Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 70: 49


$\lim\limits_{r \to 0^{+}}F(r)$ = +$\infty$

Work Step by Step

According to Newton’s Law of Universal Gravitation: F(r) = G$\frac{Mm}{r²}$, where r is the distance between the centers of the masses. G, M and m are constants. So as they get closer and closer together, $r\to0$: $\lim\limits_{r \to 0^{+}}F(r)$ = $\lim\limits_{r \to 0^{+}}G\frac{Mm}{r²}$ = +$\infty$. That means the force increases becoming very large. Of course, r is the distance between the centers of the masses, so the distance can't be 0. In a distance r $\gt$ 0 the masses will be touching each other.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.