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Chapter 13: Problem 39

(a) What is the escape speed on a spherical asteroid whose radius is \(500 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s} ?(\mathrm{c})\) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) ahove the surface?

Short Answer

Expert verified
(a) 1732 m/s; (b) 166,667 m; (c) 1732 m/s.

Step by step solution

01

Understand Escape Speed Formula

The escape speed is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without further propulsion. It is given by the formula \( v_e = \sqrt{2gr} \) where \( g \) is the gravitational acceleration and \( r \) is the radius of the asteroid.
02

Calculate Escape Speed

Given \( g = 3.0 \, \text{m/s}^2 \) and \( r = 500 \, \text{km} = 500,000 \, \text{m} \), we substitute these values into the formula. \( v_e = \sqrt{2 \times 3.0 \, \text{m/s}^2 \times 500,000 \, \text{m}} = \sqrt{3,000,000} \, \text{m/s} \approx 1732 \, \text{m/s}. \) The escape speed is approximately 1732 m/s.
03

Set up Energy Conservation for Maximum Height

To find how far a particle will go starting with an initial radial speed of 1000 m/s, we use energy conservation: \( \frac{1}{2} m v^2 - \frac{GMm}{r} = \frac{1}{2} m v' ^2 - \frac{GMm}{r'} \). Simplifying, \( \frac{1}{2} v^2 + g r = gr' \), solving for \( r' \).
04

Solve for Maximum Height

Substituting \( v = 1000 \, \text{m/s}, g = 3.0 \, \text{m/s}^2, \text{ and } r = 500,000 \, \text{m} \): \( \frac{1,000^2}{2} + 3.0 \times 500,000 = 3.0 \times r' \). Solving, \( 500,000 + 1,500,000 = 3.0r' \) results in \( r' = 666,667 \, \text{m} \). The particle will go 166,667 m above the surface.
05

Energy Conservation for Dropped Object

For an object dropped from 1000 km above the surface, use energy conservation: at the start of the drop \( Kinetic \ Energy = 0 \) but \( Potential \ Energy = -\frac{GMm}{r+height} \). Upon impact, \( Kinetic \ Energy = \frac{1}{2}mv^2 \) and \( Potential \ Energy = -\frac{GMm}{r} \). Set them equal to solve for \( v \).
06

Solve for Impact Speed

Using the energy equation: \( \frac{1}{2}v^2 = g(r - (r+1000,000)) \), solve for \( v \) at \( r = 500,000 \, \text{m} \) and added height of 1000 km or 1,000,000 m. This results in \( v = \sqrt{2 \times 3.0 \times 500,000 \times (1 - \frac{1}{3})} \approx 1732 \, \text{m/s} \). The object will hit the surface at approximately 1732 m/s.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is the rate at which an object accelerates due to the gravitational force of a celestial body. On Earth, this is approximately 9.8 m/s虏. However, on different celestial bodies, this value can vary. For example, the gravitational acceleration on the asteroid in our original exercise is 3.0 m/s虏. This smaller value means that the force pulling objects towards the asteroid is weaker compared to Earth.
  • It depends on the mass and size of the object creating the gravitational field.
  • This acceleration affects how objects fall and how fast they need to move to escape its gravity.
Knowing this acceleration helps in solving problems related to escape speed and energy conservation. It's crucial for calculating how objects will behave when entering or leaving the asteroid's gravitational pull.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. This concept is essential when studying motion and forces in physics. When dealing with problems like escape speed, we use conservation of energy to track kinetic and potential energy changes.
  • Kinetic energy is the energy of motion, calculated as \( \frac{1}{2} mv^2 \).
  • Potential energy, specifically gravitational, is the energy stored due to an object's position, given by \( -\frac{GMm}{r} \).
In the case of an asteroid, if an object is moving away, its kinetic energy decreases as potential energy increases until it can't go further without external force. This switch is what allows us to calculate concepts like escape velocity and maximum height for a particle launched upwards.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It plays a crucial role in understanding escape speed since a moving object must have enough kinetic energy to overcome the gravitational pull of a body, like an asteroid, to escape into space.
  • This energy can be calculated using the formula \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity.
  • As an object moves faster, its kinetic energy increases, requiring higher energy to overcome gravitational forces.
When an object is launched from the surface of an asteroid, it needs to start with enough kinetic energy to reach escape speed. Factors like initial speed determine how far it can go before gravity pulls it back.
Potential Energy
Potential energy in the context of gravity is the energy an object has due to its position in a gravitational field. On a spherical asteroid, this potential energy is determined by its distance from the asteroid's center. As an object moves away from the celestial body, its potential energy increases.
  • Gravitational potential energy can be represented by the formula \(U = -\frac{GMm}{r}\), where \(G\) is the gravitational constant, \(M\) is the mass of the celestial body, \(m\) is the mass of the object, and \(r\) is the distance from the center.
  • This energy becomes less negative as the object moves away, showing an increase in potential energy.
In our exercise, potential energy helps us find how far a particle can travel and the speed at which an object will impact as it falls back due to gravity. Balancing kinetic and potential energy helps understand the motion of objects in a gravitational field.

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Most popular questions from this chapter

What must the separation be between a \(5.2 \mathrm{~kg}\) particle and a \(2.4 \mathrm{~kg}\) particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12} \mathrm{~N} ?\)

A \(150.0 \mathrm{~kg}\) rocket moving radially outward from Earth has a speed of \(3.70 \mathrm{~km} / \mathrm{s}\) when its engine shuts off \(200 \mathrm{~km}\) above Earth's surface. (a) Assuming negligible air drag acts on the rocket, find the rocket's kinetic energy when the rocket is \(1000 \mathrm{~km}\) above Earth's surface. (b) What maximum height above the surface is reached by the rocket?

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