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Chapter 18: Problem 64

The masses of the earth and moon are \(5.98 \times 10^{24}\) and \(7.35 \times 10^{22} \mathrm{kg}\) , respectively. Identical amounts of charge are placed on each bouly, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?

Short Answer

Expert verified
The charge is approximately \(1.1 \times 10^{13} \text{ C}\).

Step by step solution

01

Understanding the Problem

We need to find the amount of charge placed on the Earth and Moon such that the net force (gravitational and electrical) between them is zero. This means that the gravitational attraction is equal in magnitude to the electrostatic repulsion.
02

Express Gravitational Force

The gravitational force between two masses, Earth (\(m_1 = 5.98 \times 10^{24} \text{ kg}\)) and Moon (\(m_2 = 7.35 \times 10^{22} \text{ kg}\)), separated by a distance \(r\), is given by Newton's law of universal gravitation: \[F_g = G \frac{m_1 m_2}{r^2}\] where \(G = 6.674 \times 10^{-11} \text{ Nm}^2\text{/kg}^2\) is the gravitational constant.
03

Express Electrical Force

The electrical force between two equal charges \(q\) on the Earth and Moon, separated by a distance \(r\), is given by Coulomb's law: \[F_e = k \frac{q^2}{r^2}\] where \(k = 8.988 \times 10^9 \text{ Nm}^2\text{/C}^2\) is Coulomb's constant.
04

Equate Forces for Zero Net Force

To achieve zero net force, we set the magnitude of gravitational force equal to electrical force: \[G \frac{m_1 m_2}{r^2} = k \frac{q^2}{r^2}\]. Simplifying, this equation shows that \[ G m_1 m_2 = k q^2 \].
05

Solve for Charge

Rearrange the equation to solve for charge \(q\):\[ q^2 = \frac{G m_1 m_2}{k} \].Calculation:\( q = \sqrt{\frac{(6.674 \times 10^{-11}) (5.98 \times 10^{24}) (7.35 \times 10^{22})}{8.988 \times 10^9}} \) C.\( q \approx 1.1 \times 10^{13} \text{ C} \).
06

Conclusion

The magnitude of the charge placed on each body, Earth and Moon, to make the net force zero, is approximately \(1.1 \times 10^{13} \text{ C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrical Force
The electrical force is a type of force that arises when there are charges present. It acts between charged objects due to their electric charge. When two objects have charges, they can either attract or repel each other. This behavior is governed by Coulomb's Law, which defines how the electrical force works at any given distance.

The electrical force is crucial in the problem of making the net force zero between the Earth and the Moon. If they both have the same charge, the electrical force between them will act as a repulsion. This electrically induced repulsion can be adjusted so that it completely cancels out the gravitational attraction between the two celestial bodies.
  • Similar charges: Repel each other
  • Opposite charges: Attract each other
  • Force decreases with increased distance
Newton's law of universal gravitation
Newton's law of universal gravitation is a fundamental principle that explains how every mass exerts an attractive force on another mass. This force of attraction is known as gravitational force, and it is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

This law is represented by the equation:\[F_g = G \frac{m_1 m_2}{r^2}\]where:
  • \(F_g\) = gravitational force
  • \(G\) = gravitational constant \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\)
  • \(m_1\) and \(m_2\) are the masses of the two objects
  • \(r\) = the distance between the centers of the two masses
In the exercise, this gravitational pull is one of the forces we aim to balance by introducing an equal and opposite electrical force. This brings us to the condition where the net force is zero.
Coulomb's law
Coulomb's law describes the force between two electrically charged objects. According to this law, the electrical force between two charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's law is:\[F_e = k \frac{q^2}{r^2}\]where:
  • \(F_e\) = electrical force
  • \(k\) = Coulomb's constant \(8.988 \times 10^9 \text{ Nm}^2/\text{C}^2\)
  • \(q\) = magnitude of the charge on each object (assuming identical charges)
  • \(r\) = distance between the charges
In our problem, we want this electrical force to precisely balance out the gravitational force, allowing us to equate the two to find the necessary charge.
Net Force Balance
Achieving a net force balance involves ensuring that the sum of all forces acting on an object is zero. In this scenario, the challenge is to balance the gravitational attraction between the Earth and the Moon with an equal but opposite electrical force.
To achieve zero net force, we set the gravitational force equal to the electrical force:\[G \frac{m_1 m_2}{r^2} = k \frac{q^2}{r^2}\]This equation simplifies to:\[G m_1 m_2 = k q^2\]In simple terms, achieving a net force balance is like balancing a see-saw, where one side is the gravitational force pulling them together and the other side is the electrical force pushing them apart. Solving for the charge \(q\), we are able to find how much charge is needed on each body to ensure they perfectly counteract each other's gravitational pull, resulting in a net force balance.

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Most popular questions from this chapter

Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force.

Two spherical objects are separated by a distance that is \(1.80 \times 10^{-3} \mathrm{m}\) . The objects are initially electrically neutral and are very small comparcd to the distance bctwcen them. Each objcct acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of \(4.55 \times 10^{-21} \mathrm{N}\) . How many electrons did it take to produce the charge on one of the objects?

The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface 1 has an area of \(1.7 \mathrm{m}^{2},\) while surface 2 has an area of 3.2 \(\mathrm{m}^{2}\) The clectric ficld \(\overrightarrow{\mathbf{E}}\) in the drawing is uniform and has a drawing is uniform and has a magnitude of 250 \(\mathrm{NC}\) . Find the magnitude of the electric flux through (a) surface 1 and \((\mathrm{b})\) surface 2 .

There are four charges, each with a magnitude of 2.0 C. Two are positive and two are negative. The charges are fixed to the corners of a 0.30-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is 3.50 109 kg. (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?
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