Question

The acceleration due to gravity, g, varies with height above the surface of the earth, in a certain way. If you go down below the surface of the earth, g varies in a different way. It can be shown that g is given by GMr/R^3 for r=R where R is the radius of the earth, M is the mass of earth, G is the gravitational constant, and r is the distance to the center of the earth. a) what would the graph g of r look like? b) is g a continuous function of r? explain. c) is g a differentiable function of r? explain.

Answer 1

From where have you found g function, below the Earth surface, as GMr/R^3?

Answer 2

y = GMx/R^3 if you're on or in the earth

y = GM/x^2 if your further than the surface of the earths radius

pretty straightforward, so:
G = 6.674×10^−11 N m2 kg (Newtons gravitational constant)
M = 5.9736 x 10^24 kg (mass of earth)
R = 6,371, 000 m (average radius of earths surface)
x= r = variable (radius of object to centre of earth)
y = a = gravity

so if x > R
\[y = \frac{ (6.674×10^{-11})\times(5.9736 \times 10^24)\times x }{ (6371000)^3 }\]
\[y = 1.5417\times10^{-6} \times x\]

but if x < R
\[y = \frac{ 6.674×10^{−11}\times 5.9736 x 10^{24} }{ x^2 }\]
\[y = \frac{ 3.98678\times 10^{14} }{ x^2 }\]

Answer 3

the above attached pics are probably the best i can do dude, buy it clearly shows that gravity rises up to a maximum of around 10 as you move from a position at the core to a position on the surface of the earth...
from there the value of gravity falls off as you move further away from the surface of the earth

Answer 4

@Jack1

Answer 5

interesting, but the op states a different scenario for r<R by using GMr/R^3... so I was only going off what was given, not what is real. Also, earth doesn't have a uniform density, it has a molten rotating iron/heavy element core....