Consider the Earth-Moon system. If we were to place equal charges on the Earth and the Moon, how large would they need to be for the electrostatic repulsion to counteract the gravitational attraction?
  1. 5.1×1013 C5.1\times 10^{13}\textrm{ C}
  2. 5.7×1013 C5.7\times 10^{13}\textrm{ C}
  3. 6.7×1013 C6.7 \times 10^{13}\textrm{ C}
  4. 3.3×1027 C3.3\times 10^{27} \textrm{ C}
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OpenStax College Physics for AP® Courses, Chapter 33, Problem 5 (Test Prep for AP® Courses)

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This is College Physics Answers with Shaun Dychko. We want to know how much charge we should put on the Earth and the moon such that the electrostatic repulsion between them will equal the gravitational attraction between them? So this formula here represents the gravitational attraction; that's the gravitational constant times the mass of the moon times the mass of the Earth divided by the distance between them squared. And this is the coulomb force of repulsion there's the coulomb's constant times the charge on the moon and the charge on the Earth divided by the separation between them. I did not put a subscript on the r because it's the same distance in each case— the Earth-moon distance— and I didn't need to put a subscript on the charges either because we are told that we are putting an equal amount of charge on both the Earth and moon so we'll just call it q. So this can be rewritten as k q squared replacing q 1 and q 2 each with letter q and the r's cancel because you can multiply both sides by r squared and it cancels on both sides and we have k q squared equals gravitational constant times the product of the masses of the moon and Earth. So to solve for the charge q, we divide both sides by coulomb's constant k and then take the square root of both sides and we have that the charge is square root of the gravitational constant divided by the coulomb's constant times the mass of the moon times the mass of the Earth. So we lookup all these numbers in data tables and then we end up with 5.7 times 10 to the 13 coulombs and so the answer is (b).