Question

The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is

\(F\left( r \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{GMr}}{{{R^{\bf{3}}}}}}&{{\bf{if}}\,\,\,r < R}\\{\frac{{GM}}{{{r^{\bf{2}}}}}}&{{\bf{if}}\,\,\,r \ge R}\end{array}} \right.\)

Where M is the mass of Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r?

Short answer

F is a continuous function of r.

Step-by-step solution

Check the left-hand limit of \(F\left( r \right)\) at \(r = R\)

Check the continuity at \(r = R\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to {R^ - }} F\left( r \right) &= \mathop {\lim }\limits_{x \to {R^ - }} \frac{{GMr}}{{{R^3}}}\\ &= \frac{{GM\left( R \right)}}{{{R^3}}}\\ &= \frac{{GM}}{{{R^2}}}\end{aligned}\)

Check the right-hand limit of \(F\left( r \right)\) at \(r = R\)

Check the continuity at \(r = R\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to {R^ - }} F\left( r \right) &= \mathop {\lim }\limits_{x \to {R^ - }} \frac{{GMr}}{{{R^3}}}\\ &= \frac{{GM\left( R \right)}}{{{R^3}}}\\& = \frac{{GM}}{{{R^2}}}\end{aligned}\)

As the left-hand limit and right-hand limitare equal at \(r = R\), then the function is continuous in \(r = R\).

Thus, \(\mathop {\lim }\limits_{x \to R} F\left( r \right) = \frac{{GM}}{{{R^2}}}\).