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Chapter 14: Problem 5

Split into Two \(A\) mass \(M\) is split into two parts, \(m\) and \(M-m\), which are then separated by a certain distance. What ratio \(m / M\) maximizes the magnitude of the gravitational force between the parts?

Short Answer

Expert verified
The ratio \( \frac{m}{M} \) that maximizes the gravitational force is \( \frac{1}{2} \).

Step by step solution

01

Write the Gravitational Force Equation

The gravitational force between two masses is given by Newton's Law of Gravitation: \[ F = G \frac{m_1 m_2}{r^2} \] where \( F \) is the gravitational force, \( m_1 \) and \( m_2 \) are the two masses, \( r \) is the distance between the masses, and \( G \) is the gravitational constant.
02

Substitute the Masses

Let \( m_1 = m \) and \( m_2 = M - m \). Substitute these into the equation: \[ F = G \frac{m (M - m)}{r^2} \]
03

Simplify the Force Equation

The equation becomes: \[ F = G \frac{m (M - m)}{r^2} \] As \( r \) and \( G \) are constants, focus on maximizing the term \( m (M - m) \).
04

Set Up the Maximization Problem

Define a function \( f(m) = m (M - m) \) to represent the product of the two masses.
05

Find the Derivative

Differentiate \( f(m) \) with respect to \( m \): \[ f'(m) = M - 2m \]
06

Solve for Critical Points

Set the derivative equal to zero and solve for \( m \): \[ M - 2m = 0 \] \[ m = \frac{M}{2} \]
07

Determine the Ratio

To find the ratio that maximizes the force, use \( m = \frac{M}{2} \). The ratio is: \[ \frac{m}{M} = \frac{\frac{M}{2}}{M} = \frac{1}{2} \]

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Key Concepts

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

Newton's Law of Gravitation
Newton's Law of Gravitation forms the foundation of the given problem. This law states that any two masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The equation representing this law is \[ F = G \frac{m_1 m_2}{r^2} \] where:
  • \( F \) is the gravitational force.
  • \( m_1 \) and \( m_2 \) are the two masses.
  • \( r \) is the distance between the masses.
  • \( G \) is the gravitational constant.
This equation is essential for understanding how forces between masses work and plays a crucial role in optimizing gravitational force in the given exercise.
Optimization in Physics
Optimization is a key concept in physics when we want to make something as effective as possible. In our exercise, we aim to maximize the gravitational force between two parts of a mass. Optimization involves setting up a mathematical problem that represents the physical situation, and then finding the maximum or minimum values. Here, we create a function \[ f(m) = m (M - m) \] to represent the product of two masses, which is part of the gravitational force equation. Our goal is to find the values of \( m \) that make this function as large as possible.
Critical Point Analysis
To find the maximum value of a function, we perform critical point analysis. Critical points are where the derivative of the function is zero or undefined. In our case, the function is \[ f(m) = m (M - m) \] and its derivative with respect to \( m \) is \[ f'(m) = M - 2m \]. By setting this derivative equal to zero, we identify critical points. Solving \( M - 2m = 0 \) gives us \( m = \frac{M}{2} \). This is the critical point where we evaluate the function to maximize the gravitational force.
Derivative Calculation
Derivatives help us understand how a function changes as its input changes. To maximize the gravitational force, we need to calculate the derivative of our function \[ f(m) = m (M - m) \]. Using basic differentiation rules, we find the derivative: \[ f'(m) = M - 2m \]. This derivative tells us the rate of change of the function. Setting \( f'(m) = 0 \) allows us to find the critical points that can be potential maximum or minimum values. Once we solve \( M - 2m = 0 \) for \( m \), we obtain \( m = \frac{M}{2} \),indicating a critical point.
Force Equation Simplification
Simplifying the force equation makes it easier to identify the factors affecting the gravitational force. Starting from: \[ F = G \frac{m (M - m)}{r^2} \], we notice that constants like \( G \) and \( r \) do not change. So, our goal is to simplify and maximize the term \[ m (M - m) \], which depends only on \( m \). We define a function \[ f(m) = m (M - m) \] and proceed by finding its critical points via differentiation. This simplifies the problem and focuses on maximizing the interaction between the two masses split from the original mass \( M \).

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Most popular questions from this chapter

In deep space, sphere \(A\) of mass \(20 \mathrm{~kg}\) is located at the origin of an \(x\) axis and sphere \(B\) of mass \(10 \mathrm{~kg}\) is located on the axis at \(x=0.80 \mathrm{~m} .\) Sphere \(B\) is released from rest while sphere \(A\) is held at the origin. (a) What is the gravitational potential energy of the two-sphere system as \(B\) is released? (b) What is the kinetic energy of \(B\) when it has moved \(0.20 \mathrm{~m}\) toward \(A\) ?

Zero, a hypothetical planet, has a mass of \(5.0 \times 10^{23} \mathrm{~kg}\), a radius of \(3.0 \times 10^{6} \mathrm{~m}\), and no atmosphere. A \(10 \mathrm{~kg}\) space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of \(5.0 \times 10^{7} \mathrm{~J}\), what will be its kinetic energy when it is \(4.0 \times 10^{6} \mathrm{~m}\) from the center of Zero? (b) If the probe is to achieve a maximum distance of \(8.0 \times 10^{6} \mathrm{~m}\) from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

A body is suspended from a spring scale in a ship sailing along the equator with speed \(v .\) (a) Show that the scale reading will be very close to \(W_{0}(1 \pm 2 \omega v / g)\), where \(\omega\) is the rotational speed of Earth and \(W_{0}\) is the scale reading when the ship is at rest. (b) Explain the \(\pm\) sign.

Is Newton's Law of Gravity Wrong? A professional scientist (not a physicist) stops you in the hall and says: "I can prove Newton's theory of gravity is wrong. The Sun is 320,000 times as massive as the Earth, but only 400 times as far from the Moon as is the Earth. Therefore, the force of the Sun's gravity on the Moon should be twice as big as the Earth's and the Moon should go around the Sun instead of around the Earth. Since it doesn't, Newton's theory of gravity must be wrong!" What's the matter with this reasoning?

A rocket is accelerated to speed \(v=\) \(2 \sqrt{g R_{E}}\) near Earth's surface (where Earth's radius is \(R_{E}\) ), and it then coasts upward. (a) Show that it will escape from Earth. (b) Show that very far from Earth its speed will be \(v=\sqrt{2 g R_{E}}\).
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