Problem 41 - Fundamentals of Physics

Question

(a) What equal positive charges would have to be placed En Earth and on the Moon to neutralize their gravitational attraction? (b) Why don't you need to know the lunar distance to solve this problem? (c) How many kilograms of hydrogen ions (that is, protons) would be needed to provide the positive charge calculated in (a)?

Step-by-Step Solution

Answer

(a) The charge \(q = \sqrt{\frac{G M_e M_m}{k}}\). (b) Because \(r\) cancels out. (c) Calculate protons' mass using charge/proton charge.

Step 1: Understanding the Problem

We need to find the positive charge that, when placed on the Earth and the Moon, will counterbalance the gravitational force between them, making the net force zero.

Step 2: Constructing the Equation for Gravitational Force

The gravitational force between the Earth and the Moon is given by:\[F_g = \frac{G M_e M_m}{r^2}\]where \(G\) is the gravitational constant, \(M_e\) is the mass of the Earth, \(M_m\) is the mass of the Moon, and \(r\) is the distance between the Earth and the Moon.

Step 3: Constructing the Equation for Electrostatic Force

The electrostatic force between two charges \(q\) on Earth and the Moon is:\[F_e = \frac{k q^2}{r^2}\]where \(k\) is Coulomb's constant and \(q\) is the charge on each body.

Step 4: Setting Forces Equal to Neutralize Gravitational Attraction

To neutralize the gravitational attraction, set \(F_g = F_e\):\[\frac{G M_e M_m}{r^2} = \frac{k q^2}{r^2}\].The \(r^2\) cancels out, leaving:\[G M_e M_m = k q^2\].

Step 5: Solving for the Charge

Rearrange the equation to solve for \(q\):\[q = \sqrt{\frac{G M_e M_m}{k}}\].Plug in the known values to find \(q\).

Step 6: Explaining Why Distance Isn't Needed

Notice that the distance \(r\) cancels out during the computation. Therefore, the charge is independent of \(r\).

Step 7: Calculating the Number of Protons

The charge of one proton is approximately \(1.602 \times 10^{-19}\) C. Divide the total calculated charge \(q\) by this value to find the number of protons required. Then, multiply the number of protons by the mass of one proton (approximately \(1.67 \times 10^{-27}\) kg) to find the total mass in kilograms.

Key Concepts

Gravitational force

Gravitational force is the force of attraction between two masses. This fundamental force is significant when considering celestial bodies like Earth and the Moon. The equation to calculate this force is:\[ F_g = \frac{G M_e M_m}{r^2} \]where:

• \( F_g \) is the gravitational force,
• \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \),
• \( M_e \) and \( M_m \) are the masses of the Earth and Moon respectively,
• \( r \) is the distance from the center of the Earth to the center of the Moon.

This force plays a crucial role in the motion and interaction of planetary bodies, keeping the Moon in orbit around Earth. Even though it's weaker compared to electromagnetic forces, its effects are widely observable due to the massive sizes of celestial bodies.

Coulomb's law

Coulomb's law helps us understand the electrostatic force between two electric charges. This force is expressed mathematically as:\[ F_e = \frac{k q_1 q_2}{r^2} \]where:

• \( F_e \) is the electrostatic force,
• \( k \) is Coulomb's constant \( (8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2) \),
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• \( r \) is the distance between the charges.

In the context of the Earth and Moon, equal charges \( q \) are placed on both, leading to a repulsive force. To neutralize the gravitational pull, this electrostatic force has to be equal and opposite to the gravitational force. Hence, by setting these two forces equal, the problem can be solved by finding the charge \( q \) that satisfies this condition.

Proton charge

A proton, a subatomic particle found in the nucleus of an atom, carries a positive electric charge. The charge of a single proton is approximately \( 1.602 \times 10^{-19} \) Coulombs. Understanding this is crucial when calculating the total charge needed to counterbalance gravitational attraction between the Earth and the Moon.To determine how many protons are necessary to provide a specific charge \( q \), we divide \( q \) by the charge of one proton. Once this number is determined, the next step is to calculate the mass of these protons. Each proton has a mass of approximately \( 1.67 \times 10^{-27} \) kg. Multiply the number of protons by this mass to find the total mass in kilograms required, effectively translating electrical concepts into physical properties.

Neutralization of forces

Neutralization of forces refers to the balancing act of equating opposing forces so that their net effect is zero. In this exercise, the concept involves setting the electrostatic force equal to the gravitational force acting between the Earth and the Moon, effectively canceling each other out.By applying the formula derived from setting \( F_g = F_e \):\[ \frac{G M_e M_m}{r^2} = \frac{k q^2}{r^2} \]we simplify it to:\[ G M_e M_m = k q^2 \]The distance \( r \) drops out of the equation, illustrating that the effective charge \( q \) required for neutralization is independent of the separation distance between the Earth and the Moon. Neutralization in this sense allows us to shift focus from spatial concerns to the intrinsic properties of mass and charge involved.