Question

Secondary SchoolPhysics 5 points If g = GM/r2 then where will the value of g be high at Goa Beach or on top of Mount Everest?

Answer 1

answer : acceleration due to gravity at Goa beach is same as acceleration due to gravity on earth's surface. depth of Goa beach not so much high comparison to 6400km { earth's radius.}
so, $g=g_0\left(1-\frac{d}{R}\right)$
d/R <<< 1
so, 1 - d/R ≈ 1

hence, $g=g_0$ = 9.8 m/s²

but acceleration due to gravity on the top of Mount Everest will be lower than acceleration due to gravity on earth's surface.
as you know, height of mount Everest = 8852m
so, g = $\frac{g_0}{\left(1+\frac{h}{R}\right)^2}$
here, h = 8.852km , R = 6400km

so, g = 9.8/(1 + 8.852/6400)²

= 9.7729 m/s² < 9.8 m/s²

hence, it is clear that g will be high at Goa beach.

Answer 2

$\huge\textbf{Answer}$

g = $\dfrac{GM}{{r}^{2}}$

If value of G increases then value of r decreases.

Similarly if value of G decreases then value of r increases.

Here r is the distance (we can also said it that the distance between two objects)

Means value of G will be more (high) at Goa beach and less (small) at Mountain Everest.

Similarly value of r will be high at Mount Everest and less at Goa Beach.