Suppose the moon has a radius of R miles and a payload weighs P pounds at the surface of the moon (at a distance of R miles from the center of the moon). When the payload is x miles from the center of the moon (x ≥ R), the force required to overcome the gravitational attraction between the moon and the payload is given by the following relation: required force = f(x) = R2P x2 pounds For example, the amount of work done raising the payload from the surface of the moon (i.e., x = R) to an altitude of R miles above the surface of the moon (i.e., x = 2R) is work = b f(x) dx a = 2R R2P x2 dx R = RP 2 mile-pounds How much work would be needed to raise the payload from the surface of the moon (i.e., x = R) to the "end of the universe"? work = mile-pounds

[tex]work=\int\limits^{x=\infty}_{x=R} {f(x)} \, dx \\\\=\int\limits^\infty_R {\frac{R^2P}{x^2}} \, dx \\\\=R^2P\int\limits^\infty_R {\frac{1}{x^2}} \, dx \\\\=R^2P[{-\frac{1}{x^2}}]\limits^\infty_R\\\\=R^2P[-0+\frac{1}{R}]\\\\=R^2P[\frac{1}{R}]\\\\=PR[/tex]

where R= radius of moon; P= weight of payload are constant

MrRoyal MrRoyal · Answer 2 · 2021-12-01T20:02:09+01:00

The radius of the moon is simply the distance from the center of the moon to its circumference.

The amount of work needed to raise the payload from the surface of the moon is -RP

The integral that represents the amount of work done is represented as:

[tex]\mathbf{\int\limits^{\infty}_R f(x)}[/tex]

Where

[tex]\mathbf{f(x) = \frac{R^2P}{x^2}}[/tex]

So, the integral function becomes

[tex]\mathbf{\int\limits^{\infty}_R f(x) = \int\limits^{\infty}_R\frac{R^2P}{x^2}}[/tex]

Rewrite the integral function as:

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \int\limits^{\infty}_R\frac{1}{x^2}}[/tex]

Further, rewrite as:

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \int\limits^{\infty}_R x^{-2}}[/tex]

Integrate

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \times (-x^{-1})|\limits^{\infty}_R}[/tex]

Expand

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \times (-R^{-1} - -\infty^{-1})}[/tex]

Simplify the expression above

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \times (-R^{-1} +0)}[/tex]

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \times (-R^{-1})}[/tex]

Remove bracket

[tex]\mathbf{\int\limits^{\infty}_R f(x) = R^2P \times -R^{-1}}[/tex]

Evaluate the product

[tex]\mathbf{\int\limits^{\infty}_R f(x) = -RP}[/tex]

Hence, the amount of work needed to raise the payload is -RP