(a) The potential that depends only on the distance r from P to the center of the sphere. Therefore, it suffices to compute V ( P ) for a point P = ( 0 , 0 , r ) on the z − a x i s (with r ≠ R ) using symmetry. (b) Proof of V ( 0 , 0 , r ) is equal to − m G 4 π ∫ 0 π ∫ 0 2 π sin ϕ d θ d ϕ R 2 + r 2 − 2 R r cos ϕ using spherical co-ordinates. (c) The proof of V ( 0 , 0 , r ) = − m G 2 R r ( | R + r | − | R − r | ) using the substitution u = R 2 + r 2 − 2 R r cos ϕ . (d) The verification of equation V ( P ) = − G m r and V ( P ) = − G m R for r > R and R > r respectively.

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Answer 1

Question

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

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Answer 2

Question

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

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Answer 3

Question

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

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Answer 4

Question

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

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Answer 5

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

Answer 6

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

Answer 7

Chapter 17.4, Problem 48E

To determine

(a) The potential that depends only on the distance $r$ from $P$ to the center of the sphere. Therefore, it suffices to compute $V (P)$ for a point $P = (0, 0, r)$ on the $z - a x i s$ (with $r \neq R$ ) using symmetry.

(b) Proof of $V (0, 0, r)$ is equal to $- \frac{m G}{4 π} \int_{0}^{π} \int_{0}^{2 π} \frac{\sin ϕ d θ d ϕ}{\sqrt{R^{2} + r^{2} - 2 R r \cos ϕ}}$ using spherical co-ordinates.

(c) The proof of $V (0, 0, r) = - \frac{m G}{2 R r} (| R + r | - | R - r |)$ using the substitution $u = R^{2} + r^{2} - 2 R r \cos ϕ$ .

(d) The verification of equation $V (P) = - \frac{G m}{r}$ and $V (P) = - \frac{G m}{R}$ for $r > R$ and $R > r$ respectively.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

Solution Summary: The author explains the potential that depends only on the distance r from P to the center of the sphere.

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