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Physics / Chapter 13: Gravitation / Section 13-5

Gravitation Potential Energy — Question 9

Section 13-5 · Physics pages 1988–1989

Question 9 A ring potential provides an independent check on its axial field A thin uniform ring has mass M and radius R. Every element is distance √ R2 + z2 from a point on its axis. Take potential zero at infinity. z R Ring-axis geometry; the axial destination is one ring radius from the center. (a) Find the axial potential and recover the field by differentiation. (b) Find the work to move a test mass from z = 0 to z = R quasistatically.
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Question 9 - Solution A ring potential provides an independent check on its axial field z R (a) Derivation and calculation Because every ring element is equally distant from the axial point, Φ(z) = − ∫ G, dm √ R2 + z2 = − GM √ R2 + z2 . Then gz = −dΦ dz = −GMz(R2 + z2)−3/2 = − GMz (R2 + z2)3/2 . (b) Result and interpretation Quasistatic work equals potential-energy increase: W = m[Φ(R) −Φ(0)] = m [ −GM √ 2R + GM R ] . W = GMm R ( 1 −1 √ 2 ) ≃.292893GMm R > 0. A vanishing initial field does not eliminate the work along the rest of the path. Physical check The potential is even in z, the axial field is odd, and both recover point-source behavior far away.

Question 9

A ring potential provides an independent check on its axial field

A thin uniform ring has mass MM and radius RR. Every element is distance R2+z2\sqrt{R^2+z^2} from a point on its axis. Take potential zero at infinity.

Diagram shown in the original SVG worksheet.

  1. Find the axial potential and recover the field by differentiation.

  2. Find the work to move a test mass from z=0z=0 to z=Rz=R quasistatically.

Question 9 - Solution

A ring potential provides an independent check on its axial field

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Because every ring element is equally distant from the axial point, Φ(z)=−∫G,dmR2+z2=−GMR2+z2.\Phi(z)=-\int\frac{G,dm}{\sqrt{R^2+z^2}} =\boxed{-\frac{GM}{\sqrt{R^2+z^2}}}. Then gz=−dΦdz=−GMz(R2+z2)−3/2=−GMz(R2+z2)3/2.g_z=-\frac{d\Phi}{dz} =-GMz(R^2+z^2)^{-3/2} =\boxed{-\frac{GMz}{(R^2+z^2)^{3/2}}}.

(b) Result and interpretation

Quasistatic work equals potential-energy increase: W=m[Φ(R)−Φ(0)]=m[−GM2R+GMR].W=m[\Phi(R)-\Phi(0)] =m\left[-\frac{GM}{\sqrt2R}+\frac{GM}R\right]. W=GMmR(1−12)≃.292893GMmR>0.\boxed{W=\frac{GMm}R\left(1-\frac1{\sqrt2}\right) \simeq.292893\frac{GMm}R}>0. A vanishing initial field does not eliminate the work along the rest of the path.

Physical check

The potential is even in zz, the axial field is odd, and both recover point-source behavior far away.

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