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Potential of a Uniform Spherical Shell As an example, let us find the potential function for a uniform spherical shell. Solution: we have Phi =-Gint (dM)/(s)=-Gint (2pi rho R^2sinTheta dTheta )/(s) From the relation between s and Othat we used in Equation 6.2.5, we find that the pre- ceding equation may be simplified to read Phi =-G(2pi rho R^2)/(rR)int _(r-R)^r+Rds=-(GM)/(r) where M is the mass of the shell . This is the same potential function as that of a single particle of mass M located at O. Hence , the gravitational field outside the shell is the same as if the entire mass were concentrated at the center. It is left as a problem to show that, with an appropriate change of the integral and its limits, the potential inside the shell is constant and , hence, that the field there is zero.

Question

Potential of a Uniform Spherical Shell As an example, let us find the potential function for a uniform spherical shell. Solution: we have Phi =-Gint (dM)/(s)=-Gint (2pi rho R^2sinTheta dTheta )/(s) From the relation between s and Othat we used in Equation 6.2.5, we find that the pre- ceding equation may be simplified to read Phi =-G(2pi rho R^2)/(rR)int _(r-R)^r+Rds=-(GM)/(r) where M is the mass of the shell . This is the same potential function as that of a single particle of mass M located at O. Hence , the gravitational field outside the shell is the same as if the entire mass were concentrated at the center. It is left as a problem to show that, with an appropriate change of the integral and its limits, the potential inside the shell is constant and , hence, that the field there is zero.

Potential of a Uniform Spherical Shell
As an example, let us find the potential function for a uniform spherical shell.
Solution:
we have
Phi =-Gint (dM)/(s)=-Gint (2pi rho R^2sinTheta dTheta )/(s)
From the relation between s and Othat we used in Equation 6.2.5, we find that the pre-
ceding equation may be simplified to read
Phi =-G(2pi rho R^2)/(rR)int _(r-R)^r+Rds=-(GM)/(r)
where M is the mass of the shell . This is the same potential function as that of a single
particle of mass M located at O. Hence , the gravitational field outside the shell is the
same as if the entire mass were concentrated at the center. It is left as a problem to show
that, with an appropriate change of the integral and its limits, the potential inside the
shell is constant and , hence, that the field there is zero.

Solution

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AaliyahVeteran · Tutor for 9 years

Answer

$\Phi = -\frac{GM}{r}$
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