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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.

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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.

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Most popular questions from this chapter

Four identical masses of 8.00 kg each are placed at the corners of a square whose side length is 2.00 m. What is the net gravitational force (magnitude and direction) on one of the masses, due to the other three?

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Two uniform spheres, each with mass \(M\) and radius \(R\), touch each other. What is the magnitude of their gravitational force of attraction?

At a certain instant, the earth, the moon, and a stationary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 \(\times\) 10\(^5\) km in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? Ignore any gravitational effects due to the other planets or the sun.

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