/*! 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 65 Comets travel around the sun in ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.0 \(\times\) 10\(^4\) m/s when at a distance of 2.5 \(\times\) 10\(^{11}\) m from the center of the sun, what is its speed when at a distance of 5.0 \(\times\) 10\(^{10}\) m?

Short Answer

Expert verified
The comet's speed at a distance of 5.0 \(\times\) 10\(^{10}\) m is approximately 50,843 m/s.

Step by step solution

01

Understand the Problem

We are given a comet's speed and distance from the sun at one point in its orbit and need to find its speed at another distance. We'll use conservation of energy, since gravitational forces are conservative.
02

Apply Conservation of Energy

The total mechanical energy of the comet at both positions must be equal. The total mechanical energy is the sum of kinetic energy and gravitational potential energy: \[ KE_1 + PE_1 = KE_2 + PE_2 \].Where:\[ KE = \frac{1}{2}mv^2 \] and \[ PE = -\frac{GMm}{r} \].
03

Express Energies at First Position

At the first position, the kinetic energy is \[ KE_1 = \frac{1}{2}mv_1^2 \] and gravitational potential energy is \[ PE_1 = -\frac{GMm}{r_1} \].
04

Express Energies at Second Position

At the second position, the kinetic energy is \[ KE_2 = \frac{1}{2}mv_2^2 \] and gravitational potential energy is \[ PE_2 = -\frac{GMm}{r_2} \].
05

Set Up the Conservation Equation

Equate the total mechanical energies at the two positions: \[ \frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GMm}{r_2} \].
06

Cancel the Mass

The mass of the comet \( m \) cancels out from the equation, simplifying it to:\[ \frac{1}{2}v_1^2 - \frac{GM}{r_1} = \frac{1}{2}v_2^2 - \frac{GM}{r_2} \].
07

Solve for the Unknown Speed

Rearrange the equation to solve for \( v_2^2 \): \[ v_2^2 = v_1^2 + 2GM\left(\frac{1}{r_2} - \frac{1}{r_1}\right) \].Substitute \( v_1 = 2.0 \times 10^4 \) m/s, \( r_1 = 2.5 \times 10^{11} \) m, \( r_2 = 5.0 \times 10^{10} \) m, and \( GM = 1.327 \times 10^{20} \) \( \text{m}^3/\text{s}^2 \).
08

Calculate \( v_2 \)

Input the values into the equation:\[ v_2^2 = (2.0 \times 10^4)^2 + 2\times 1.327 \times 10^{20}\left(\frac{1}{5.0 \times 10^{10}} - \frac{1}{2.5 \times 10^{11}}\right) \].Calculating gives \( v_2^2 \approx 2.584 \times 10^9 \), so \( v_2 \approx 50843 \) m/s.

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.

Orbital Mechanics
Orbital mechanics is the study of the motions of celestial bodies under the influence of gravitational forces. This field explains how objects like planets, comets, and satellites move in space. One key aspect is the shape and nature of orbits. Most celestial bodies travel in elliptical orbits, which are oval-shaped paths, with varying speeds at different points. Understanding these paths and the forces involved allows scientists to predict the future positions and velocities of these objects.

When analyzing the motion of comets around the Sun, we consider celestial mechanics principles, which tell us that motion is governed by gravity. This makes things like speed and distance from other mass points (like the Sun for a comet) crucial in understanding their orbits. Kepler's laws of planetary motion provide a framework for understanding these movements. However, to calculate specific parameters, like the speed of a comet at different points in its orbit, we utilize the law of conservation of energy. This ensures energy is neither lost nor gained, only transformed between potential and kinetic forms.
Gravitational Potential Energy
Gravitational potential energy (PE) is energy an object possesses due to its position in a gravitational field. This energy is important in celestial mechanics because it changes as the object moves closer or farther from another massive object like the Sun. The equation for gravitational potential energy is:
  • PE = -\( \frac{GMm}{r} \), where:
  • \( G \) is the gravitational constant (approximately \( 6.674 \times 10^{-11} \) N(m/kg)² ),
  • \( M \) is the mass of the large body (e.g., the Sun),
  • \( m \) is the mass of the smaller body (e.g., the comet),
  • \( r \) is the distance between the centers of the two bodies.

The negative sign indicates that this energy is lower when the two masses are farther apart. As a comet moves in its orbit, gravitational potential energy will change. Closer to the Sun, it will have lower potential energy, which in turn affects its speed. By calculating this potential energy at different points in an orbit, we can make predictions about changes in speed necessary to maintain orbital motion.
Kinetic Energy
Kinetic energy (KE) is the energy associated with the motion of an object and is given by the equation:
  • KE = \( \frac{1}{2}mv^2 \), where:
  • \( m \) is the mass of the object,
  • \( v \) is the velocity of the object.

Kinetic energy is vital in the study of orbital mechanics as it fluctuates with changes in velocity. When comets or other celestial bodies travel through their elliptical orbits, they speed up or slow down depending on their distance to the Sun, thus affecting their kinetic energy.

As per the conservation of energy principle, the total energy in a closed system remains constant. For a comet orbiting the Sun, this means that if its potential energy decreases (as it gets closer to the Sun), its kinetic energy must increase to keep the total energy constant, and vice versa. This interplay between kinetic and potential energy provides insights into the velocity of celestial objects at various points along their orbits and is key in answering questions about their dynamic behavior in space.

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

Some scientists are eager to send a remote-controlled submarine to Jupiter's moon Europa to search for life in its oceans below an icy crust. Europa's mass has been measured to be 4.80 \(\times\) 10\(^{22}\) kg, its diameter is 3120 km, and it has no appreciable atmosphere. Assume that the layer of ice at the surface is not thick enough to exert substantial force on the water. If the windows of the submarine you are designing each have an area of 625 cm\(^2\) and can stand a maximum inward force of 8750 N per window, what is the greatest depth to which this submarine can safely dive?

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surface, the ship's orbital speed is 4900 m/s. By observing the planet, you determine its radius to be 4.48 \(\times\) 10\(^6\) m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 12.6 m/s at an angle of 30.8\(^\circ\) above the horizontal. If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile?

On July 15, 2004, NASA launched the \(Aura\) spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface. Assume a circular orbit. (a) How many hours does it take this satellite to make one orbit? (b) How fast (in km/s) is the \(Aura\) spacecraft moving?

A uniform sphere with mass 50.0 kg is held with its center at the origin, and a second uniform sphere with mass 80.0 kg is held with its center at the point \(x =\) 0, \(y =\) 3.00 m. (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 kg placed at the point \(x =\) 4.00 m, \(y =\) 0? (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at 30,000 km/s. (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?

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.