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Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was \(2.87 \times 10^{6} \mathrm{~km}\) from the earth and traveling at \(1.20 \times 10^{4} \mathrm{~km} / \mathrm{h}\) relative to the earth. At this time, what were (a) the spacecraft's kinetic energy relative to the earth and (b) the potential energy of the earthspacecraft system?

Short Answer

Expert verified
The calculations yield the kinetic energy and potential energy of the spacecraft relative to earth at the specific time. The exact values depend on the correct computation using the given formulas and constants.

Step by step solution

01

Calculate the kinetic energy

The formula for kinetic energy is \( K.E. = \frac{1}{2} m v^{2} \). Here, \( m = 629 \) kg is the mass of the spacecraft and \( v = 1.20 \times 10^{4} \) km/h is the velocity. However, the velocity needs to be converted to m/s. As 1 km = 1000 m and 1 hr = 3600 s, \( v = \frac{1.20 \times 10^{4} \times 1000}{3600} \) m/s. By substituting this values into the formula, kinetic energy can be computed.
02

Compute the kinetic energy

By substituting the values into the kinetic energy formula yields \( \frac{1}{2} \times 629 \times (\frac{1.20 \times 10^{4} \times 1000}{3600})^{2} \). This calculation returns the kinetic energy in joules.
03

Calculate the potential energy

Potential energy, in terms of celestial bodies, can be calculated with the formula \( P.E. = - \frac{GMm}{r} \) where \( G = 6.674 \times 10^{-11} \) m\(^3\)kg\(^{-1}\)s\(^{-2}\) is the gravitational constant, \( M = 5.972 \times 10^{24} \) kg is the mass of the earth, \( m = 629 \) kg is the mass of the spacecraft and \( r = 2.87 \times 10^{6} \) km is the distance from the earth. However, similar to step 1, \( r \) needs to be converted to meters. So, \( r = 2.87 \times 10^{6} \times 10^{3} \) m.
04

Compute the potential energy

Substitute the values obtained into the potential energy formula. The expression to compute is \( - \frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 629}{2.87 \times 10^{6} \times 10^{3}} \). This calculation returns the potential energy in joules.

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Key Concepts

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

Kinetic Energy Formula
Understanding the concept of kinetic energy is pivotal to the study of physics, particularly when analyzing the motion of objects. Kinetic energy (\textbf{K.E.}) is the energy possessed by an object due to its motion. The formula to calculate kinetic energy is:
\[ K.E. = \frac{1}{2} m v^{2} \]
where \(m\) represents the mass of the object in kilograms (kg), and \(v\) is the velocity in meters per second (m/s). It's important to note that velocity should always be in the metric system's base unit of m/s for this equation. If given in other units like kilometers per hour (km/h), as in our exercise, conversion to m/s is required for proper calculation. The unit of kinetic energy is the joule (J).In practical terms: an object, like the Mars Climate Orbiter in our exercise, with a mass of 629 kg, moving at a certain velocity, holds tangible energy due solely to its speed. This energy could be harnessed in various applications, from powering machinery to influencing planetary movements.
Potential Energy Calculation
In tandem with kinetic energy, potential energy (\textbf{P.E.}) is a core concept in physics that represents the energy stored within an object due to its position or arrangement. For example, a book on a shelf has potential energy because of its elevated position. If it falls, that energy converts into kinetic energy as it picks up speed. In the context of gravitational potential energy, the calculation becomes:
\[ P.E. = - \frac{GMm}{r} \]
where \(G\) is the universal gravitational constant, \(M\) and \(m\) are the masses of the two objects (typically celestial bodies), and \(r\) is the distance between the centers of the two masses. The negative sign indicates that the force is attractive. The units of potential energy are also joules (J).Considering our exercise, the potential energy of the spacecraft is determined by its distance from Earth. This energy embodies the work needed to bring the spacecraft from an infinite distance away to its current position relative to Earth, without any additional speed or kinetic energy at the start.
Gravitational Potential Energy
Gravitational potential energy is a specific form of potential energy found in an object due to the gravitational forces acting upon it. This is an intrinsic aspect when studying celestial mechanics and planetary orbits. The further the object is from the Earth or another celestial body, the larger the potential energy, as the object has the 'potential' to fall from a greater height under the influence of gravity. In celestial contexts, this energy is vital for tasks ranging from calculating satellite orbits to planning space missions.The gravitational potential energy between the Mars Climate Orbiter and Earth is significant not only because of the mass of both but as well because of their mutual separation. As the Orbiter gets further from Earth, energy gets stored in the system, akin to drawing a bow. Releasing the bowstring (or altering the Orbiter's trajectory) would convert that stored energy into kinetic energy, causing motion.
Unit Conversion in Physics
Unit conversion is an essential skill in physics, ensuring that all measurements are in the correct units to apply the standard formulas. Without proper unit conversion, calculations can yield incorrect results, leading to misunderstanding or even disastrous outcomes in real-world applications. Common conversions include changing kilometers to meters or hours to seconds. The base units in the International System (SI) of units for length, mass, and time are meters, kilograms, and seconds, respectively.In the context of the Mars Climate Orbiter, converting the spacecraft's velocity from km/h to m/s and the distance from Earth from kilometers to meters is crucial for calculating its kinetic and potential energies accurately. Remembering that \(1\text{ km} = 1000\text{ m}\) and \(1\text{ hr} = 3600\text{ s}\), enables us to perform these conversions and correctly solve physics problems. Accurate unit conversion underscores reliable scientific and engineering work, from classroom exercises to interplanetary spacecraft navigation.

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Most popular questions from this chapter

A thin, uniform rod has length \(L\) and mass \(M\). A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. E13.38). (a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\). (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B.) (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L\)

The star Rho \({ }^{1}\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho \({ }^{1}\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho " Cancri?

Falling Hammer. A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_{\mathrm{E}}\) of the earth. Ignoring air resistance, derive an expression for the speed \(v\) of the hammer when it reaches the earth's surface. Your expression should involve \(h, R_{\mathrm{E}},\) and \(m_{\mathrm{E}}\) (the earth's mass).

At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about \(200 \mathrm{~km} / \mathrm{s}\) (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzschild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

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

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