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Chapter 5: Problem 16

The distance between the centres of the Moon and the earth is \(D .\) The mass of the earth is 81 times the mass of the Moon. At what distance from the centre of the earth, the gravitational force will be zero? (1) \(\frac{D}{2}\) (2) \(\frac{2 D}{3}\) (3) \(\frac{4 D}{3}\) (4) \(\frac{9 D}{10}\)

Short Answer

Expert verified
The gravitational force is zero at \( \frac{9D}{10} \) from the Earth's center.

Step by step solution

01

Understanding the Gravitational Force Balance

We need to find the point at which the gravitational forces by the Earth and the Moon on an object will cancel out to be zero. Let this point be at a distance \( x \) from the Earth's center, meaning the distance from the Moon's center is \( D - x \). The gravitational force can be given as \( F = \frac{G \, m_1 \, m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are masses, and \( r \) is the distance.
02

Set Up the Gravitational Force Equation

The gravitational force exerted by the Earth \( F_E \) and the Moon \( F_M \) will balance each other. The masses of the Earth \( M_E = 81 M_M \) and Moon \( M_M \) play a role here. Set the forces equal to each other: \[ \frac{GM_EM}{x^2} = \frac{GM_MM}{(D-x)^2} \]with \( M_E = 81 M_M \).
03

Simplify the Expression

We can cancel \( G \) and the mass \( M \) from both sides, simplifying the equation to: \[ \frac{81 M_M}{x^2} = \frac{M_M}{(D-x)^2} \]which reduces further to: \[ \frac{81}{x^2} = \frac{1}{(D-x)^2} \].
04

Solve for \( x \)

By cross-multiplying, we obtain \[ 81 (D-x)^2 = x^2 \]. Expanding the equation, we have \[ (D-x)^2 = \frac{x^2}{81} \]leading to \[ D^2 - 2Dx + x^2 = \frac{x^2}{81} \].
05

Rearrange and Factor

Rearrange to form: \[ 81 (D^2 - 2Dx + x^2) = x^2 \].Multiply through to get \[ 81D^2 - 162Dx + 81x^2 = x^2 \].Simplify by combining like terms: \[ 81D^2 - 162Dx = 0 \] by setting \[ 80x^2 \] on the left-hand side.
06

Solve for \( x \)

Since \( 81D^2 = 162Dx + 80x^2 \) implies zero of the quadratic free term, we simplify to solve \[ x = \frac{9D}{10} \]. This solution indicates that the point where gravitational force is zero is at \( \frac{9D}{10} \) from the Earth's center.

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Key Concepts

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

Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another. In our exercise, we are dealing with the gravitational forces between the Earth and the Moon. These forces are described by Newton's Law of Universal Gravitation. According to this law, the gravitational force that two masses exert on each other is:
  • Directly proportional to the product of their masses, and
  • Inversely proportional to the square of the distance between their centers.
This law is represented mathematically by the formula: \[ F = \frac{G \, m_1 \, m_2}{r^2} \]Here, \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.
In order to determine the point where the gravitational forces from the Earth and the Moon cancel each other out, we need to find the balance of these forces. This is done by setting the gravitational force exerted by the Earth on an object equal to the gravitational force exerted by the Moon on the same object.
Distance Calculation
The distance calculation in this exercise revolves around finding the exact location where the gravitational pull of the Earth and the Moon on an object is perfectly balanced. As the problem states, the distance between the centers of the Earth and the Moon is given as \( D \). We need to calculate the specific point \( x \) from the Earth where the gravitational forces cancel each other out. In the exercise solution, this is approached through setting up the equations of gravitational force exerted by Earth and Moon and then solving for \( x \).
Remove \( G \) and any common factors to simplify the equation and solve:
  • Gravitational force from Earth is proportional to \( \frac{81}{x^2} \)
  • Gravitational force from Moon is proportional to \( \frac{1}{(D-x)^2} \)
By solving these equivalencies, we find that at \( x = \frac{9D}{10} \), the forces are equal, making the gravitational force on an object zero at this point.
Mass Ratio
Understanding the mass ratio is crucial to determining the point of gravitational equilibrium. In our scenario, the mass of the Earth is 81 times that of the Moon. This mass ratio impacts how gravitational forces are distributed between the two celestial bodies. The mass comparison emphasizes how much stronger Earth's gravitational influence is compared to that of the Moon, due to its larger mass.
The formula from the gravitational force equation can be simplified because of this known mass ratio. When set to balance each other, the mass ratio tells us that the stronger gravitational pull from the Earth must be at a greater distance compared to the Moon's pull. These relationships are expressed as:
  • Mass of Earth \( M_E = 81 \times \) mass of Moon \( M_M \)
  • Balance is located closer to the Moon but still significantly influenced by Earth's mass.
In finding the zero gravitational force point, the ratio helps isolate the variable \( x \) in terms of the distance \( D \), allowing us to pinpoint exactly where the effective gravitational force is neutral.

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