91Ó°ÊÓ

We will use Newton's law of universal gravitation to find the force between the Earth and the Moon. The formula for gravitational force is: \(F_g = G\frac{m_1m_2}{r^2}\) Where: - \(F_g\) is the gravitational force - \(G\) is the gravitational constant, \(6.674 \times 10^{-11} N\cdot m^2 / kg^2\) - \(m_1\) is the mass of the Earth, \(5.972 \times 10^{24} kg\) - \(m_2\) is the mass of the Moon, \(7.342 \times 10^{22} kg\) - \(r\) is the distance between the Earth and the Moon, \(3.844 \times 10^8 m\) Now, we will substitute these values into the equation: \(F_g = 6.674 \times 10^{-11} \frac{(5.972 \times 10^{24})(7.342 \times 10^{22})}{(3.844 \times 10^8)^2}\) Calculate \(F_g\) to get the gravitational force between the Earth and the Moon.

The question asks for the charge that will produce an electrostatic force equal to 1% of the gravitational force. First, we need to find 1% of the gravitational force: \(F_e = 0.01 \times F_g\) Calculate \(F_e\) using the value of \(F_g\) obtained in Step 1.

We will use Coulomb's law to find the charge required to produce the desired electrostatic force. The formula for electrostatic force is: \(F_e = k\frac{q_1q_2}{r^2}\) Since the question states that the charges on the Earth and the Moon are equal, we can simplify the equation to: \(F_e = k\frac{q^2}{r^2}\) Where: - \(F_e\) is the electrostatic force - \(k\) is the electrostatic constant, \(8.987 \times 10^9 N\cdot m^2 / C^2\) - \(q\) is the charge on the Earth and the Moon - \(r\) is the distance between the Earth and the Moon, \(3.844 \times 10^8 m\) Now, rearrange the equation to solve for \(q\): \(q^2 = \frac{F_e \cdot r^2}{k}\) Substitute the values of \(F_e\), \(r\), and \(k\) into the equation and solve for \(q^2\). Finally, take the square root of both sides to find \(q\). This value is the charge required to produce an electrostatic force equal to 1% of the gravitational force between the Earth and the Moon.