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29 May, 06:45

Suppose we placed a positive charge Q on the Moon and an equal positive charge Q on the Earth. What value of Q would be needed to neutralize the gravitational attraction of the Moon and the Earth?

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  1. 29 May, 09:23
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    Answer: q=5.70 x 10^13 C

    Explanation:

    gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 * 10^24 kg m is moon's mass = 7.342*10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 / C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt (6.67 x 10^-11 x 5.972 x 10^24 * 7.342*10^22 / 9 x10^9) = 5.70 x 10^13 C
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