# In general, astronomical objects are not exactly electrically neutral. Suppose the Earth and the Moon each carry a charge of –1.27·106 C.?

In general, astronomical objects are not exactly electrically neutral. Suppose the Earth and the Moon each carry a charge of –1.27·106 C (this is approximately correct; a more precise value is identified in Chapter 22). What is the ratio of the resulting electrostatic repulsion with the gravitational attraction between the Moon and the Earth? Look up any necessary data.

I've tried a couple things and looked up a couple things but I just can't seem to get it. Any help would be greatly appreciated.

### 2 Answers

- 3 months agoFavorite Answer
Coulomb's law:

F = k.q1.q2 /r²

Newton's gravitational:

Fn = G.m1.m2 /r²

ratio F/Fn = (k.q1.q2 /r²) / (G.m1.m2 /r²) {1/r² cancels out}

(k.q1.q2 ) / (G.m1.m2 )

k = 8.99×10⁹ N·m²·C⁻²

q1 = q2 = -1.27 x 10^6 C

G = 6.67 × 10^-11 N·m²·kg⁻²

m1 = 5.97 × 10^24 kg {earth}

m2 = 7.35 × 10^22 kg {moon}

F/Fn = (8.99 × 10^9 × (1.27 × 10^6)² ) / (6.67 × 10^-11 × 5.97 × 10^24 × 7.35 × 10^22)

= 4.95 × 10-16

- billrussell42Lv 73 months ago
–1.27·106 C ??? what does this mean ?

in any case

Fc/Fg = (kQ²/r²) / (Gm₁m₂/r²) = kQ² / Gm₁m₂

look up the values for mass of earth and moon, and plug in the numbers.

earth mass 5.974e24 kg

moon mass 7.35e22 kg

k = 8.99e9 Nm²/C²

G = 6.674e-11 m³/kgs²

Coulomb's law, force of attraction/repulsion

F = kQ₁Q₂/r²

Q₁ and Q₂ are the charges in coulombs

F is force in newtons

r is separation in meters

k = 8.99e9 Nm²/C²

Gravitational attraction in newtons

F = G m₁m₂/r²

G = 6.674e-11 m³/kgs²

m₁ and m₂ are the masses of the two objects in kg

r is the distance in meters between their centers

Thanks Dr.Z. I don't know if it's the site or my new browser that won't let me expand the answer window to see more than the bottom few lines any more.