Question

Suppose the Earth and the Moon carried positive charges of equal magnitude. How large would the charge need to be to produce an electrostatic repulsion equal to \(1.00 \%\) of the gravitational attraction between the two bodies?

Short answer

Answer: The charge required for the electrostatic repulsion to be equal to 1% of the gravitational attraction between the Earth and the Moon is approximately \(3.4 \times 10^4 C\).

Step-by-step solution

Write down the gravitational force formula

The gravitational force between two bodies is given by the formula: $$ F_{gravitational} = G * \frac{m_1 * m_2}{r^2} $$ where \(F_{gravitational}\) is the gravitational force, \(G\) is the gravitational constant (\(6.674 \times 10^{-11} N m^2/kg^2\)), \(m_1\) and \(m_2\) are the masses of two bodies, and \(r\) is the distance between their centers.

Write down the electrostatic force formula

The electrostatic force between two bodies is given by the formula: $$ F_{electrostatic} = k * \frac{q_1 * q_2}{r^2} $$ where \(F_{electrostatic}\) is the electrostatic force, \(k\) is the electrostatic constant (\(8.99 \times 10^9 N m^2/C^2\)), and \(q_1\) and \(q_2\) are the charges of the two bodies.

Set up the equation for the condition given

We are given that the electrostatic repulsion is 1% of the gravitational attraction. So, we can write the following equation: $$ F_{electrostatic} = 0.01 * F_{gravitational} $$

Substitute the values for Earth and Moon in the equation

For the Earth, we have \(m_1 = 5.97 \times 10^{24} kg\) and for the Moon, we have \(m_2 = 7.35 \times 10^{22} kg\). The distance between the Earth and the Moon is approximately \(r = 3.84 \times 10^8 m\). The charges on both bodies are equal, so \(q_1 = q_2 = q\). Now, we can substitute these values in our equation: $$ k * \frac{q^2}{r^2} = 0.01 * G * \frac{m_1 * m_2}{r^2} $$

Solve the equation for q

The \(r^2\) term in the equation cancels out. We can now solve for \(q\): $$ q^2 = 0.01 * \frac{G * m_1 * m_2}{k} $$ Plug in the given values: $$ q^2 = 0.01 * \frac{(6.674 \times 10^{-11}) * (5.97 \times 10^{24}) * (7.35 \times 10^{22})}{8.99 \times 10^9} $$ Calculate the value of \(q^2\): $$ q^2 \approx 1.159 \times 10^{9} C^2 $$ Now, we can find the value of \(q\) by taking the square root of \(q^2\): $$ q \approx \sqrt{1.159 \times 10^{9}} \approx 3.4 \times 10^{4} C $$ The charge required for the electrostatic repulsion to be equal to 1% of the gravitational attraction between the Earth and the Moon is approximately \(3.4 \times 10^4 C\).

Key concepts

Gravitational Force Formula

Understanding the gravitational force formula is crucial for comprehending the interactions between masses in the universe. Simply put, every object with mass attracts every other object with mass. The formula for gravitational force, according to Newton's universal law of gravitation, is:

\[\begin{equation}F_{gravitational} = G * \frac{m_1 * m_2}{r^2}\end{equation}\]
Where:

• G is the gravitational constant ( 6.674 \times 10^{-11} N m^2/kg^2 ), a measure of the strength of gravity.
• m_1 and m_2 are the masses of the two objects experiencing mutual attraction.
• r is the distance between the centers of the two masses.

This formula shows that the gravitational force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.

In application, this means that larger masses or smaller distances will result in stronger gravitational forces. The concept plays an essential role in everything from celestial mechanics to predicting orbital paths.

Electrostatic Force Formula

Electrostatic force, much like gravitational force, illustrates the interaction between two charges. The quantitative expression for the electrostatic force comes from Coulomb’s law and is given by:

\[\begin{equation}F_{electrostatic} = k * \frac{q_1 * q_2}{r^2}\end{equation}\]
Where:

• k is the electrostatic constant ( 8.99 \times 10^9 N m^2/C^2 ), symbolizing the force between charges separated by one meter.
• q_1 and q_2 represent the magnitudes of the two charges.
• r is the separation between the charges.

Electrostatic force can be both attractive and repulsive, depending on the signs of the charges: like charges repel, while opposite charges attract. The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Understanding this relationship is especially helpful in predicting the behavior of charged particles and is fundamental in fields such as electromagnetism and electronics.

Coulomb's Law

Coulomb's law provides a foundation for electrostatics, analogous to how Newton's law underpins gravity. The law defines the strength of the electrical force between two point charges. Named after Charles-Augustin de Coulomb, who first formulated the law in 1785, it can be stated succinctly as:

\[\begin{equation}F_{electrostatic} = k * \frac{q_1 * q_2}{r^2}\end{equation}\]

Key Points of Coulomb's Law:

• The force is inversely proportional to the square of the distance between two charges, implying the force diminishes greatly as distance increases.
• It is directly proportional to the product of the magnitudes of the charges, indicating larger charges exert a stronger force.
• The direction of the force is along the line joining the two charges and it can be attractive or repulsive.
• It takes into account the medium between the charges as it affects the value of the electrostatic constant ( k).

Coulomb's law is pivotal to designing electric fields and understanding how charges will interact under varying conditions. It assists in calculating forces in electrostatic experiments and forms the basis for electric charge measurements in various applications.