Question

Suppose the Sun and the Earth were each given an equal amount of charge of the same sign, just sufficient to cancel their gravitational attraction. How many times the charge on an electron would that charge be? Is this number a large fraction of the number of charges of either sign in the Earth?

Short answer

Answer: The amount of charge required on both the Sun and the Earth to cancel their gravitational attraction is extremely large, but not a significant fraction of the total charges present in the Earth.

Step-by-step solution

Finding the gravitational force between the Sun and the Earth

The gravitational force between two objects can be calculated using the formula: \[F_g = G \frac{m_1 m_2}{r^2}\] where \(F_g\) is the gravitational force, \(G\) is the gravitational constant (approximately \(6.674 \times 10^{-11} N m^2/kg^2)\), \(m_1\) and \(m_2\) are the masses of the objects, and \(r\) is the distance between their centers. For the Sun-Earth system, we have the following values: - Mass of the Earth, \(m_1 = 5.972 \times 10^{24} kg\) - Mass of the Sun, \(m_2 = 1.989 \times 10^{30} kg\) - Distance between Earth and Sun, \(r = 1.496 \times 10^{11} m\) Now, let's calculate the gravitational force: \[F_g = \left(6.674 \times 10^{-11}\right) \frac{\left(5.972 \times 10^{24}\right) \left(1.989 \times 10^{30}\right)}{\left(1.496 \times 10^{11}\right)^2}\]

Calculating the electrostatic force

In order for the electrostatic force to cancel the gravitational force, both forces should be equal in magnitude: \[F_e = F_g\] The electrostatic force between two objects can be calculated using Coulomb's law: \[F_e = k \frac{q_1 q_2}{r^2}\] where \(F_e\) is the electrostatic force, \(k\) is the electrostatic constant (approximately \(8.987 \times 10^9 N m^2/C^2)\), \(q_1\) and \(q_2\) are the charges on the objects, and \(r\) is the distance between them. Since the same amount of charge is given to both the Earth and the Sun, we have \(q_1=q_2=q\). We can now set the gravitational force equal to the electrostatic force, and solve for the charge: \[k \frac{q^2}{r^2} = G \frac{m_1 m_2}{r^2}\]

Solving for the charge

We can simplify the equation and solve for the charge, \(q\): \[q^2 = \frac{G m_1 m_2}{k}\] \[q = \sqrt{\frac{G m_1 m_2}{k}}\] \[q = \sqrt{\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1.989 \times 10^{30})}{8.987 \times 10^9}}\]

Comparing the charge to the charge on an electron

The charge on an electron is approximately \(-1.602 \times 10^{-19} C\). We will now find out how many times the charge on an electron would that charge be: \[n = \frac{q}{e}\] where \(n\) is the number of times the charge of an electron and \(e = 1.602 \times 10^{-19} C\). Substitute the value of \(q\) from the previous step: \[n = \frac{\sqrt{\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1.989 \times 10^{30})}{8.987 \times 10^9}}}{1.602 \times 10^{-19}}\]

Analyzing the result

After calculating the number of times the charge of an electron, we observe that the number is extremely large, as both bodies would need a huge number of charges to cancel their gravitational attraction. This number is not a large fraction of the total number of charges in the Earth, as the number of charges inside the Earth is enormous and even larger than the amount of charge needed. In conclusion, the amount of charge required to cancel the gravitational attraction between the Earth and the Sun is an extremely large value but still not a significant fraction of the total charges present in the Earth.