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The acceleration due to gravity, g, is given by g=GMr2, where M is the mass of the Earth, r is the distance from the center of the Earth, and G is the uniform gravitational constant.
(a) Suppose that we increase from our distance from the center of the Earth by a distance Δr=x. Use a linear approximation to find an approximation to the resulting change in g, as a fraction of the original acceleration: Δg≈ g× (Your answer will be a function of x and r.)
(b) Is this change positive or negative? Δg is (Think about what this tells you about the acceleration due to gravity.)
(c) What is the percentage change in g when moving from sea level to the top of Mount Elbert (a mountain over 14,000 feet tall in Colorado; in km, its height is 4.29 km; assume the radius of the Earth is 6400 km)? percent change =1

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dexterevi

Answer:

(a) [tex]g'  = g{(1 -\frac{2x}{r})[/tex]

(b) The change in acceleration is negative.

(c) 99.87%

Step-by-step explanation:

(a) Given that the acceleration due to gravity, g, is given by

          [tex]g = \frac{GM}{r^{2} } ----------------------------- Equation 1[/tex]

where M is the mass of the Earth,

           r is the distance from the center of the Earth, and

           G is the uniform gravitational constant.

If we increase from our distance from the center of the Earth by a distance Δr=x. Equation 1 becomes;

       [tex]g' = \frac{GM}{(r + x)^{2}} ---------------- Equation 2[/tex]

where g' is the acceleration due to gravity at the new height (r + Δr) or (r + x)

Dividing equation 2 by 1

       [tex]\frac{g'}{g}  = \frac{GM}{(r + x)^{2}} * \frac{r^{2} }{GM}[/tex]

       [tex]\frac{g'}{g}  = \frac{r^{2} }{(r + x)^{2}}[/tex]

       Dividing through by [tex]r^{2}[/tex]

       [tex]\frac{g'}{g}  = \frac{1}{(1 + \frac{x}{r}) ^{2} }[/tex]

       [tex]\frac{g'}{g}  = {(1 + \frac{x}{r}) ^{-2} }[/tex]

Using Binomial approximation (Linear approximation),

        [tex]\frac{g'}{g}  = {(1 -\frac{2x}{r})[/tex]

        [tex]g'  = g{(1 -\frac{2x}{r})[/tex]

       

(b) The change in acceleration is negative. This implies that the higher we move from the earth surface, the smaller the acceleration due to gravity becomes.

(c)  x = 4.29 km

     r = 6400 km

The percentage change in g is given as:

       [tex]\frac{g'}{g} * 100  = {(1 -\frac{2x}{r}) * 100[/tex]

       [tex]= {(1 -\frac{2 * 4.29}{6400}) * 100[/tex]

       [tex]= {(1 - 0.001340625) * 100[/tex]

       [tex]= {(0.9987) * 100[/tex]

       = 99.87%

       

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