Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 376: 70

Answer

See the detailed answer below.

Work Step by Step

a) We can see, from the figure below, that the $y$-component of the gravitational potential between $dm$'s and $m$ are equal and are in opposite directions so they cancel each other. The same thing happens for the unseen axis $z$-axis (perpendicular to the page), the $z$-components are also having the same magnitude and opposite directions. This means that the net gravitational potential energy between the ring $M$ and the mass $m$ is in the positive $x$-direction. Since the mass is uniformly distributed along the ring, its density is given by $$\rho=\dfrac{M}{V}=\dfrac{M}{AL}$$ where $L$ here is the circumference of the ring, so $L=2\pi R$ $$\rho =\dfrac{M}{2\pi R A}$$ Now the small piece of mass $dm$ has a length of $dL$ and its density is given by $$\rho=\dfrac{dm}{dV}=\dfrac{dm}{AdL}$$ From the last two formulas, $$\dfrac{dm}{ \color{red}{\bf\not}AdL}=\dfrac{M}{2\pi R \color{red}{\bf\not}A}$$ $$dm=\dfrac{MdL}{2\pi R}\tag 1$$ So the gravitational potential between the ring and the mass is given by $$dU=\dfrac{-Gm dm}{r}$$ where $r$ is the hypotenuse and is given by $r=\sqrt{x^2+R^2}$ $$dU=\dfrac{-Gm dm}{\sqrt{x^2+R^2}}$$ Plugging from (1); $$dU=\dfrac{-Gm }{\sqrt{x^2+R^2}}\dfrac{MdL}{2\pi R}$$ Taking the integral for both sides; $$\int_0^UdU=\int\dfrac{-Gm }{\sqrt{x^2+R^2 } }\dfrac{MdL}{2\pi R}$$ Noting that $R$ and $x$ are constants. $$ U=\dfrac{-Gm }{\sqrt{x^2+R^2 } }\dfrac{M}{2\pi R}\int\limits_{\rm ring}dL$$ $$ U=\dfrac{-Gm }{\sqrt{x^2+R^2 } }\dfrac{M}{2\pi R}L\bigg|_0^L$$ where $L$ is the circumference of the ring and is given by $2\pi R$ $$ U=\dfrac{-Gm }{\sqrt{x^2+R^2 } }\dfrac{M}{2\pi R} 2\pi R$$ Thus, $$ \boxed{U=\dfrac{-Gm M}{\sqrt{x^2+R^2 } }} $$ --- b) We know that the force is given by $$F=\dfrac{-dU}{dx}$$ Plugging from the boxed formula above; $$F=\dfrac{-d}{dx}\left(\dfrac{-Gm M}{\sqrt{x^2+R^2 } }\right)$$ $$F=GMm\cdot \dfrac{d}{dx} (x^2+R^2 )^{-\frac{1}{2}} $$ $$F=GMm\cdot \left(\dfrac{-1}{2} (2x)(x^2+R^2 )^{-\frac{3}{2}} \right)$$ $$ \boxed{F=\dfrac{Gm Mx}{(x^2+R^2 )^{\frac{3}{2}} }} $$
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