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Gravitational equilibrium between earth and sun

  1. Aug 2, 2008 #1
    Can someone tell me at what mileage point a rocket launched from earth toward the sun would have its primary gravitational draw shift from the earth to the sun?
     
  2. jcsd
  3. Aug 2, 2008 #2
    About 160,725.935 miles. Closer even than the moon. The moon feels 4 times as much force from the sun as the earth.
     
  4. Aug 3, 2008 #3
    Here's my question then. All I ever hear negative about nuclear power is the problem with disposal of the spent fuel rods. If jettisoned to the gravitational draw of the sun idicated above wouldn't that be the perfect solution?
     
  5. Aug 3, 2008 #4

    uart

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    Yes a one way trip to the Sun is a feasible solution to the permanent disposal of any type of toxic waste. This is something that's been suggested many times in the past however it does have one major problem.

    The cost would be quite high so presumably it would be most suitable to high level waste. Now here's the issue, if the rocket explodes on takeoff before it leaves the atmosphere then it spreads huge amounts of high level waste over a very wide area. The bottom line is that it's just too risky.
     
    Last edited: Aug 3, 2008
  6. Aug 3, 2008 #5

    D H

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    No, it is not. If you want to dispose of toxic waste in space it is much cheaper to send it out of the solar system entirely than to send it into the Sun.
     
  7. Aug 3, 2008 #6

    D H

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    The moon feels 2.2 times the force from the Sun than from the Earth, not 4 times. While true, this is a misleading answer. A better way to look at this is to look at the point where the Sun can no longer be treated as a third body effect, and that is much further out than the Moon. The Earth's Hill sphere has a radius of about 0.01 AU, or nearly four times further from the Earth than is the Moon.
    No. Just because you make something escape Earth's gravity well does not mean it will magically fall into the Sun. All you will have accomplished is to create a very bad kind of near Earth object. Those spent fuel rods will eventually reencounter the Earth, and all of the nuclear waste will be dispersed into the atmosphere.
     
  8. Aug 3, 2008 #7
    Found these looking around for stuff about nuclear reactors


    (Clipped from the Abstract) - This is interesting

    http://www.iop.org/EJ/abstract/1063-7869/46/7/A05

    Progress in particle accelerator technology makes it possible to use a proton accelerator to produce energy and to destroy nuclear waste efficiently. The energy amplifier (EA) proposed by Carlo Rubbia and his group is a subcritical fast neutron system driven by a proton accelerator. It is particularly attractive for destroying, through fission, transuranic elements produced by presently operating nuclear reactors. The EA could also efficiently and at minimal cost transform long-lived fission fragments using the concept of adiabatic resonance crossing (ARC), recently tested at CERN with the TARC experiment. The ARC concept can be extended to several other domains of application (production of radioactive isotopes for medicine and industry, neutron research applications, etc.).



    Another...
    http://www.iht.com/articles/1995/05/16/atomen.ttt.php
     
  9. Aug 3, 2008 #8

    uart

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    How do you come to that conclusion? By my calcuations the energy required to escape the Earths gravity (even without any help from the Sun) is less then one tenth of that required to escape the Suns gravity.

    No not by magic certainly, but by calculating the correct trajectory to send it off it does.



    BTW, I'm not supporting the idea, I've already ruled it out for reasons of cost and lanch safety.
     
    Last edited: Aug 3, 2008
  10. Aug 3, 2008 #9

    russ_watters

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    Don't worry - the waste disposal problem isn't real anyway. Nuclear waste is recyclable.
     
  11. Aug 3, 2008 #10

    D H

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    The Earth is in a nearly circular orbit. Achieving escape velocity from a circular orbit requires a 41.4% change in velocity. Diving into the Sun requires a canceling nearly 100% of the orbital velocity: nearly 2.5 times the change in velocity compared to escaping the solar system.

    Energy is a false metric because spacecraft need to carry the fuel needed to change velocity. Change in velocity, or delta V, is a much more meaningful metric. That is the metric that the world's space agencies use to determine the amount of fuel needed to accomplish some mission.
     
  12. Aug 3, 2008 #11

    Janus

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    In addition, the amount of fuel needed increases faster than the change in delta v does. That same 2.5 times difference in delta v requires 4.5 times as much fuel.
     
  13. Aug 3, 2008 #12

    tiny-tim

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    Welcome to PF!

    Hi cokerrm ! Welcome to PF! :smile:

    The trouble is that, although it will then be attracted more strongly to the sun than to the earth, it will still have roughly the same tangential velocity as the earth (18 miles per second).

    And that tangential velocity will prevent it from falling into the sun. :cry:
     
  14. Aug 3, 2008 #13

    D H

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    It might well (and probably will) take a lot more fuel than that. The rocket equation is very demanding.
     
  15. Aug 3, 2008 #14

    uart

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    Thanks for the info DH. Yes I was thinking in terms of the KE being one half the magnetude of of the negative PE. The delta V arguement makes sense.

    BTW just so that cokerrm knows exactly what we're talking about, the trajectory to launch into the Sun essentually means accelerating back in the opposite direction to what the Earth is travelling around the Sun until (almost) all of that 18 mile/sec tangential velocity is cancelled out. At this point the object would be on a trajectory right into the Sun.
     
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