/*! 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 36 An object of mass \(m\) is dropp... [FREE SOLUTION] | 91Ó°ÊÓ

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An object of mass \(m\) is dropped from height \(h\) above a planet of mass \(M\) and radius \(R\). Find an expression for the object's speed as it hits the ground.

Short Answer

Expert verified
The object's speed as it hits the ground is given by the expression \(v = \sqrt{2GM/R}\).

Step by step solution

01

Identify the Energy Conversion

In this exercise, there is a conversion of gravitational potential energy into kinetic energy. The gravitational potential energy of the object at the beginning is converted into kinetic energy when the object hits the ground.
02

Define the Energy Conservation Law

The Law of Energy Conservation states: \[ \text{Total Energy Initial} = \text{Total Energy Final} \]. From Step 1, initially the object has potential energy, \(PE_i\), and final kinetic energy, \(KE_f\), due to movement. So, \(PE_i = KE_f\).
03

Write Down the Potentional and Kinetic Energy

The formula for gravitational potential energy is \(- GMm / R\) where \(G\) is the gravitational constant, \(M\) is the mass of planet, \(m\) is the mass of the object and \(R\) is the radius of planet. The formula for kinetic energy is \(1/2 mv^2\) where \(v\) is the speed of the object.
04

Substitute the Formulas into the Energy Conservation Equation

Substitute these energy expressions into our energy equation, we get \(- GMm / R = 1/2 mv^2\).
05

Solve for the Object's Speed (\(v\))

To solve for \(v\), multiply both sides by \(-2/R\) and then take the square root of both sides. This yields \(v = \sqrt{2GM/R}\).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a form of energy that an object possesses because of its position in a gravitational field. It's the energy that an object has due to its height above the ground.
The formula is given by \[ PE = -\frac{GMm}{R} \] where
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of the planet,
  • \( m \) is the mass of the object, and
  • \( R \) is the radius of the planet.
When objects are moved from a distance to a closer point within the gravitational field, this energy can be converted to other forms of energy, such as kinetic energy. The negative sign in the formula indicates that gravitational potential energy is inversely related to the distance from the center of the planet.
This energy transformation is crucial when objects fall towards a planet, as it converts into kinetic energy, moving from potential to motion.
Kinetic Energy
Kinetic energy is the energy of motion. When an object moves, it harnesses kinetic energy, which is calculated using the formula: \[ KE = \frac{1}{2} mv^2 \] Here,
  • \( m \) is the mass of the object, and
  • \( v \) is its speed.
In our gravitational problem, as the object falls, its gravitational potential energy decreases while its kinetic energy increases. This is due to the conservation of energy; the potential energy that the object loses is transformed into kinetic energy, causing the object to speed up. When the object finally hits the ground, all of its initial gravitational potential energy has been converted into kinetic energy. Understanding this concept helps us solve various physics problems involving motion and energy transformations, making it a key component of studying physics.
Physics Problem Solving
Physics problem solving involves using known principles to find unknown values, which is accomplished through systematic approaches like applying the law of energy conservation. The core idea here is the exchange between energies. When faced with a question like the initial exercise, start by identifying energy forms at the start and end of the object's journey.In our exercise, we learned that initially, the object has gravitational potential energy due to its height. As it falls, this is transformed into kinetic energy just before impact. This energy exchange can be handled with the conservation equation: \[\text{Total Energy Initial} = \text{Total Energy Final} \]
Follow steps like:
  • Identifying known quantities and formulas,
  • Substituting these into the energy conservation equation, and
  • Solving algebraically for the unknowns.
This approach allows solving for quantities like speed at impact, ensuring robust understanding of energy transformations. This method underscores the beauty of physics—connecting concepts to observe how nature operates.

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Most popular questions from this chapter

Large stars can explode as they finish burning their nuclear fuel, causing a supernova. The explosion blows away the outer layers of the star. According to Newton's third law, the forces that push the outer layers away have reaction forces that are inwardly directed on the core of the star. These forces compress the core and can cause the core to undergo a gravitational collapse. The gravitational forces keep pulling all the matter together-tighter and tighter, crushing atoms out of existence. Under these extreme conditions, a proton and an electron can be squeezed together to form a neutron. If the collapse is halted when the neutrons all come into contact with each other, the result is an object called a neutron star, an entire star consisting of solid nuclear matter. Many neutron stars rotate about their axis with a period of \(\approx 1\) s and, as they do so, send out a pulse of electromagnetic waves once a second. These stars were discovered in the 1960 s and are called pulsars. a. Consider a neutron star with a mass equal to the sun, a radius of \(10 \mathrm{km},\) and a rotation period of \(1.0 \mathrm{s}\). What is the speed of a point on the equator of the star? b. What is \(g\) at the surface of this neutron star? c. A stationary \(1.0 \mathrm{kg}\) mass has a weight on earth of \(9.8 \mathrm{N}\). What would be its weight on the star? d. How many revolutions per minute are made by a satellite orbiting \(1.0 \mathrm{km}\) above the surface? e. What is the radius of a geosynchronous orbit about the neutron star?

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?

In 2000 . NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of \(1.0 \times 10^{16} \mathrm{kg}\) and a radius of \(8.8 \mathrm{km}\) a. What is the speed of a satellite orbiting \(5.0 \mathrm{km}\) above the surface? b. What is the escape speed from the asteroid?

Suppose that on earth you can jump straight up a distance of \(50 \mathrm{cm} .\) Can you escape from a \(4.0-\mathrm{km}\) -diameter asteroid with a mass of \(1.0 \times 10^{14} \mathrm{kg} ?\)

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