The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is F(r) = GMr R3 if r < R GM r2 if r ≥ R where M is the mass of Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r? Why/why not?

Respuesta :

Answer:

Yes the function is continuous.

Step-by-step explanation:

Given

[tex]F(r)=\frac{GMr}{R^{3}}(r< R)[/tex] and [tex]F(r)=\frac{GM}{r^{2}}(r\geq R)[/tex]

For a function to be continuous we should have

[tex]\lim_{r\rightarrow R^{-}}F(r)=\lim_{r\rightarrow R^{+}}F(r)[/tex]

Thus

[tex]\lim_{r\rightarrow R^{-}}F(r)=\frac{GMR}{R^{3}}=\frac{GM}{R^{2}}[/tex]

Similarly we have

[tex]\lim_{r\rightarrow R^{+}}F(r)=\frac{GM}{R^{2}}[/tex]

thus we can see that

[tex]\lim_{r\rightarrow R^{-}}F(r)=\lim_{r\rightarrow R^{+}}F(r)[/tex]

Hence the function is continuous.

By evaluating both pieces of the function in r = R, we will see that the function is continue.

Is F a continuous function of r?

The force F will only be continuous if in the limit r = R (where the piecewise function changes, let's say) we have the same value in both pieces of the function.

In this case, we have:

F(r) = (G*M r/R^3)    if r < R

F(r) = G*M/r^2         if r > R

To evaluate the continuity, we need to evaluate both of these in r = R and see if we get the same thing.

For the first one we have:

F(R) = G*M*R/R^3 = G*M/R^2

For the second one we have:

F(R) = G*M/R^2

So yes, we got the same thing in both cases, meaning that F is a continuous function of r.

if you want to learn more about continuity, you can read

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