/*! 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 63 (a) Determine the amount of work... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Determine the amount of work (in joules) that must be done on a \(100-\mathrm{kg}\) payload to elevate it to a height of \(1000 \mathrm{km}\) above the Earth's surface. (b) Determine the amount of additional work that is required to put the payload into circular orbit at this elevation.

Short Answer

Expert verified
The work done to elevate the payload is \(9.8 * 10^8 Joules\) and the additional work required to put the payload into circular orbit at this elevation is \(4.9 * 10^8 Joules\). Therefore, the total work is \(14.7 * 10^8 Joules\).

Step by step solution

01

Calculate the Work Done Lifting the Payload

Work done against gravity, also known as gravitational potential energy, is given by \(PE = m * g * h\), where \(m = 100kg\) is the mass, \(g = 9.8m/s^2\) is the acceleration due to gravity near Earth's surface and \(h = 1000km = 1.0 * 10^6m\) is the height. The value of \(g\) decreases as one moves further from the center of earth, but given that the altitude provided is significantly smaller than the radius of earth (Earth's radius is about 6,400km) we can use the approximation. This gives \(PE = 100 * 9.8 * 1.0 * 10^6 = 980 * 10^6 = 9.8 * 10^8 Joules\).
02

Calculate the Additional Work Required for Circular Orbit

The minimum kinetic energy needed to maintain circular orbit is equal to half the magnitude of the gravitational potential energy at that altitude, mathematically expressed as \( KE = 0.5 * PE \). Thus, \( KE = 0.5 * 9.8 * 10^8 = 4.9 * 10^8 Joules \). Therefore, the total work required to lift the object and place it in the orbit is sum of potential and kinetic energy which is \( W = PE + KE = 9.8 * 10^8 + 4.9 * 10^8 = 14.7 * 10^8 Joules\).

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 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. It plays an essential role when an object moves, accelerating from rest to a desired speed or orbit. When discussing a payload launched into space, kinetic energy is all about getting it to travel at the right velocity to enter and maintain a circular orbit. To achieve this needs energy in the form of work done by engines or thrusters.

For an object in space, kinetic energy (\( KE \)) can be calculated with the formula: \[KE = \frac{1}{2} m v^2\]where \( m \) is the mass of the object and \( v \) is its velocity.

In our example of putting a payload into a circular orbit, the kinetic energy needed is half of the gravitational potential energy at the designated altitude. This ensures it can stay in orbit, balancing the gravitational pull with its speed, remaining in perpetual motion without falling back to Earth. So, understanding kinetic energy helps us calculate the hard work necessary to lift the payload and sustain it in the orbit.
Work Against Gravity
Work against gravity refers to the energy required to lift an object against the Earth's gravitational pull. This is often expressed in terms of gravitational potential energy, calculated using the formula:\[PE = m * g * h\]where\( m \) is the mass of the object,\( g \) is the gravitational acceleration (approximately 9.8 \( m/s^2 \) near the Earth's surface), and\( h \) is the height the object is lifted.

It's important to note that as the object moves away from the Earth, the gravitational force acting on it weakens. However, for altitudes relatively small compared to Earth's radius, \( g \) can be taken as constant. This simplification allows us to calculate the gravitational potential energy and the work done to elevate the payload without complex adjustments for varying gravity.

In the exercise provided, raising a 100 kg payload 1000 km above Earth's surface requires substantial energy. This energy accounts for overcoming the Earth's gravity, enabling the payload to reach the desired height.
Circular Orbit
A circular orbit is achieved when an object revolves around a planet in a path that forms a circle. For such an orbit, the speed must be exact: too slow, and the object will fall back to Earth; too fast, and it will escape into space.

Reaching a circular orbit involves not just lifting a payload to a specific altitude but also giving it enough horizontal speed to maintain its path around the Earth. This is where the importance of kinetic energy comes in. The formula \( KE = 0.5 * PE \) showcases how kinetic energy relates to the gravitational potential energy: half the potential energy is dedicated to maintaining the orbit's motion.

When calculating the total energy required for orbit, it's important to consider both the gravitational pull an object needs to overcome and the additional energy needed for the object to travel horizontally at the right speed. Achieving a circular orbit thus involves combining both potential and kinetic energies, forming a complete and balanced energy equation essential for sustained orbital path.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A satellite of mass \(m,\) originally on the surface of the Earth, is placed into Earth orbit at an altitude \(h\). (a) With a circular orbit, how long does the satellite take to complete one orbit? (b) What is the satellite's speed? (c) What is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet's daily rotation. At what location on the Earth's surface and in what direction should the satellite be launched to minimize the required energy investment? Represent the mass and radius of the Earth as \(M_{E}\) and \(R_{E}\).

The maximum distance from the Earth to the Sun (at our aphelion) is \(1.521 \times 10^{11} \mathrm{m},\) and the distance of closest approach (at perihelion) is \(1.471 \times 10^{11} \mathrm{m} .\) If the Earth's orbital speed at perihelion is \(3.027 \times 10^{4} \mathrm{m} / \mathrm{s}\), determine (a) the Earth's orbital speed at aphelion, (b) the kinetic and potential energies of the Earth-Sun system at perihelion, and (c) the kinetic and potential energies at aphelion. Is the total energy constant? (Ignore the effect of the Moon and other planets.)

A comet of mass \(1.20 \times 10^{10} \mathrm{kg}\) moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between \(0.500 \mathrm{AU}\) and \(50.0 \mathrm{AU},\) (a) What is the eccentricity of its orbit? (b) What is its period? (c) At aphelion what is the potential energy of the comet-Sun system? Note: \(1 \mathrm{AU}=\) one astronomical unit \(=\) the average distance from Sun to Earth \(=1.496 \times 10^{11} \mathrm{m}\).

The \(Explorer\) VIII satellite, placed into orbit November 3 , \(1960,\) to investigate the ionosphere, had the following orbit parameters: perigee, \(459 \mathrm{km} ;\) apogee, \(2289 \mathrm{km}\) (both distances above the Earth's surface); period, 112.7 min. Find the ratio \(v_{p} / v_{a}\) of the speed at perigee to that at apogee.

As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of \(3.64 \times 10^{9} \mathrm{kg} / \mathrm{s} .\) During the \(5000-\mathrm{yr}\) period of recorded history, by how much has the length of the year changed due to the loss of mass from the Sun? Suggestions: Assume the Earth's orbit is circular. No external torque acts on the Earth-Sun system, so its angular momentum is conserved. If \(x\) is small compared to \(1,\) then \((1+x)^{n}\) is nearly equal to \(1+n x\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.