permanent thrust. Look at the probes to Mars and Saturn - the engine is shut off while the vehicles were moving millions of kilometers. The same physics are working if they would have gone into the inner of the earthian orbit.

Next - it is known that the gas giants do modify and change the orbits of asteroids and comets significantly. And there are comets that have fallen into the sun - so I don't see any reason why it shouldn't be possible to give a package a course leading it into the sun without engine or drive theoretically at least - the comets that fell into the sun didn't have engines or drives too. I was arguing that no Jupiter gravity assist is required because I'm supposing that the velocity that can be got by Jupiter isn't required necessaryly. If this amount of velocity is to high to get it without Jupiter this would be an argument and I would have to think it over - but that would be an argument by number not by physics themselves.

To repeat it - please point directly to the point where is a logical error, a mathematical error, an error of thoughts or where are incorrect knowledges of physics involved in my argumentation. Then I would have to look into my books to find the source of the errors or mistakes. I invite you to go through my argumentation step by step together with me.



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Post    Posted on: Thu Dec 23, 2004 2:23 pm
The error or mistake you have made seen from my point of view is that your estimate of HOW MUCH THRUST would be required is WAY TOO SMALL. You have never specified any numbers, only made unsupported claims that a “small” thrust would be enough to send a package of radioactive waste on a path spiraling into the Sun. How much thrust is “small”? Or more importantly, what velocity would the package have after this “small” thrust were applied?

Your ability to do math is not in question.

We agree on the effect of rocket thrust on a space craft.

We disagree on the speed and direction that an unpowered package of radioactive waste leaving the vicinity of the Earth would need to hit the Sun.

I am stating that a package of radioactive waste would need to leave the space craft with a speed of 2681 meters per second relative to the Sun or 26,809 meters per second relative to the Earth, directed tangent to the Earth’s orbital path to reach the sun.

You have stated that some much smaller velocity in some other direction would be all that is needed without giving any numbers.

Please state your numbers.

What velocity and in what direction would a package of radioactive waste need after is released from the space craft to reach the Sun? State the numbers relative to either the Sun or Earth.


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Post    Posted on: Thu Dec 23, 2004 2:38 pm
Allright - up to now we have been discussing mathematics to be used. This is the reason why I didn't state numbers yet - we first had to agree that the mathematics are right.

You agree that my matematics can be used and I have started to do some calculations a few days ago - I hope to find time within the next three days to post results here.

The first value I'm going to use is around 7 km/s - I will use others later. I have in mind several cases.



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Post    Posted on: Thu Dec 23, 2004 2:54 pm
7 km/s or 7,000 meters per second. Is that relative to the Sun or Earth? And in what direction?


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