Answer to Question #104852 in Calculus for Kassim

Question

The acceleration due to gravity, g, is given by
g =
GM
r
2
,
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 cen￾ter of the Earth by a distance ∆r = x. Use a linear approximation
to find an approximation to the resulting change in g, as a frac￾tion of the original acceleration:
∆g ≈ g×
(Your answer will be a function of x and r.)
(b) Is this change positive or negative?
∆g is [?/positive/negative]
(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 4.35 km; assume the
radius of the Earth is 6400 km)?

Answer 1

$g =GM/r^2$

Let for $r'=r+ \Delta r=r+x$, acceleration due to gravity be $g'=GM/(r+x)^2=g(1+x/r)^{-2}$

Using binomial expansion, we get;

$(1+x/r)^{-2} =[1-2x/r+...] \approx [1-2x/r]$

(For small changes, x may be small in comparison to the original distance r, thus , $x/r<<1,$ hence higher powers of $x/r$ can be neglected in the expansion.)

a) $g'=g(1+x/r)^{-2}\approx g(1-2x/r)$

$\Delta g =g'-g =(-2x/r)g$

b) This change is dependent on the sign of x(direction of displacement), the change is negative if we increase the distance, and positive if we decrease the distance.

This tells us that, g is inversely proportional to distance from centre of the earth, it decreases as we move away from the earth's centre.

c) Given, $r=6400km,x=4.35km$

Thus, $\Delta g/g*100 =\% (\Delta g/g)$

$=(-2x/r)*100=-8.7/64=-0.136 \%$