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 Here R is the radius of the earth, M is the mass of the earth, G is the gravitational constant, and r is the distance to the center of the earth.

Sketch a graph of g against r, and use it to answer the following questions:

A. Is g a continuous function of r? why?


B. Is g a differentiable function of r? why?

Answer 1

A. can you make the denominator = 0? if so, g(r) is not continuous. otherwise, it is

B. is g(r) continuous over a certain interval? if so, it's differentiable over that same interval.

i do not remember the equation, but i think it is highly likely that you're dealing with \[\left\{ r | r \in \mathbb{R}, r >0\right\}\] that means that g(r) is continous over (0, infinity) and that it is differentiable over the same interval.

Answer 2

This is the equation for the force, F(r)
\[ F(r) = -GMm/r^2 \text{ , for } r \geq R_{earth} \]
\[ F(r) = - \frac{r}{R_{earth}} GMm/R_{earth}^2 \text{ , for } r \leq R_{earth} \]
Note that when r = R_earth, the two equations match and hence the function is continuous at that point. However, it is not differentiable there.

It is easy to see that for r > R_earth or for r < R_earth the function is differentiable.

Answer 3

\[\iint\limits_S\vec{g}\cdot d\vec{S}=-4\pi GM_S\]\[1.\quad r\leq R\quad,\quad M_S=M\frac{r^3}{R^3}\]\[-g(r)4\pi r^2=-4\pi GM\frac{r^3}{R^3}\]\[g(r)=GM\frac{r}{R^3}\]
\[2.\quad r\geq R\quad,\quad M_S=M\]\[-g(r)4\pi r^2=-4\pi GM\]\[g(r)=GM\frac{1}{r^2}\]
\[A.\quad g(R-)=GM\frac{1}{R^2}=g(R+)\]\[B.\quad g'(R-)=GM\frac{1}{R^3}\quad,\quad g'(R+)=-2GM\frac{1}{R^3}\quad,\quad g'(R-)\neq g'(R+)\]

Answer 4

(What I took particular exception to in what alexmiles said is that a continuous function is differentiable. That is not true, however the opposite is.)