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Chapter 6: Problem 50

(III) Two identical particles, each of mass \(m,\) are located on the \(x\) axis at \(x=+x_{0}\) and \(x=-x_{0}\). (a) Determine a formula for the gravitational field due to these two particles for points on the \(y\) axis; that is, write \(\vec{g}\) as a function of \(y, m,\) \(x_{0},\) and so on. \((b)\) At what point (or points) on the \(y\) axis is the magnitude of \(\vec{g}\) a maximum value, and what is its value there? [Hint: Take the derivative \(d \overrightarrow{\mathrm{g}} / d y .\)

Short Answer

Expert verified
The point of maximum gravitational field magnitude is at \(y = \pm \frac{x_0}{\sqrt{2}}\) and the magnitude is \(|\vec{g}| = \frac{4Gm}{3\sqrt{3}x_0^2}\).

Step by step solution

01

Define the Problem

We need to find the gravitational field at a point on the y-axis due to two identical particles with mass \(m\) located symmetrically at positions \(x = +x_0\) and \(x = -x_0\) on the x-axis. We shall express the gravitational field \(\vec{g}\) in terms of \(y, m,\) and \(x_0\).
02

Express Distance and Gravitational Field

The distance from a point on the y-axis at \((0, y)\) to either particle located at \((x_0, 0)\) or \((-x_0, 0)\) is \(r = \sqrt{x_0^2 + y^2}\). The gravitational field magnitude due to one particle is \(g = \frac{Gm}{r^2}\), where \(G\) is the gravitational constant.
03

Determine Direction of Gravitational Field

Each particle exerts a gravitational field with a horizontal component along \(x\) and a vertical component along \(y\). The horizontal components cancel each other, while the vertical components add up, resulting in a net field in the y-direction.
04

Combine Gravitational Fields

The net gravitational field at \((0, y)\) is twice the vertical component of the field from one particle: \(\vec{g} = 2 \cdot g \cdot \frac{y}{r}\). Substitute \(g = \frac{Gm}{r^2}\):\[ \vec{g} = 2 \cdot \frac{Gm}{(x_0^2 + y^2)} \cdot \frac{y}{\sqrt{x_0^2 + y^2}} \hat{j} \]Simplify to:\[ \vec{g} = \frac{2Gmy}{(x_0^2 + y^2)^{3/2}} \hat{j} \]
05

Find Maximum Magnitude of \(\vec{g}\)

To find the maximum, set the derivative of the magnitude \(|\vec{g}|\) with respect to \(y\) to zero. Compute:\[ \frac{d}{dy} \left( \frac{2Gmy}{(x_0^2 + y^2)^{3/2}} \right) = 0 \]Simplifying:\[ y^2 = \frac{x_0^2}{2} \Rightarrow y = \pm \frac{x_0}{\sqrt{2}} \]
06

Calculate \(|\vec{g}|\) at Maximum

Substitute \(y = \pm \frac{x_0}{\sqrt{2}}\) into the equation for \(|\vec{g}|\):\[ |\vec{g}| = \frac{2Gm \frac{x_0}{\sqrt{2}}}{\left(x_0^2 + \left(\frac{x_0}{\sqrt{2}}\right)^2\right)^{3/2}} = \frac{2Gm \frac{x_0}{\sqrt{2}}}{\left(\frac{3}{2} x_0^2\right)^{3/2}} \]Solving:\[ |\vec{g}| = \frac{4Gm}{3\sqrt{3}x_0^2} \]

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Key Concepts

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

Gravitational Force
The gravitational force is a fundamental interaction in the universe that attracts two masses toward each other. Every object with mass exerts a gravitational pull on every other mass. The strength of this force depends on the masses involved and the distance between them.

In our exercise, we consider two identical particles, each with mass \(m\), positioned symmetrically on the x-axis. Their gravitational influence is exerted along the y-axis at any point \((0, y)\).

  • The distance \(r\) from a particle to a point on the y-axis is given by the Pythagorean theorem: \(r = \sqrt{x_0^2 + y^2}\).
  • The gravitational force exerted by one particle is \(\frac{Gm}{r^2}\), where \(G\) is the gravitational constant.
  • Since the particles are identical and symmetrically placed, their horizontal gravitational pulls cancel, leaving only vertical forces.
The combined gravitational field along the y-axis, therefore, doubles the vertical component from one particle, aligning the field purely in the positive or negative y-direction, depending on the location.
Symmetry in Physics
Symmetry is a powerful concept in physics, often leading to simplifications in the analysis of physical problems. Symmetry can manifest in various forms, such as spatial symmetry, which involves geometric arrangements that remain unchanged under certain transformations.

In this exercise, symmetry arises because we have two particles placed symmetrically on either side of the y-axis.

  • Each particle induces a gravitational field component towards itself.
  • The horizontal components, due to their opposing directions, cancel each other out due to this symmetry.
  • The result is a net gravitational field directed along the y-axis, which simplifies the calculation significantly.
This symmetrical property enables us to focus solely on the vertical aspects of the gravitational force, effectively reducing the complexity of the problem and allowing a more manageable mathematical solution.
Derivatives in Physics
Derivatives are a core tool in physics, enabling us to understand how physical quantities change with respect to one another. They are especially useful in finding rates of change and determining maxima or minima of functions.

In the context of this exercise, we utilize derivatives to find the points on the y-axis where the gravitational field \(|\vec{g}|\) is maximized.

  • We begin by expressing the gravitational field as a function of \(y\): \(\frac{2Gmy}{(x_0^2 + y^2)^{3/2}}\).
  • The next step involves taking the derivative of this expression with respect to \(y\) to find when it equals zero. This indicates maximum or minimum values.
  • Solving \(\frac{d}{dy} \left( \frac{2Gmy}{(x_0^2 + y^2)^{3/2}} \right) = 0\) helps identify the value \(y = \pm \frac{x_0}{\sqrt{2}}\), where \(|\vec{g}|\) reaches its maximum.
This process demonstrates how derivatives facilitate the discovery of crucial insights within physics problems, showcasing their indispensable nature in analysis and comprehension of changing systems.

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Most popular questions from this chapter

(II) Calculate the period of a satellite orbiting the Moon, \(120 \mathrm{~km}\) above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is \(1740 \mathrm{~km}\).

A geologist searching for oil finds that the gravity at a certain location is 2 parts in \(10^{7}\) smaller than average. Assume that a deposit of oil is located \(2000 \mathrm{~m}\) directly below. Estimate the size of the deposit, assumed spherical. Take the density (mass per unit volume) of rock to be \(3000 \mathrm{~kg} / \mathrm{m}^{3}\) and that of oil to be \(800 \mathrm{~kg} / \mathrm{m}^{3}\).

(II) Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is \(4.9 \times 10^{22} \mathrm{~kg}\) and making the assumption that its density is the same as Earth's.

(II) Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. A "near-Earth" orbit is at a height above the surface of the Earth that is very small compared to the radius of the Earth. [Hint: You may take the acceleration due to gravity as essentially the same as that on the surface.] Does your result depend on the mass of the satellite?

(II) Two equal-mass stars maintain a constant distance apart of \(8.0 \times 10^{11} \mathrm{~m}\) and rotate about a point midway between them at a rate of one revolution every 12.6 yr. \((a)\) Why don't the two stars crash into one another due to the gravitational force between them? (b) What must be the mass of each star?
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