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

Question: Compare the electrostatic repulsion force and gravitational attraction force between the Earth and the Moon and discuss their effects on the size, shape, and stability of the Moon's orbit around the Earth. Answer: The electrostatic repulsion force between the Earth and the Moon is calculated to be approximately \(6.13 \times 10^{20} N\), whereas the gravitational attraction force is approximately \(1.982 \times 10^{20} N\). While the electrostatic force is larger, the difference in magnitudes is not significant enough to substantially affect the size, shape, and stability of the Moon's orbit around the Earth. This is because the electrostatic force partially cancels out the effect of the gravitational force, resulting in a stable orbit.

Step-by-step solution

Calculate the electrostatic repulsion force

To calculate the electrostatic repulsion force between Earth and Moon, we can use Coulomb's law: \(F_{e} = \frac{k \cdot q1 \cdot q2}{r^2}\), where \(F_{e}\) is the electrostatic force, \(k \approx 8.99 \times 10^9 \frac{N \cdot m^2}{C^2}\) is the electrostatic constant, \(q1\) and \(q2\) are the charges of the Earth and the Moon, and \(r\) is the distance between their centers. Given that both Earth and Moon carry a charge of \(-1.00 \cdot 10^6 \mathrm{C}\), we have \(q1 = q2 = -1.00 \cdot 10^6\mathrm{C}\). The average distance between the Earth and the Moon is about \(r= 3.84 \times 10^8 m\). Plugging in the given values, we can calculate the electrostatic force as: \(F_{e} = \frac{(8.99 \times 10^9)(-1.00 \cdot 10^6)^2}{(3.84 \times 10^8)^2}\) \(\approx 6.13 \times 10^{20} N\).

Calculate the gravitational attraction force

To calculate the gravitational attraction force between the Earth and Moon, we can use Newton's law of universal gravitation: \(F_{g} = G \frac{m1 \cdot m2}{r^2}\), where \(F_{g}\) is the gravitational force, \(G = 6.674 \times 10^{-11} \frac{m^3}{kg \cdot s^2}\) is the gravitational constant, \(m1\) is the mass of the Earth, \(m2\) is the mass of the Moon, and \(r\) is the distance between their centers. The mass of the Earth is approximately \(m1 = 5.972 \times 10^{24} kg \) and the mass of the Moon is approximately \(m2 = 7.342 \times 10^{22} kg\). Using the average distance between the Earth and the Moon, \(r= 3.84 \times 10^8 m\), We can calculate the gravitational force as: \(F_{g} = (6.674 \times 10^{-11}) \frac{(5.972 \times 10^{24}) (7.342 \times 10^{22})}{(3.84 \times 10^8)^2}\) \(\approx 1.982 \times 10^{20} N\).

Compare the resulting forces

Now that we have calculated both the electrostatic repulsion force (\(F_{e} \approx 6.13 \times 10^{20} N\)) and the gravitational attraction force (\(F_{g} \approx 1.982 \times 10^{20} N\)), we can compare them. The electrostatic force is larger than the gravitational force (\(F_{e} > F_{g}\)). However, this difference in magnitudes is not significant enough to substantially affect the size, shape, and stability of the Moon's orbit around the Earth.

Effects on the Moon's orbit

The effects of the electrostatic force on the size, shape, and stability of the Moon's orbit around the Earth are minimal due to the following reasons: 1. The difference in magnitudes of the electrostatic and gravitational forces is not significant enough to substantially alter the gravitational attraction between the Earth and the Moon. 2. The electrostatic force acts in opposition to the gravitational force, partially canceling out its effects. As a result, the Moon's orbit around the Earth remains stable, and the size and shape of the orbit are not significantly affected. In conclusion, while there is a difference in the magnitudes of the two forces, it is not significant enough to have any substantial impact on the size, shape, and stability of the Moon's orbit around the Earth.