Question

In general, astronomical objects are not exactly electrically neutral. Suppose the Earth and the Moon each carry a charge of \(-1.00 \cdot 10^{6} \mathrm{C}\) (this is approximately correct; a more precise value is identified in Chapter 22 ). a) Compare the resulting electrostatic repulsion with the gravitational attraction between the Moon and the Earth. Look up any necessary data. b) What effects does this electrostatic force have on the size, shape, and stability of the Moon's orbit around the Earth?

Short answer

Short Answer: The electrostatic force between the Earth and the Moon is approximately \(10^{-13}\) times weaker than the gravitational force and is repulsive. Although the electrostatic force slightly increases the size and eccentricity of the Moon's orbit, its effects are negligible due to the overwhelming dominance of the gravitational force.

Step-by-step solution

Calculate the gravitational force between the Earth and the Moon

Let's start by calculating the gravitational force between the Earth and the Moon. The formula to calculate the gravitational force is given by: \(F_g = G \frac{M_Em_M}{r^2}\) Where \(F_g\) is the gravitational force, \(G\) is the gravitational constant (\(6.674 \cdot 10^{-11} \,\mathrm{N m^2/kg^2}\)), \(M_E\) is the mass of the Earth (\(5.972 \cdot 10^{24} \,\mathrm{kg}\)), \(m_M\) is the mass of the Moon (\(7.342 \cdot 10^{22} \,\mathrm{kg}\)), and \(r\) is the distance between the Earth and the Moon (\(3.844 \cdot 10^8 \,\mathrm{m}\)). Plugging these values, we get: \(F_g = \frac{6.674 \cdot 10^{-11} \,\mathrm{N m^2/kg^2} \cdot 5.972 \cdot 10^{24} \,\mathrm{kg} \cdot 7.342 \cdot 10^{22} \,\mathrm{kg}}{(3.844 \cdot 10^8 \,\mathrm{m})^2}\) \(F_g \approx 1.982 \cdot 10^{20} \,\mathrm{N}\)

Calculate the electrostatic force between the Earth and the Moon

Now, let's calculate the electrostatic force between the Earth and the Moon. The formula to calculate the electrostatic force is given by Coulomb's law: \(F_e = k \frac{q_Eq_M}{r^2}\) Where \(F_e\) is the electrostatic force, \(k\) is the electrostatic constant (\(8.9875 \cdot 10^{9} \,\mathrm{N m^2/C^2}\)), \(q_E\) is the charge of the Earth (\(-1.00 \cdot 10^{6} \,\mathrm{C}\)), \(q_M\) is the charge of the Moon (\(-1.00 \cdot 10^{6} \,\mathrm{C}\)), and \(r\) is the distance between the Earth and the Moon (\(3.844 \cdot 10^8 \,\mathrm{m}\)). Plugging these values, we get: \(F_e = \frac{8.9875 \cdot 10^{9} \,\mathrm{N m^2/C^2} \cdot (-1.00 \cdot 10^{6} \,\mathrm{C})^2}{(3.844 \cdot 10^8 \,\mathrm{m})^2}\) \(F_e \approx -2.175 \cdot 10^7 \,\mathrm{N}\) Note that the electrostatic force is actually repulsive since both the Earth and the Moon carry negative charges.

Compare the forces and discuss the effects on the Moon's orbit

To compare the gravitational and electrostatic forces, we can calculate the ratio between these forces: \(\frac{F_e}{F_g} = \frac{-2.175 \cdot 10^7 \,\mathrm{N}}{1.982 \cdot 10^{20} \,\mathrm{N}} \approx - 1.097 \cdot 10^{-13}\) The electrostatic force is approximately \(10^{-13}\) times weaker than the gravitational force. The negative sign indicates that the electrostatic force is repulsive. Since the electrostatic force is much weaker than the gravitational force, the effects on the Moon's orbit will be small. The electrostatic repulsion would slightly increase the size and eccentricity of the Moon's orbit, but would have negligible effects on its stability due to the overwhelming dominance of the gravitational force.