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How much charge would the Earth and Sun need to knock the Earth out of orbit

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data

    How much charge would the Earth and Sun need to knock the Earth out of orbit. Then, assuming the distance from one proton to another and the length of one electron to another is 197 pm, find out how large of a block you would need of protium located on Earth and how large of a block of pure electrons located on the Sun you would need to accomplish the aforementioned result. Ignore the fact that a block of protium cannot exist since all of the protons would be repulsed by each other. Same for electrons. I made up this question.


    2. Relevant equations

    F = gMm
    F = kqq

    3. The attempt at a solution

    For the force of gravity the Earth's orbit around the Sun creates I have

    6.7*10^-11 * 1.99 * 10^30 * 5.97 * 10^24 = 7.96 * 10^44 Nm^2

    So to find out how much charge you need:

    7.96 * 10^44 = kqq = kq^2 = (8.99 * 10^9)q^2

    = 2.98 * 10^17

    To find out how many protons you would need you divide 2.98 * 10^17 by the charge of one proton, which is 1.61 * 10^-19 which equals 1.85 * 10^36 protons.

    I think 197 pm is 1.97 * 10^-10

    First I multiply them

    1.97 * 10^-10 * 1.85 * 10^36 = 3.64 * 10^26

    but then to get it in three dimensions I'm pretty sure you need to take the cube root.

    (3.64 * 10^26)^1/3 = 7.14 * 10^8 m^3 = 7.14 * 10^5 m^3

    On the one hand the volume of the Earth is 1.08 * 10^12 km^3. So the block would be something 6 * 10^-5 % of the Earth's volume which is small. But on the other hand the block would be bigger than the United States which is large.

    Let me know if you see any mistakes in my calculations.
     
  2. jcsd
  3. Jul 16, 2012 #2

    Andrew Mason

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    Homework Helper

    What about the r^2 term?

    Are you trying to see how much charge on the earth and sun is needed in order to 1. make the earth fall into the sun (ie its orbit < radius of the sun) or 2. to send the earth out of the solar system? It sounds like it is the former case.

    What is your reasoning for equating the electrical and gravitational forces? How does that make the earth crash into the sun?

    If you are trying to figure out the electrical force needed for the earth to crash into the sun, it is a bit complicated, because you have to calculate the force needed to make the earth's orbit such that on its closest point to the sun it hits the sun. That distance (ie. effective distance from the solar centre to its "surface") is a bit difficult to determine so you will have to state your assumptions.

    Suppose that you posed the second question: sending the earth out of the solar system. If the repulsive Coulomb force is equal in magnitude to the attractive gravitational force, what will happen?

    AM
     
  4. Jul 17, 2012 #3
    In the equation gmm/r^2 and kqq/r^2 the r^2 is the same for both of them, so I figured it does not factor in.

    Being repelled out of orbit or being attracted to the sun it would be the same number of charges, just the sign on the charge would be different, so either way it takes the same number of charges.

    I'm trying to find the force and both the gravitational force and the electric force are measured in Nm^2.

    Correct me if I'm wrong but if the electrical force were stronger than the gravitational force that would knock it out of orbit.
     
  5. Jul 17, 2012 #4

    Ibix

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    Your proposed formulae will work (although the quantity you are calculating is not a force) to calculate the charge needed to neutralise the gravitational attraction of the Sun-Earth system. However, this will not cause the Earth to crash into the Sun - quite the opposite. You also need to calculate the extra charge needed to counter the kinetic energy of the Earth.

    Equal sized spheres touching each other tend to adopt a configuration called face-centered cubic. If you look that up, and work out the volume of the cube and the number of spheres inside it (careful - look up fencepost error too) that will give you a volume per proton/electron. Take it from there.

    An alternative strategy would be dimensional analysis, which might give you a hint where your volume calculation has gone wrong. What units have you got? What should they be?
     
    Last edited: Jul 17, 2012
  6. Jul 17, 2012 #5
    Why not? By opposite do you mean it will expel the Earth from the solar system? If opposite electric charges attract, then why would a positively charged sun not attract a negatively charged Earth?

    See, I was wondering if the velocity of the Earth through space matters. I'm pretty sure it's roughly 21 km/s. I would think it would be harder to knock a fast object out of orbit than a slow object.

    So are you saying I should add the gravitational energy + the kinetic energy to get the total energy?

    I'll do this later.


    the units are meters cubed and I think that's what they should be.
     
  7. Jul 17, 2012 #6

    Ibix

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    From your calculation equating the electrostatic force to the gravitational force, I assumed you were trying to calculate the repulsion necessary to exactly counter the gravitational attraction of the Sun. If you did that, it would be analogous to swinging a ball round your head on a string and then cutting the string. The ball (the Earth) would go flying off, not spiral in to your head.

    If you want to know what force is necessary to cause the Earth to spiral in to the Sun, I think that you need to
    • calculate the angular momentum of the Earth as it moves around the Sun
    • calculate the centripetal force needed to maintain an orbit with that same angular momentum in an orbit just grazing the Sun
    • subtract the gravitational force at that distance
    • calculate the charge you need to provide the remaining force at that distance.

    With regard to the volume, you do not have a number in cubic meters, although you are correct that that is what you want. Your figure [itex]3.64\times10^{26}[/itex] is the length (in meters) of all of the necessary protons laid out in a straight line. Taking the cube root of that gives you a number ([itex]7.14\times10^8[/itex]) with associated unit [itex]m^{1/3}[/itex], which is not what you want. The simplest way to get the number you are looking for is to calculate the volume occupied by one proton and multiply by the number of protons. That's a slightly simplified variant on what I suggested with the face-centered cubes. Since the situation you are describing isn't really physically possible, worrying about crystallography is probably redundant.
     
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