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A \(1.0 \mathrm{kg}\) object is released from rest \(500 \mathrm{km}\) ( \(\sim 300\) miles) above the earth. a. What is its impact speed as it hits the ground? Ignore air resistance. b. What would the impact speed be if the earth were flat? c. By what percentage is the flat-earth calculation in error?

Short Answer

Expert verified
a. The impact speed when the object hits the ground on a round Earth would be calculated using the conservation of energy concept in physics. b. If the Earth were flat, the impact speed would be calculated using the constant gravitational acceleration and the height. c. The error between the two calculations would be calculated using the percentage error formula.

Step by step solution

01

Calculate the impact speed on round Earth

First, use the conservation of energy, where the change in potential energy equals the change in kinetic energy. The initial potential energy is \( GMEm / r_{E} \) and the final kinetic energy is \( 0.5 * m * v^{2} \), where 'G' is the gravitational constant, 'ME' is the mass of Earth, 'RE' is the radius of the Earth, and 'v' is the velocity or impact speed. Solve the equation for 'v': \[ v = \sqrt{ \frac{ 2*GM_{E}}{R_{E}} } \]
02

Calculate the impact speed on a flat Earth

Now calculate the impact speed if the Earth were flat. In this model, the gravitational force is assumed constant. Therefore, use the equation \( v_{F} = \sqrt{2gH} \), where 'g' is the acceleration due to gravity (approximated to 9.8 m/s²) and 'H' is the height from which the object fell.
03

Calculate the percentage error

Error can be calculated by using the formula \( error \% = \frac{|predicted-actual|}{actual} * 100 \) . Insert the 'v' (actual impact speed on a round Earth) and 'v_{F}' (predicted impact speed on a flat Earth) into the formula.

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Key Concepts

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

Conservation of Energy
When we talk about the conservation of energy, we refer to a fundamental principle of physics. This principle states that energy within a closed system remains constant—energy cannot be created or destroyed, only transformed from one form to another.

In the context of our problem, we're concerned with gravitational potential energy transforming into kinetic energy as an object falls toward Earth. Potential energy is the stored energy an object possesses because of its position in a gravitational field. For our object released high above Earth, its potential energy is eventually converted into kinetic energy as it accelerates downward.
  • Initial Energy: The gravitational potential energy when the object is at height.
  • Final Energy: The kinetic energy just before it strikes the ground.
Recognize that this principle allows us to set up an equation where the initial potential energy equals the final kinetic energy. This provides the basis for solving problems like calculating impact speed.
Kinetic Energy
Kinetic energy is all about motion. It refers to the energy an object possesses due to its velocity. The faster an object moves, the greater its kinetic energy.

Kinetic energy is mathematically described by the formula \( KE = \frac{1}{2} mv^2 \) where:
  • \( m \): Mass of the object (in kilograms).
  • \( v \): Velocity of the object (in meters per second).
This equation becomes crucial in our problem as it relates to how potential energy transforms into kinetic energy during the fall. As potential energy decreases, kinetic energy increases, leading to the calculation of the object's speed right before impact.
Understanding kinetic energy not only helps in these calculations but also in grasping how objects interact with forces and motion in physics.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a key component in Newton's law of universal gravitation. It helps quantify the gravitational attraction between two masses. Importantly, \( G \) is a constant value, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).

In the calculation of gravitational potential energy, the formula \( U = \frac{GM_E m}{R_E} \) involves this constant, where:
  • \( M_E \): Mass of the Earth.
  • \( m \): Mass of the object.
  • \( R_E \): Radius of the Earth.
This facilitated our ability to calculate the energy conversion from gravitational potential energy to kinetic energy as the object falls from the significant height above Earth toward the ground. Understanding \( G \) specializes your comprehension of gravitational physics and its constant influence on bodies in the universe.
Impact Speed Calculation
Calculating impact speed involves understanding and utilizing the principles of energy conversion. Based on the conservation of energy, when an object falls, its gravitational potential energy is converted to kinetic energy.

In our specific exercise, two approaches are utilized:
  • Round Earth Calculation: Utilize the conservation of energy by equating potential energy to kinetic energy. The formula used is \( v = \sqrt{ \frac{2GM_E}{R_E} } \), which helps to obtain the speed just before the object hits the ground considering Earth's curvature.
  • Flat Earth Calculation: This simplified model assumes constant gravitational force, leading to the formula \( v = \sqrt{2gH} \). Here, \( g \) is the average gravitational acceleration, and \( H \) is the height of free fall.
The comparison of these speeds brings us to the calculation of percentage error, highlighting deviations under different modeling assumptions. Impact speed significantly illustrates the practical application of concepts like energy conservation, potential, and kinetic energy.

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Most popular questions from this chapter

Two spherical asteroids have the same radius \(R\). Asteroid 1 has mass \(M\) and asteroid 2 has mass \(2 M .\) The two asteroids are released from rest with distance \(10 R\) between their centers. What is the speed of each asteroid just before they collide? Hint: You will need to use two conservation laws.

You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of earth but the free-fall acceleration at the surface is only one-fourth as large. a. What is the planet's radius? b. To get back to earth, you need to escape the planet. What minimum speed does your rocket need?

Planet \(Z\) is \(10,000 \mathrm{km}\) in diameter. The free-fall acceleration on Planet \(Z\) is \(8.0 \mathrm{m} / \mathrm{s}^{2}\) a. What is the mass of Planet Z? b. What is the free-fall acceleration \(10,000 \mathrm{km}\) above Planet \(Z\) 's north pole?

The solar system is 25.000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in one year at a speed of \(3.0 \times 10^{8} \mathrm{m} / \mathrm{s}\). Astronomers have delermined that the solar system is orbiting the center of the galaxy at a speed of \(230 \mathrm{km} / \mathrm{s}\). a. Assuming the orbit is circular, what is the period of the solar system's orbit? Give your answer in years. b. Our solar system was formed roughly 5 billion years ago. How many orbits has it completed? c. The gravitational force on the solar system is the net force due to all the matter inside our orbit. Most of that matter is concentrated near the center of the galaxy. Assume that the matter has a spherical distribution, like a giant star. What is the approximate mass of the galactic center? d. Assume that the sun is a typical star with a typical mass. If galactic matter is made up of stars, approximately how many stars are in the center of the galaxy? Astronomers have spent many years trying to determine how many stars there are in the Milky Way. The number of stars seems to be only about \(10 \%\) of what you found in part d. In other words, about \(90 \%\) of the mass of the galaxy appears to be in some form other than stars. This is called the dark matter of the universe. No one knows what the dark matter is. This is one of the outstanding scientific questions of our day.

In Problems 65 through 68 you are given the equation(s) used to solve a problem. For each of these, you are to a. Write a realistic problem for which this is the correct equation(s). b. Draw a pictorial representation. c. Finish the solution of the problem. $$\begin{array}{l} \left(2.0 \times 10^{30} \mathrm{kg}\right) \mathrm{v}_{\mathrm{n}}+\left(4.0 \times 10^{30} \mathrm{kg}\right) v_{0}=0 \\ \frac{1}{2}\left(2.0 \times 10^{30} \mathrm{kg}\right) \mathrm{v}_{\mathrm{n}}^{2}+\frac{1}{2}\left(4.0 \times 10^{30} \mathrm{kg}\right) \mathrm{v}_{\mathrm{R}}^{2} \\ -\frac{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(2.0 \times 10^{30} \mathrm{kg}\right)\left(4.0 \times 10^{30} \mathrm{kg}\right)}{1.0 \times 10^{9} \mathrm{m}} \\ =0+0 \\ -\frac{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(2.0 \times 10^{10} \mathrm{kg}\right)\left(4.0 \times 10^{20} \mathrm{kg}\right)}{1.0 \times 10^{12} \mathrm{m}} \end{array}$$

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