/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 129 In a later chapter, you will fin... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 129

In a later chapter, you will find that the weight of a particle varies with altitude such that \(w=\frac{m g r_{0}^{2}}{r^{2}}\) where \(r_{0}\) is the radius of Earth and \(r\) is the distance from Earth's center. If the particle is fired vertically with velocity \(v_{0}\) from Earth's surface, determine its velocity as a function of position \(r\). (Hint: use \(a d r=v d v,\) the rearrangement mentioned in the text.)

Short Answer

Expert verified
The velocity of the particle as a function of its position \(r\) is given by: \(v(r) = \sqrt{v_{0}^2 - 2g(r_0 - \frac{r_0^2}{r})}\)

Step by step solution

01

Write down the given information

We are given the following information: - Weight as a function of altitude: \(w = \frac{m g r_{0}^{2}}{r^{2}}\) - Radius of Earth: \(r_{0}\) - Initial velocity: \(v_{0}\) - The hint: \(a dr = v dv\) Our goal is to find the velocity of the particle as a function of position \(r\).
02

Find acceleration using the weight formula

We know that weight is the force acting on a mass due to gravity and can be expressed as \(w = m a\). So, we can write the acceleration as: \(a = \frac{w}{m} = \frac{g r_{0}^2}{r^2}\)
03

Substitute the acceleration formula into the hint

Now we can substitute the acceleration expression into the given hint: \(\frac{g r_{0}^2}{r^2} dr = v dv\)
04

Separate variables and integrate

We will now separate the variables and integrate both sides of the equation: \(\int \frac{g r_{0}^2}{r^2} dr = \int v dv\)
05

Integration

Now integrate both sides of the equation: \(- \frac{g r_{0}^2}{r} = \frac{1}{2} v^2 + C\) Here, "C" is the integration constant.
06

Find the constant C

Recall that the initial velocity of the particle is given as \(v_{0}\). Hence, when \(r = r_0\), the velocity \(v = v_{0}\). Substituting these values into our equation, we can find the constant C: \(- \frac{g r_{0}^2}{r_0} = \frac{1}{2} v_{0}^2 + C\) Solving for C: \(C = - \frac{g r_{0}}{2} + \frac{1}{2} v_{0}^2\)
07

Write the velocity as a function of position r

Now substitute the value of C back into our equation: \(- \frac{g r_{0}^2}{r} = \frac{1}{2} v^2 - \frac{g r_{0}}{2} + \frac{1}{2} v_{0}^2\) Solve for the velocity v: \(v^2 = v_{0}^2 - 2g(r_0 - \frac{r_0^2}{r})\) Now, we have successfully found the velocity of the particle as a function of its position \(r\): \(v(r) = \sqrt{v_{0}^2 - 2g(r_0 - \frac{r_0^2}{r})}\)

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Key Concepts

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

Gravitational Force
Understanding the concept of gravitational force is key to solving problems involving motion under the influence of gravity. It is the attractive force that occurs between any two masses, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Sir Isaac Newton formulated this phenomenon in his law of universal gravitation.

In the context of our exercise, as a particle moves away from Earth's surface, the gravitational force it experiences decreases, which is modeled by the weight equation provided. This fundamental concept explains why the particle's acceleration due to gravity changes with altitude. Remember, this force is the source of the acceleration we're analyzing in the kinematics equations.
Kinematics Equations
Kinematics is the study of motion without considering the forces that cause this motion. The kinematics equations allow us to describe an object's motion in terms of displacement, velocity, acceleration, and time. These equations are powerful tools in physics鈥攖hey predict where and when a moving object will be.

However, in our exercise, time is not an explicit factor. Instead, we investigate how velocity changes with position, requiring a modification of the standard equations. By using the given hint, we link acceleration and velocity in a way that can be solved through integration, leading us to a unique solution tailoring the general kinematic formulation to a scenario where gravity's effect varies with altitude.
Integration in Physics
Integration is a mathematical technique that plays a critical role in physics, particularly in the fields involving continuous quantities, such as calculating areas under curves, finding the accumulated change, or determining the distribution of particles over a given space.

In our exercise, integration allows us to determine the velocity of the particle over the distance it has traveled. By separating variables, we can integrate both sides of our equation with respect to their respective variables鈥攄istance for gravitational acceleration, and velocity for itself鈥攁nd arrive at an expression that characterizes the motion of the particle from its initial position to any given distance from Earth's center.
Acceleration Due to Gravity
Acceleration due to gravity, usually denoted as 'g', is the acceleration gained by an object because of the gravitational force exerted by a massive body like Earth. It's worth noting that 'g' is not a constant throughout the universe, or even across different altitudes on Earth鈥攊t decreases as we move away from Earth's surface, as demonstrated in our exercise.

The typical value of 'g' near Earth's surface is approximately 9.8 m/s2, but as the altitude increases, the value of 'g' effectively decreases because the gravitational force weakens with distance, as per the inverse-square law. This change in 'g' is crucial when calculating the variable velocity of an object moving in a gravitational field, such as our problem with a vertically fired particle, where we need to determine how its velocity evolves as it moves to higher altitudes.

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Most popular questions from this chapter

A child has mass 6.0 kg and slides down a 35" incline with constant speed under the action of a \(34-\mathrm{N}\) force acting up and parallel to the incline. What is the coefficient of kinetic friction between the child and the surface of the incline?

Stokes' law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes' law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density \(7.8 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) diameter \(3.0 \mathrm{mm}\) ) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.

Shown below is a 10.0 -kg block being pushed by a horizontal force \(\overrightarrow{\mathbf{F}}\) of magnitude \(200.0 \mathrm{N}\). The coefficient of kinetic friction between the two surfaces is 0.50. Find the acceleration of the block.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/ s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

An elevator filled with passengers has a mass of \(1.70 \times 10^{3} \mathrm{kg} .\) (a) The elevator accelerates upward from rest at a rate of \(1.20 \mathrm{m} / \mathrm{s}^{2}\) for \(1.50 \mathrm{s}\). Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of \(0.600 \mathrm{m} / \mathrm{s}^{2}\) for \(3.00 \mathrm{s}\). What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?
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