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Chapter 12: Problem 12

Can the expression for gravitational potential energy \(U_{\mathrm{g}}(y)=m g y\) be used to analyze high-altitude motion? Why or why not?

Short Answer

Expert verified
Explain your answer. Answer: No, the expression \(U_{\mathrm{g}}(y) = mgy\) is not suitable for analyzing high-altitude motion as it relies on the assumption of a constant gravitational field, which breaks down at high altitudes. Instead, the general potential energy formula, \(U(r)=-\frac{GMm}{r}\), should be used for high-altitude motion analysis, as it accounts for the varying gravitational field with distance from the Earth's center.

Step by step solution

01

Understand the expression for gravitational potential energy

The gravitational potential energy of an object is the energy that it possesses due to its position in the gravitational field. The formula for gravitational potential energy is given by \(U_{\mathrm{g}}(y) = mgy\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(y\) is the height of the object from the reference point, often Earth's surface.
02

Assumptions in the expression for gravitational potential energy

The formula \(U_{\mathrm{g}}(y) = mgy\) is derived under certain assumptions. The primary assumption is that the Earth's gravitational field is uniform, i.e., the value of \(g\) is constant throughout the region of interest. This assumption holds true for motion near the Earth's surface where the changes in the gravitational field are negligible. However, at very high altitudes, the Earth's gravitational field is no longer uniform as gravity weakens with distance from the Earth's center.
03

Considering high-altitude motion

High-altitude motion refers to motion at great distances from the Earth's surface, where the gravitational field is not uniform. The expression \(U_{\mathrm{g}}(y) = mgy\) is based on the assumption of a constant gravitational field. At high altitudes, it becomes crucial to account for the varying gravitational field by using the general potential energy formula, \(U(r)=-\frac{GMm}{r}\), where \(M\) is the Earth's mass, \(r\) is the distance from the object to the center of the Earth, and \(G\) is the gravitational constant.
04

Conclusion

Therefore, the expression \(U_{\mathrm{g}}(y) = mgy\) cannot be used to analyze high-altitude motion, as it relies on the assumption of a constant gravitational field. This assumption breaks down at high altitudes where changes in the gravitational field become significant. Instead, the general potential energy formula that considers the changes in the gravitational field should be used for high-altitude motion analysis.

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Most popular questions from this chapter

A plumb bob located at latitude \(55.0^{\circ} \mathrm{N}\) hangs motionlessly with respect to the ground beneath it. \(A\) straight line from the string supporting the bob does not go exactly through the Earth's center. Does this line intersect the Earth's axis of rotation south or north of the Earth's center?

The more powerful the gravitational force of a planet, the greater its escape speed, \(v,\) and the greater the gravitational acceleration, \(g\), at its surface. However, in Table 12.1 , the value for \(v\) is much greater for Uranus than for Earth - but \(g\) is smaller on Uranus than on Earth! How can this be?

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