The masses of the earth and moon are 5.98 × 1024 and 7.35 × 1022 kg, respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?
To find the magnitude of the charge placed on each body, we need to set the gravitational force equal to the electrical force and solve for the charge.
The gravitational force between the Earth and Moon is given by Newton's law of universal gravitation:
F_grav = G * (m1 * m2) / r²
Where:
F_grav is the gravitational force
G is the gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
m1 and m2 are the masses of the Earth and Moon, respectively
r is the distance between the centers of the Earth and Moon (approximately 3.84 × 10^8 meters)
In this problem, we are given the masses of the Earth (m1 = 5.98 × 10^24 kg) and the Moon (m2 = 7.35 × 10^22 kg).
The electrical force between two charged objects is given by Coulomb's law:
F_elec = k * (q1 * q2) / r²
Where:
F_elec is the electrical force
k is the electrostatic constant (approximately 9 × 10^9 N m²/C²)
q1 and q2 are the magnitudes of the charges on the Earth and Moon, respectively
r is the distance between the centers of the Earth and Moon
In this problem, we are looking for the magnitude of the charges on the Earth and Moon. Let's assume the charge on each body is q.
Given that the net force (gravitational plus electrical) on each body is zero, we can set the gravitational force equal to the electrical force and solve for q.
G * (m1 * m2) / r² = k * (q * q) / r²
Canceling out the common terms:
G * m1 * m2 = k * q²
Now we can solve for q:
q² = (G * m1 * m2) / k
q = sqrt((G * m1 * m2) / k)
Plugging in the values:
q = sqrt((6.674 × 10^-11 N m²/kg² * 5.98 × 10^24 kg * 7.35 × 10^22 kg) / (9 × 10^9 N m²/C²))
Calculating this expression will give us the magnitude of the charge placed on each body.