Question

you have only one attempt for this question. the gravitational force exerted by earth on a unit mass (i.e., a mass of 1 kilogram) at a distance r (in meters) from the center of the planet is $f(r)=\begin{cases}\frac{gmr}{r^{3}}& \text{if }0 < r < r\\frac{gm}{r^{2}}& \text{if }rgeq rend{cases}$ here $g$ is the so - called \gravitational constant\, $r$ is the radius of the earth (in meters), and $m$ is the mass of the earth (in kilograms). is $f$ continuous at $r = r$? answer: select answer

Explanation:

Step1: Recall the definition of continuity

A function $y = F(r)$ is continuous at $r = a$ if $\lim_{r
ightarrow a^{-}}F(r)=\lim_{r
ightarrow a^{+}}F(r)=F(a)$. Here $a = R$.

Step2: Calculate the left - hand limit

For $r
ightarrow R^{-}$, $F(r)=\frac{GMr}{R^{3}}$. Then $\lim_{r
ightarrow R^{-}}F(r)=\lim_{r
ightarrow R^{-}}\frac{GMr}{R^{3}}=\frac{GM\cdot R}{R^{3}}=\frac{GM}{R^{2}}$.

Step3: Calculate the right - hand limit

For $r
ightarrow R^{+}$, $F(r)=\frac{GM}{r^{2}}$. Then $\lim_{r
ightarrow R^{+}}F(r)=\lim_{r
ightarrow R^{+}}\frac{GM}{r^{2}}=\frac{GM}{R^{2}}$.

Step4: Evaluate the function at $r = R$

When $r = R$, using the second - part of the piece - wise function (since $r = R$ satisfies $r\geq R$), $F(R)=\frac{GM}{R^{2}}$.

Answer:

Yes