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Chapter 6: Problem 726

The Gravitational P.E. of a body of mass \(\mathrm{m}\) at the earth's surface is \(-\mathrm{mgRe}\). Its gravitational potential energy at a height \(\operatorname{Re}\) from the earth's surface will be \(=\ldots \ldots \ldots\) here (Re is the radius of the earth) (A) \(-2 \mathrm{mgRe}\) (B) \(2 \mathrm{mgRe}\) (C) \((1 / 2) \mathrm{mg} \mathrm{Re}\) (D) \(-(1 / 2) \mathrm{mg} \operatorname{Re}\)

Short Answer

Expert verified
The gravitational potential energy of the body at a height Re from the earth's surface is \(-\frac{1}{2}mgRe\). The correct answer is (D) \(-\frac{1}{2}mgRe\).

Step by step solution

01

Convert the given potential energy formula to GmM/r form

First, let's convert -mgRe to the standard P.E. formula: \(-mgRe = -\frac{GmM}{Re}\) Here, g is the acceleration due to gravity near the surface of the earth, which can be written as: \(g = \frac{GM}{Re^2}\) Combine these two equations to get: \(-mgRe = -\frac{GmM}{Re}\)
02

Find the potential energy at height Re

Now we find the potential energy at a height Re above the earth's surface. In this case, the distance from the center of the earth will be the sum of the radius of the earth and the height: \(r = Re + Re = 2Re\) The potential energy at this height is given by: \(P.E. = -\frac{GmM}{2Re}\)
03

Convert the potential energy to the given Re form

Now let's convert this potential energy formula back to the given Re form. By using the earlier mentioned relationship: \(g = \frac{GM}{Re^2}\) We get: \(P.E. = -\frac{mgRe^2}{2Re} = -\frac{1}{2}mgRe\)
04

Write the final answer

The gravitational potential energy of the body at a height Re from the earth's surface is: \(-\frac{1}{2}mgRe\) Comparing this with the given options, we find that the correct answer is: (D) \(-\frac{1}{2}mgRe\)

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Most popular questions from this chapter

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)

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