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%
