What equal charges would have to be placed on earth and moon to neutralize their gravitational force of attraction? Given that mass of earth = $10^{25}$ kg and mass of moon = $10^{23}$ kg.

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$5.45 x 10^{16} C$

$5.45 x 10^{13} C$

$8.57 x 10^{13} C$

$8.57 x 10^{16} C$

answer is B.

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Detailed Solution

If q is charge on the earth and the moon, then

$\begin{array}{l} \frac{1}{4 {πε}_{o}} . \frac{q \times q}{r^{2}} = G \frac{M_{earth} \times M_{moon}}{r^{2}} or \\ q = \sqrt{4 {πε}_{o} M_{earth} M_{moon}} \\ setting G = 6 .6 \times 10^{- 11} {Nm}^{- 2}; M_{earth} = 10^{25} kg \\ and M_{moon} = 10^{23} kg, we get \\ q = 8 .57 \times 10^{13} C \end{array}$