Problems 10.13. Explain the steps necessary to arrive at Eq. (10.89). 10.14. Consider a uniform hemispherical...

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Problems 10.13. Explain the steps necessary to arrive at Eq. (10.89). 10.14. Consider a uniform hemispherical shell of radius R and m axis be its symmetry axis. Calculate the gravitational pote the Z-axis. Graph V(r) and g(r). How do these compare w 10.15. Consider a uniform solid hemisphere of radius R and mas axis be its symmetry axis. Calculate the gravitational poto e results JR which gives V(r) = - 27GP(R; – R}) = constant, %3D for r<R, (10.88) Thus the potential inside (r < R¡) the shell is constant and is independent of the position. The potential inside (R, <r<R) the shell is a little bit tricky to calculate. But an easy approach to this problem is to change the lower limit by replacing R, by rin Eq. (10.87) for V(r) for r<R,, and to change the upper limit by replacing R2 byr in Eq. (10.83) for V(r) for r> Rz Thus, combining the two gives the potential inside the shell: 4 TGp V(R, <r< R,) = -2mGp(R} - r²) - 3r That is, V(R, <r< R) = -4 GP for R, <r<R, (10.89) 3r 61 The field intensity vector g can be calculated from the relation g = -dVldr for each of the three regions by using Eqs. (10.86), (10.89), and (10.88). That is, GM ,r> (10.90) g(r) = %3D da@e ( -) R, <r< R, <r<R3 4 Gp (10.91) g(r) 3 (10.92) g(r) = 0, r< R, %3D The plots of V(r) using Eqs. (10.86), (10.88), and (10.89) and of g(r) using Eqs. (10.90), (10.91), and (10 92) are shown in Fig. 10.13. Let us make some important observations. The potential Problems 10.13. Explain the steps necessary to arrive at Eq. (10.89). 10.14. Consider a uniform hemispherical shell of radius R and m axis be its symmetry axis. Calculate the gravitational pote the Z-axis. Graph V(r) and g(r). How do these compare w 10.15. Consider a uniform solid hemisphere of radius R and mas axis be its symmetry axis. Calculate the gravitational poto e results JR which gives V(r) = - 27GP(R; – R}) = constant, %3D for r<R, (10.88) Thus the potential inside (r < R¡) the shell is constant and is independent of the position. The potential inside (R, <r<R) the shell is a little bit tricky to calculate. But an easy approach to this problem is to change the lower limit by replacing R, by rin Eq. (10.87) for V(r) for r<R,, and to change the upper limit by replacing R2 byr in Eq. (10.83) for V(r) for r> Rz Thus, combining the two gives the potential inside the shell: 4 TGp V(R, <r< R,) = -2mGp(R} - r²) - 3r That is, V(R, <r< R) = -4 GP for R, <r<R, (10.89) 3r 61 The field intensity vector g can be calculated from the relation g = -dVldr for each of the three regions by using Eqs. (10.86), (10.89), and (10.88). That is, GM ,r> (10.90) g(r) = %3D da@e ( -) R, <r< R, <r<R3 4 Gp (10.91) g(r) 3 (10.92) g(r) = 0, r< R, %3D The plots of V(r) using Eqs. (10.86), (10.88), and (10.89) and of g(r) using Eqs. (10.90), (10.91), and (10 92) are shown in Fig. 10.13. Let us make some important observations. The potential