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Newton's Law of Gravitation can be used to show that if an object weighs w pounds on the surface of the earth, then its weight at distance x from the center of the earth is W(x) = wR2 x2 (for x ≥ R) where R = 3,960 miles is the radius of the earth. Estimate the weight change, at altitude h, per mile of altitude gained for a 120-lb pilot.

Respuesta :

Answer:

W'(x) = [tex]\frac{-240R^2}{h^3}[/tex]

Here, the negative sign depicts the loss in weight

Step-by-step explanation:

Data provided in the question :

Function:

W(x) = [tex]\frac{wR^2}{x^2}[/tex]

Here,

R = 3,960 miles is the radius of the earth

w = Weight of the pilot = 120 lb

x is the distance from the center of the earth

Therefore,

Rate of change of weight

W'(x) = [tex]\frac{d(\frac{wR^2}{x^2})}{dx}[/tex]

or

W'(x) = [tex]\frac{(-2)wR^2}{x^3}[/tex]

on substituting the respective values, we get

W'(x) = [tex]\frac{(-2)(120)R^2}{(h)^3}[/tex]

or

W'(x) = [tex]\frac{-240R^2}{h^3}[/tex]

Here, the negative sign depicts the loss in weight

MrRoyal

Equations can be used to model real life situations.

The weight change at altitude h is: [tex]\mathbf{ -3763584000h^{-3}}[/tex]

The equation is given as:

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

Differentiate with respect to x

[tex]\mathbf{W'(x) = -\frac{2wR^2}{x^3}}[/tex]

From the question, we have:

[tex]\mathbf{R=3960\ miles}[/tex]

[tex]\mathbf{w =120-lb}[/tex]

[tex]\mathbf{x = h}[/tex]

So, we have:

[tex]\mathbf{W'(x) = -\frac{2 \times 120 \times 3960^2}{x^3}}[/tex]

[tex]\mathbf{W'(x) = -\frac{3763584000}{x^3}}[/tex]

Substitute h for x

[tex]\mathbf{W'(h) = -\frac{3763584000}{h^3}}[/tex]

Rewrite as:

[tex]\mathbf{W'(h) = -3763584000h^{-3}}[/tex]

Hence, the weight change at altitude h is: [tex]\mathbf{ -3763584000h^{-3}}[/tex]

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