91影视

(a) Calculate how much work is required to launch a spacecraft of mass \(m\) from the surface of the earth (mass \(m_{\mathrm{E}},\) radius \(R_{\mathrm{E}} )\) and place it in a circular low earth orbit - that is, an orbit whose altitude above the earth's surface is much less than \(R_{\mathrm{E}}\) . (As an example, the International Space Station is in low earth orbit at an altitude of about \(400 \mathrm{km},\) much less than \(R_{\mathrm{E}}=6380 \mathrm{km} .\) You can ignore the kimetic energy that the spacecraft has on the ground due to the earth's rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement: "In terms of energy, low earth orbit is halfway to the edge of the universe."

Step by step solution

01

Understanding the Problem for Part (a)

We need to find the work required to launch a spacecraft from the surface of the Earth into a low Earth orbit (LEO). We'll use the gravitational potential energy difference between the surface and the orbit to calculate this. The altitude of LEO is much smaller than the Earth's radius.

02

Calculate Gravitational Potential at Earth's Surface

The gravitational potential energy (U) at Earth's surface is given by the formula: \[ U_{ ext{surface}} = -\frac{G m m_E}{R_E} \]where \(G\) is the gravitational constant, \(m\) is the mass of the spacecraft, \(m_E\) is the mass of Earth, and \(R_E\) is the Earth's radius.

03

Calculate Gravitational Potential in Low Earth Orbit

In LEO, the gravitational potential energy is slightly different due to the altitude \(h\). It is given by:\[ U_{ ext{LEO}} = -\frac{G m m_E}{R_E + h} \]Since \(h\) is much less than \(R_E\), it can be approximated that \(R_E + h \approx R_E\) for calculation purposes in LEO.

04

Find Work Required to Move to LEO

The work done in moving the spacecraft to LEO is the difference in gravitational potential energy between the surface and LEO:\[ W_{ ext{LEO}} = U_{ ext{LEO}} - U_{ ext{surface}} \]Substituting the potential energies:\[ W_{ ext{LEO}} = -\frac{G m m_E}{R_E + h} + \frac{G m m_E}{R_E} \approx \frac{G m m_E h}{R_E^2} \]This approximation is valid because \(h \ll R_E\).

05

Understanding the Problem for Part (b)

For part (b), calculate the additional work needed to move the spacecraft from LEO to a very great distance (essentially escape work).

06

Calculate Work for Escape Velocity from LEO

The work to completely escape Earth's gravity from LEO is the difference between the gravitational potential energy at LEO and zero (infinity potential), where the gravitational potential energy at infinity is zero:\[ W_{ ext{escape}} = 0 - U_{ ext{LEO}} = \frac{G m m_E}{R_E + h} \approx \frac{G m m_E}{R_E} - \frac{G m m_E h}{R_E^2} \approx \frac{1}{2} \frac{G m m_E}{R_E} \]This assumes escaping from LEO requires about half the energy as launching from the surface.

07

Justifying Energy Equivalence in Low Earth Orbit

In part (c), we consider the energy required to reach LEO compared to the energy needed to escape from LEO. Launching to LEO requires energy equivalent to about half the energy needed to escape to a very great distance. Therefore, being in LEO can be thought of as 鈥渉alfway鈥� in energy terms to escaping into interstellar space.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Gravitational Potential Energy

The concept of gravitational potential energy is central in understanding how energy is stored within a gravitational field. It is the energy an object possesses because of its position relative to another mass, typically seen in relation to Earth. The gravitational potential energy at a point is given by \[ U = -\frac{G m m_E}{r} \]where:

• \( G \) is the gravitational constant,
• \( m \) is the mass of the object (e.g., the spacecraft),
• \( m_E \) is the mass of the Earth,
• \( r \) is the distance from the center of the Earth.

Gravitational potential energy is at its maximum when objects are at their most distant. In our context, when a spacecraft is at Earth's surface, its potential energy is less than when it reaches low Earth orbit. This is because it is less "deep" in Earth's gravity well. The change in this energy tells us the work done to move the spacecraft into orbit.

Work-Energy Principle

The work-energy principle is a valuable physics concept that bridges the gap between the force applied on an object and its resulting energy changes. It states that the total work done on an object is equal to the change in its kinetic energy. This principle can also involve potential energy differences, which makes it highly applicable in orbital mechanics.For example, to find how much work is required to move a spacecraft from Earth's surface to low Earth orbit, we look at the change in gravitational potential energy. The work \( W \) needed is:\[ W = \Delta U = U_{\text{LEO}} - U_{\text{surface}} \]This shows that lifting a spacecraft into orbit requires a specific amount of work to increase its potential energy. Similarly, additional work is required to move it from the orbit to a point away from Earth's gravitational field entirely鈥攁 concept crucial when discussing escape velocities.

Low Earth Orbit

Low Earth Orbit (LEO) is a region of space close to Earth's surface, typically at an altitude of 400 km to 2,000 km. It is where most satellites and the International Space Station (ISS) operate. LEO is important in several ways:

• It requires relatively low energy to place a spacecraft into orbit compared to higher orbits.
• The gravitational pull is still significant, meaning fall back to Earth is possible without additional propulsion.
• It offers a stable environment for satellites close enough for fast data transmission on Earth.

When a spacecraft is in LEO, its gravitational potential energy is higher than when on Earth鈥檚 surface but not as high as it would be in higher, more distant orbits. Therefore, LEO is said to be about halfway to the "edge of the universe" in energy terms.

Escape Velocity

Escape velocity is the minimum speed needed for an object to escape from the gravitational attraction of a celestial body without further propulsion. It is a key concept in orbital mechanics and determines how much energy a spacecraft requires to exit Earth's gravitational influence entirely.To calculate escape speed from Earth's surface, the formula is:\[ v_e = \sqrt{\frac{2G m_E}{R_E}} \]Here:

• \( v_e \) is escape velocity,
• \( G \) is the gravitational constant,
• \( m_E \) is Earth's mass,
• \( R_E \) is Earth's radius.

For a spacecraft in LEO, achieving escape velocity is more energy-efficient because it has already overcome significant gravitational pull. The work needed from LEO for escape involves an energy calculation that shows the spacecraft is halfway to having the total energy needed to "escape into the void" or the vastness of space. This concept helps justify referring to LEO as a critical milestone in space exploration.