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Question
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon,
- Calculate the value of q required to balance the gravitational attraction between Earth and Moon
- Suppose the distance between the Moon and Earth is halved, would the charge q change?
(Take mE = 5.9 x 1024 kg, mM = 7.348 x 1022 kg)
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Solution
Mass of the Earth, ME = 5.9 x 1024 kg
Mass of the Moon, MM = 7.348 x 1022 kg
Charge placed on the surface of Earth and Moon = q
(a) Required charge to balance the FG between Earth and Moon
FC = FG (or) `1/(4pi epsilon_0) "q"^2/"r"^2 = ("GM"_"E" xx "M"_"M")/"r"^2`
q2 = G × ME × MM × 4πε0 = 320.97 × 1025
q = `sqrt(320.97 xx 10^25)`= 5.665 x 1013 = 5.67 x 1013 C
(b) The distance between Moon and Earth is
`1/(4pi epsilon_0) "q"^2/("r"/2) = ("GM"_"E" xx "M"_"M")/("r"/2)`
so q = 5.67 x 1013 C
There is no change.
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