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.

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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft