Chapter 11, Problem 80P

(a)

To determine

To Calculate:The value of constant

Answer to Problem 80P

  $6.4\times {10}^8{\text{kg/m}}^{\text{2}}$

Explanation of Solution

Given data:

The density of the sphere, $\rho \left(r\right)=\frac{C}{r}$

Radius, $=5.0\text{m}$

Mass of the sphere, $M=1.0\times {10}^{11}\text{kg}$

Formula Used:

Mass = Density $\times$ Volume

Calculation:

The density of the sphere is

  $\rho \left(r\right)=\frac{C}{r}\mathrm{...........}\left(1\right)$

Here C is the constant and r is the distance.

The density of the sphere is varied by a distance so the differential element of the sphere is ,

  $\begin{array}{l}dm=\rho dV\\ =\rho \left(4\pi r^{2}dr\right)\end{array}$

Integrate within the limits 0 to R .

  $\begin{array}{l}M=\left(\frac{C}{r}\right){\displaystyle {\int }_0^R\left(4\pi r^{2}dr\right)}\\ =4\pi C{\displaystyle {\int }_0^{5.0\text{m}}rdr}\\ =4\pi C{\left[\frac{r^2}{2}\right]}_0^{5.0}\\ =\pi C{\left(50\text{m}\right)}^2\end{array}$

Therefore the constant C is,

  $C=\frac{M}{\pi \left(50{\text{m}}^{\text{2}}\right)}$

Substitute $1.0\times {10}^{11}\text{kg}$ for M and solve for C .

  $\begin{array}{l}C=\frac{\left(1.0\times {10}^{11}\text{kg}\right)}{\pi \left(50{\text{m}}^{\text{2}}\right)}\\ =6.4\times {10}^8{\text{kg/m}}^{\text{2}}\end{array}$

Conclusion:

The constant C is $6.4\times {10}^8{\text{kg/m}}^{\text{2}}$ .

(b)

To determine

The acceleration due to gravity for a distance $r>5.0\text{m}$ .

The gravitation field with in the region $r>5.0\text{m}$ .

Answer to Problem 80P

The acceleration due to gravity for a distance $r>5.0\text{m}$ is $\frac{6.7{\text{N×m}}^{\text{2}}{\text{/kg}}^{\text{2}}}{r^2}$ .

The gravitation field with in the region $r<5.0\text{m}$ is $0.27\text{ N/kg}$ .

Explanation of Solution

Given data:

The density of the sphere, $\rho \left(r\right)=\frac{C}{r}$

Radius, $=5.0\text{m}$

Mass of the sphere, $M=1.0\times {10}^{11}\text{kg}$

The constant C is $6.4\times {10}^8{\text{kg/m}}^{\text{2}}$ .

Formula used:

Gravitational field:

  $g=-\frac{GM}{r^2}\widehat{r}$

Here, G is the gravitational constant, M is the mass and r is the distance of the point from the center of the sphere.

Calculation:

The expression for the magnitude of gravitational field at a point outside $\left(r>5.0\text{m}\right)$ the sphere is,

  $g=\frac{GM}{r^2}$

Substitute $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^{\text{2}}\text{for}\text{G,}\text{1}\text{.0}\times {\text{10}}^{11}\text{kg}$ for M and solve for g .

  $\begin{array}{l}g=\frac{\left(6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^{\text{2}}\right)\left(1.0\times {10}^{11}\text{kg}\right)}{r^2}\\ =\frac{6.7{\text{N×m}}^{\text{2}}{\text{/kg}}^{\text{2}}}{r^2}\end{array}$

Therefore, the acceleration due to gravity for a distance $r>5.0\text{m}$ is $\frac{6.7{\text{N×m}}^{\text{2}}{\text{/kg}}^{\text{2}}}{r^2}$

The expression for the gravitational field at a point inside $\left(r<5.0\text{m}\right)$ the sphere is,

  $g=-\frac{GM}{r^2}\widehat{r}$

Since the density of the sphere is varying with the distance, so gravitational field is given by for

  $r<5.0\text{m}$

  $\begin{array}{l}g=G\frac{{\displaystyle {\int }_0^r\rho \left(4\pi r^2\right)dr}}{r^2}\\ =G\frac{{\displaystyle {\int }_0^r\left(\frac{C}{r}\right)\left(4\pi r^2\right)dr}}{r^2}\\ =G\frac{4\pi C{\displaystyle {\int }_0^{r}rdr}}{r^2}\\ =G\frac{4\pi C\left(\frac{r^2}{2}\right)}{r^2}\\ =2\pi GC\end{array}$

Substitute $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G, $6.4\times {10}^8{\text{kg/m}}^{\text{2}}$ for C and solve for g .

  $\begin{array}{l}g=2\pi GC\\ =2\pi \left(6.673\times {10}^{-11}{\text{N×m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(6.4\times {10}^{\text{8}}{\text{kg/m}}^{\text{2}}\right)\\ =0.27\text{N/kg}\end{array}$

Conclusion:

The acceleration due to gravity for a distance $r>5.0\text{m}$ is $\frac{6.7{\text{N×m}}^{\text{2}}{\text{/kg}}^{\text{2}}}{r^2}$ .

The gravitation field with in the region $r<5.0\text{m}$ is $0.27\text{ N/kg}$ .