1. a) What equal amount of positive charge would have to be placed on the Earth and on the Moon to neutralize their gravitational attraction

Question

1. a) What equal amount of positive charge would have to be placed on the Earth and on the Moon to neutralize their gravitational attraction? b) Why don’t you need to know the lunar distance to solve this problem? c) How many kilograms of hydrogen ions (that is, protons) would be needed to provide the positive charge calculated in part a?

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5 months 2021-08-17T02:42:58+00:00 1 Answers 4 views 0

A) q = 5.714 × 10^(13) C

B) Because from a above, F_e = F_g and so r² canceled out in the formulas

C) m_t = 5.964 × 10⁴ kg

Explanation:

For the gravitational attraction of the earth and moon to be neutralized, the electrostatic force must be equal to the gravitational force i.e F_e = F_g

Now, F_e = kq²/r² and F_g = GmM/r²

Equating them and making q the subject, we arrive at;

q = √(GmM/k)

Where;

G is gravitational constant = 6.67 × 10^(-11) m³/kg.s²

m is mass of moon = 7.36 × 10^(22) kg

M is mass of earth = 5.98 × 10^(24) kg

k is coulombs constant = 8.99 x 10^(9) N.m²/c²

q = √(6.67 × 10^(-11) × 7.36 × 10^(22) × 5.98 × 10^(24)/8.99 x 10^(9))

q = 5.714 × 10^(13) C

B) We don’t need the lunar distance because from a above, F_e = F_g and so r² canceled out in the formulas.

C) The number of protons is given by the formula;

n = q/e

Where, e is charge of the proton = 1.6 × 10^(-19) C

n = (5.714 × 10^(13))/(1.6 × 10^(-19))

n = 3.57125 × 10^(32)

Total mass of these protons is given by the formula;

m_t = nm_p

Where m_p is mass of a single proton = 1.67 × 10^(-27) kg

m_t = 3.57125 × 10^(32) × 1.67 × 10^(-27)

m_t = 5.964 × 10⁴ kg