Question

What equal amounts of positive charge would have to be placed on the Earth and on the Moon to neutralize their gravitational attraction? Do you need to know the Moon’s distance to solve this problem? Why or why not? (b) How many metric tons of hydrogen would be needed to provide the positive charge calculated in part (a)? The molar mass of hydrogen is 1.008 g/mol.

Answer 1

Answer:

Fg = G M m / R^2       gravitational attraction

Fe = k Q^2 / R^2       electric repulsion

G M m = k Q^2

Q = (G M m / k)^1/2

Since m = .0123 M

Q = (.0123 G / k)^1/2 * M

Q = (.0123 * 6.67 * 10E-11 / 9 * 10E9)^1/2 * 5.98 * 10E24

Q = 5.71 * 10E13 C       charge required on each body

n = 5.71 * 10E13 / 1.6 * 10E-19 = 3.57 * 10E32 atoms

N = 3.57 * 10E32 / 6.02 * 10E23  = 5.93 * 10E8 g-mol = 5.93 * 10E*5 kg-mol

Since 1 metric ton = 1000 kg

One would need 593 metric tons of hydrogen