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anonymous

  • 5 years ago

Any idea where to start for this problem, I really have no idea : Supose that earth and the moon have positive charges of the same magnitude, how big does the charge need to be to generate a electrostatic repulsion equal to 1.00% of the gravitational attraction between the bodies? thank you so much!

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  1. TuringTest
    • 5 years ago
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    (I'm assuming you mean 100%, though you wrote 1.00%=1%) We have two forces here that are going to be equal in magnitude: gravity and charge. The formula for gravity between two masses m1 and m2 is\[F_g=G\frac{m_1m_2}{r^2}\]The formula for electrostatic force between two charges q1 and q2 is\[F_e=k\frac{q_1q_2}{r^2}\]we are told here that the charge on the two bodies in question (the earth and moon) are equal, so q1=q2 in this case. Setting the forces equal we get\[F_g=F_e\]\[G\frac{m_{earth}m_{moon}}{r^2}=k\frac{q^2}{r^2}\]which we can solve for q easily. If your question was in fact supposed to be that the charges create 1% of the gravitational force, we would simply alter our starting equation to\[\frac{F_g}{100}=F_e\]and do the same as above.

  2. TuringTest
    • 5 years ago
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    ...the value of the constants k and G are well-known and can be found online, by the way.

  3. anonymous
    • 5 years ago
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    It is 1%, I dont know why the teacher used this notation 1.00% , so then it would just be the result of \[(k(q^2/r^2)) 100 \right?\]

  4. TuringTest
    • 5 years ago
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    can't read what you wrote in latex very well, but F_e=1% of F_g leads to\[100k\frac{q^2}{r^2}=G\frac{m_1m_2}{r^2}\]in our case. I think that is what you wrote above...

  5. anonymous
    • 5 years ago
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    hah yep, thank you so much! for r should I use the distance between them?

  6. TuringTest
    • 5 years ago
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    Normally yes, but look at the two formulas when set equal and notice that the r^2's will cancel. So it really doesn't matter here.

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