Question

(a) Suppose that \({\bf{F}}\) is an inverse square force field, that is,

\({\bf{F}}({\bf{r}}) = \frac{{c{\bf{r}}}}{{|{\bf{r}}{|^3}}}\)

for some constant \(c\), where \({\bf{r}} = x{\bf{i}} + y{\bf{j}} + z{\bf{k}}\). Find the work done by \({\bf{F}}\) in moving an object from a point \({P_1}\) along a path to a point \({P_2}\) in terms of the distances \({d_1}\) and \({d_2}\) from these points to the origin.

(b) An example of an inverse square field is the gravitational field \({\bf{F}} = - (mMG){\bf{r}}/|{\bf{r}}{|^3}\) discussed in Example \(4\) in Section \({\rm{13}}{\rm{.1}}\). Use part (a) to find the work done by the gravitational field when the earth moves from aphelion (at a maximum distance of \(1.52 \times {10^8}\;{\rm{km}}\) from the sun) to perihelion (at a minimum distance of \(1.47 \times {10^8}\;{\rm{km}}\) ). (Use the values \(m = 5.97 \times {10^{24}}\;{\rm{kg}},M = 1.99 \times {10^{30}}\;{\rm{kg}}\), and \(G = 6.67 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}\).)

(c) Another example of an inverse square field is the electric force field \({\bf{F}} = \varepsilon qQ{\bf{r}}/|{\bf{r}}{|^3}\) discussed in Example 5 in Section \({\rm{13}}{\rm{.1}}\). Suppose that an electron with a charge of \( - 1.6 \times {10^{ - 19}}{\rm{C}}\) is located at the origin. A positive unit charge is positioned a distance \({10^{ - 12}}\;{\rm{m}}\) from the electron and moves to a position half that distance from the electron. Use part (a) to find the work done by the electric force field. (Use the value \(\left. {\varepsilon = 8.985 \times {{10}^9}.} \right)\)

Short answer

a) The work done is\(c\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\).

b) The work done by the gravitational field is\(1.77 \times {10^{32}}\;{\rm{J}}\).

c) The work done by the electric force field is\(1.423 \times {10^4}\;{\rm{J}}\).

Step-by-step solution

Given data

The inverse square force field is\({\rm{F}}({\rm{r}}) = \frac{{c{\rm{r}}}}{{|{\rm{r}}{|^3}}}\), where\({\rm{r}} = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\).

The inverse square force field is\({\rm{F}}({\rm{r}}) = - mMG\frac{{\rm{r}}}{{|{\rm{r}}{|^3}}}\)or\({\rm{F}}({\rm{r}}) = - mMG\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\), where\(c = - mMG\).

\(\begin{array}{l}m = 5.97 \times {10^{24}}\;{\rm{kg}}\\M = 1.99 \times {10^{30}}\;{\rm{kg}}\\G = 6.67 \times {10^{ - 11}}{\rm{N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}n\\{d_1} = 1.52 \times {10^8}\;{\rm{km and }}{d_2} = 1.47 \times {10^8}\;{\rm{km}}\end{array}\)

The inverse square force field is\({\rm{E}} = EqQ\frac{{\rm{r}}}{{\mid {{\rm{r}}^3}}}\)or\({\rm{E}} = EqQ\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\), where\(c = EqQ.\)

\(\begin{array}{l}q = 1,Q = 1.6 \times {10^{ - 19}}{\rm{c}}\\E = 8985 \times {10^{10}}{\rm{N}}{{\rm{m}}^2}/{{\rm{c}}^2}\\{d_1} = {10^{ - 12}}\;{\rm{m and }}{d_2} = \frac{1}{2} \times {10^{ - 12}}\;{\rm{m}}\end{array}\)

Inverse square law

Inverse square law says that “the Intensity of the radiation is inversely proportional to the square of the distance”.

Find the work done by \({\bf{F}}\) in moving the object from a point \({P_1}\) to a point \({P_2}\)

a)

The inverse square force field is\({\rm{F}}({\rm{r}}) = \frac{{c{\rm{r}}}}{{|{\rm{r}}{|^3}}}\), where\({\rm{r}} = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\).

Substitute\({\rm{r}} = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\)in\({\rm{F}}({\rm{r}})\),

\({\rm{F}}(x,y,z) = \frac{{c(xi + yj + zk)}}{{{{(xi + yj + zk)}^{\frac{3}{2}}}}}\)

If there exists a function\(f\)such that\(\nabla f = {\rm{F}}\).

That is,

\(\begin{array}{c}{f_x} = \frac{{cx}}{{{{(xi + yj + zk)}^{\frac{3}{2}}}}}\\{f_x} = \frac{{cy}}{{{{(xi + yj + zk)}^{\frac{3}{2}}}}}\\{f_x} = \frac{{cz}}{{{{(xi + yj + zk)}^{\frac{3}{2}}}}}\end{array}\)

Integrate equation (1) with respect to\(x\),

\(f(x,y,z) = \frac{{ - c}}{{\sqrt {{x^2} + {y^2} + {z^2}} }} + g(y,z)\)

Here\(g(y,z)\)is constant.

Differentiate equation (4) partially with respect to\(y\),

\({f_x}(x,y,z) = \frac{{cy}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}} + {g_y}(y,z)\)

Now compare the equations (2) and (5),\({g_y}(y,z) = 0\)

Integrate on both sides,

\(g(y,z) = h(z)\)

Here\(h(z)\)is constant.

Differentiate equation (4) partially with respect to\(z\),

\({f_z}(x,y,z) = \frac{{cz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}} + {g_z}(y,z)\)

Integrate on both sides,

\(g(y,z) = k(y)\)

Here\(k(y)\)is constant.

Now compare the equations (6) and (8),

\(g(y,z) = 0\)

Hence\(f(x,y,z) = \frac{{ - c}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\).

Obtain the work done by\({\bf{F}}\)in moving the object from a point\({P_1}\left( {{x_1},{y_1},{z_1}} \right)\)to a point\({P_2}\left( {{x_2},{y_2},{z_2}} \right)\)as follows.

\(\begin{aligned}\int_c {\bf{F}} \cdot d{\bf{r}} & = \int_c \nabla f \cdot d{\bf{r}}\\ & = f\left( {{P_2} - {P_1}} \right)\\ & = \frac{{ - c}}{{\sqrt {{x_2}^2 + {y_2}^2 + z_2^2} }} + \frac{c}{{\sqrt {{x_1}^2 + {y_1}^2 + {z_1}^2} }}\\ & = c\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\end{aligned}\)

Thus, the work done is \(c\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\).

The work done by the gravitational field for the given condition

b)

The inverse square force field is\({\rm{F}}({\rm{r}}) = - mMG\frac{{\rm{r}}}{{|{\rm{r}}{|^3}}}\)or\({\rm{F}}({\rm{r}}) = - mMG\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\), where\(c = - mMG\).

\(\begin{array}{l}m = 5.97 \times {10^{24}}\;{\rm{kg}}\\M = 1.99 \times {10^{30}}\;{\rm{kg}}\\G = 6.67 \times {10^{ - 11}}{\rm{N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}n\\{d_1} = 1.52 \times {10^8}\;{\rm{km and }}{d_2} = 1.47 \times {10^8}\;{\rm{km}}\end{array}\)

Substitute the values in\({\rm{F}}({\rm{r}})\),

\(\begin{aligned}{\bf{F}}({\bf{r}}) & = - \left( {5.97 \times {{10}^{24}}} \right)\left( {1.99 \times {{10}^{30}}} \right)\left( {6.67 \times {{10}^{ - 11}}} \right){\rm{N}}{{\rm{m}}^2}\left( {\frac{1}{{1.52 \times {{10}^8}}} - \frac{1}{{1.47 \times {{10}^8}}}} \right){\rm{k}}{{\rm{m}}^{ - 1}}\\ & = - 7.924 \times {10^{44}} \times {10^{ - 11}}(0.6578 - 0.6802){\rm{Nm}}\\ & = 7 - 924 \times {10^{33}} \times 0.02247{\rm{Nm}}\\ & = 1.77 \times {10^{32}}\;{\rm{J}}\end{aligned}\)

Thus, the work done by the gravitational field is \(1.77 \times {10^{32}}\;{\rm{J}}\).

The work done by the electric force field for the given condition

c)

The inverse square force field is\({\rm{E}} = EqQ\frac{{\rm{r}}}{{\mid {{\rm{r}}^3}}}\)or\({\rm{E}} = EqQ\left( {\frac{1}{{{d_1}}} - \frac{1}{{{d_2}}}} \right)\), where\(c = EqQ.\)

\(\begin{array}{l}q = 1,Q = 1.6 \times {10^{ - 19}}{\rm{c}}\\E = 8985 \times {10^{10}}{\rm{N}}{{\rm{m}}^2}/{{\rm{c}}^2}\\{d_1} = {10^{ - 12}}\;{\rm{m and }}{d_2} = \frac{1}{2} \times {10^{ - 12}}\;{\rm{m}}\end{array}\)

Substitute the values in\({\rm{E}}\),

\(\begin{aligned}{\bf{E}} & = 1\left( {8.985 \times {{10}^{10}}} \right)\left( { - 1.6 \times {{10}^{ - 19}}} \right){\rm{N}}{{\rm{m}}^2}\left( { - 1 \times {{10}^{12}}} \right){\rm{m}}\\ & = - 1.423 \times {10^{ - 8}}\left( { - 1 \times {{10}^{12}}} \right){\rm{Nm}}\\ & = 1.423 \times {10^4}\;{\rm{J}}\end{aligned}\)

Thus, the work done by the electric force field is \(1.423 \times {10^4}\;{\rm{J}}\).