/*! 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 39 A projectile is fired vertically... [FREE SOLUTION] | 91Ó°ÊÓ

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A projectile is fired vertically upwards from the surface of the earth with a velocity \(K v_{e}\); where \(v_{e}\) is the escape velocity and \(k<1\). If \(R\) is the radius of the earth, the maximum height to which it will rise, measured from the centre of the earth, will be (neglect air resistance) (1) \(\frac{1-k^{2}}{R}\) (2) \(\frac{R}{1-k^{2}}\) (3) \(R(1-k)^{2}\) (4) \(\frac{R}{1+k^{2}}\)

Short Answer

Expert verified
The maximum height from Earth's center is \( \frac{R}{1-k^2} \). Hence, option (2) is correct.

Step by step solution

01

Identify Escape Velocity Formula

The escape velocity from the surface of the earth is given by the formula \( v_{e} = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the earth, and \( R \) is the radius of the earth. The problem states that the projectile's velocity is \( kv_{e} \).
02

Express Initial Kinetic and Potential Energy

The initial total energy of the projectile when it is fired is given by the sum of its kinetic and potential energies. The initial kinetic energy is \( \frac{1}{2} (kv_{e})^2 \) and the initial potential energy is \( -\frac{GMm}{R} \), where \( m \) is the mass of the projectile. Hence the total initial energy is \( \frac{k^2GMm}{R} - \frac{GMm}{R} \).
03

Calculate Total Energy at Maximum Height

At maximum height, the projectile's velocity is zero, so its kinetic energy is zero. Its potential energy at a distance \( R + h \) from the center of the earth (center because height is measured from the center) is \( -\frac{GMm}{R+h} \). Therefore, the total energy at maximum height is \( -\frac{GMm}{R+h} \).
04

Set Initial and Maximum Height Energy Equal

According to the conservation of energy, the initial total energy must equal the total energy at maximum height. Equating the two gives us: \( \frac{k^2GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \).
05

Solve for Maximum Height

Simplify the equation: \( \frac{(k^2 - 1) GMm}{R} = -\frac{GMm}{R+h} \). Cancel \( GMm \) from both sides and rearrange to get \( R+h = \frac{R}{1-k^2} \). Thus, \( h = \frac{R}{1-k^2} - R \). This simplifies to \( h = \frac{k^2R}{1-k^2} \). Therefore, the distance from the center is \( R + h = \frac{R}{1-k^2} \). Thus, the answer is option (2) \( \frac{R}{1-k^2} \).

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Key Concepts

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

Escape Velocity
Escape velocity is a crucial concept in understanding how objects move in space. It is the minimum speed an object must have to break free from the gravitational pull of a celestial body without further propulsion.
For Earth, this velocity is derived using the formula: \[ v_{e} = \sqrt{\frac{2GM}{R}} \] where
  • \( v_{e} \) is the escape velocity,
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of Earth,
  • \( R \) is the radius of Earth.
Understanding escape velocity helps in planning spacecraft missions and predicting the behavior of objects in space. It tells us how much energy is needed to overcome Earth's gravitational force. If an object moves with a speed equal to or greater than this velocity, it can leave Earth's gravity without any extra help, like continuous thrust from a rocket engine.
Kinetic Energy
Kinetic energy is the energy associated with the motion of an object. The faster an object moves, the more kinetic energy it has. It is calculated using the formula: \[ KE = \frac{1}{2} mv^2 \] Here,
  • \( m \) is the object's mass,
  • \( v \) is the velocity of the object.
When a projectile is fired upwards, part of the kinetic energy is initially imparted to it. This energy determines how high and fast it can travel against Earth's gravity. In our exercise, the kinetic energy was initially given as \[ \frac{1}{2} (kv_{e})^2 \]where \( k \) is a fraction less than 1, indicating it's not traveling at full escape speed. Understanding kinetic energy helps in determining how different velocity fractions impact a projectile's ability to reach different heights.
Conservation of Energy
The conservation of energy is a fundamental principle in physics, stating that energy cannot be created or destroyed but only transformed from one form to another. In the context of projectile motion, it implies that the total energy of the projectile remains constant when neglecting any external forces like air resistance.
Initially, the projectile's energy consists of kinetic and gravitational potential energy. As it rises, kinetic energy is converted into potential energy until it reaches its maximum height, where its kinetic energy is zero, and potential energy is at its peak.This balance is shown in the formula:\[ \text{Initial Energy} = \text{Maximum Height Energy} \]The understanding of this principle allows us to predict the projectile's final position and behavior under such physical constraints.
Gravitational Potential Energy
Gravitational potential energy refers to the energy an object possesses because of its position in a gravitational field. For objects on Earth or near its surface, it depends on the object's mass, the gravitational force, and the distance from the center of the Earth.It's given by the formula: \[ U = -\frac{GMm}{R} \] where
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of Earth,
  • \( m \) is the mass of the object,
  • \( R \) is the distance from the Earth's center to the object.
In this exercise, the change in potential energy is crucial for determining the max height the projectile will reach. When kinetic energy is fully converted to potential, we can calculate the maximum height in terms of Earth's radius and the projectile's speed.

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Most popular questions from this chapter

Choose the correct statements from the following: (I) The magnitude of the gravitational force between two bodies of mass \(1 \mathrm{~kg}\) each and separated by a distance of \(1 \mathrm{~m}\) is \(9.8 \mathrm{~N}\) (2) The higher the value of the escape velocity for a planet, the higher is the abundance of the lighter gases in its atmosphere. (3) The gravitational force of attraction between two bodies of ordinary mass is not noticeable because the value of the gravitational constant is extremely small. (4) Force of friction arises due to gravitational attraction.

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