Chapter 15.4, Problem 51E

a.

To determine

To find:the value of k for the LSM.

Answer to Problem 51E

  $k=4.096\times {10}^{16}N{m}^2$ .

Explanation of Solution

Given:

The weight of an object that is at a distance x from the center of earth is written as $k{x}^{-2}$ .

where k is the constant depends on the mass of the object.

The energy required to move the object from $x=a\text{to}x=b$ is the integral of its weight.

Radius of earth is given by $R=6.4\times {10}^6$ .

Calculation:

According to the question,

  $\begin{array}{l}P=k{x}^{-2}\\ \Rightarrow 1000=k{\left(6.4\times {10}^6\right)}^{-2}\\ \therefore k=\frac{1000}{{\left(6.4\times {10}^6\right)}^{-2}}=4.096\times {10}^{16}N{m}^2\end{array}$

Hence, the value of k for the LSMis $k=4.096\times {10}^{16}N{m}^2$ .

b.

To determine

To find: the energy required to lift the LSM from the earth’s surface to the moon.

Answer to Problem 51E

  $W=2.61\times {10}^{11}J$ .

Explanation of Solution

Given:

The weight of an object that is at a distance x from the center of earth is written as $k{x}^{-2}$ .

where k is the constant depends on the mass of the object.

The energy required to move the object from $x=a\text{to}x=b$ is the integral of its weight.

Calculation:

According to the question,

  $\begin{array}{l}W=4.096\times {10}^{16}{\displaystyle \underset{6.4\times {10}^6}{\overset{3.8\times {10}^8}{\int }}{x}^{-2}dx}\\ W=-4.096\times {10}^{16}{\left[x^{-1}\right]}_{6.4\times {10}^6}^{3.8\times {10}^8}\\ W=4.096\times {10}^{16}\left(3.8\times {10}^{-8}-6.4\times {10}^{-6}\right)\\ W=2.61\times {10}^{11}J\end{array}$

Hence, the energy required to lift the LSM from the earth’s surface to the moonis $W=2.61\times {10}^{11}J$ .