Suppose a charge +q on Earth's surface and another +q charge is placed on the surface of the moon. (a) Calcualte the value of a q required to balance the r=gravitional attraction between Earth and Moon (b) Suppose the distance between the moon and Earth is halved, would the charge q change?

Explanation

(Take `m_E = 5.9 xx 10^24kg, m_M = 7.9 xx 10^22 kg`)
Gravitional force
`F_g = (Gm_E xx m_M)/r^2`
G ` = 6.626 xx 10^-11 Nm^2kg^-2`
`m_E = 5.9 xx 10^24kg`
`m_M = 7.9 xx 10^22 kg`
`:.F_g = (6.626 xx 10^-11 xx 5.9 xx 10^24 xx 7.9 xx 10^22)/r^2`
`:.F_g = (308.8378 xx 10^35)/r^2` ...(1)
Electrostatic force
`F_e = 1/(4piepsilon_0)(q_1q_2)/r^2`
`1/(4piepsilon_0) = 9 xx 10^9` `q_1 = q_2 = q`
`:.F_e = (9 xx 10^9 xx q^2)/r^2` ...(2)
But r is same
`:.` From (1) and (2) we get
`9 xx 10^9 q^2 = 308.8378 xx 10^5`
`q = sqrt((308.8378 xx 10^35)/(9 xx 10^9))`
`q= sqrt(34.3153 xx 10^26)`
`q ~= 5.64 xx 10^13 C`
`q = 5.64 xx 10^13 C`
`:.q~= +5.64 xx 10^13C`
(b) Even if the distance between the Moon and Earth is halved then the distance of separation r remians same.
Hence there is no change in the value of charge .q..