MATH SOLVE

5 months ago

Q:
# 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

Accepted Solution

A:

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 kmThe 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%