Chapter 11, Problem 106P

(a)

To determine

Toshow: The ratio of the force exerted on a point particle on the surface of Earth by the Sun to that exerted by the moon is, $M_s{r}_m^2/M_m{r}_s^2$ .

Explanation of Solution

Given information :

Mass of the Sun $=M_s$

Mass of the Moon $=M_m$

Distance of the particle from Earth to Sun $=r_s$

Distance of the particle from Earth to Moon $=r_m$

Mass of the Sun $=1.99\times {10}^{30}\text{kg}$

Mass of the Moon $=7.36\times {10}^{22}\text{kg}$

Average orbital distance between earth and sun $=1.50\times {10}^{11}\text{m}$

Average orbital distance between earth and moon $=3.84\times {10}^8\text{m}$

Formula used :

Gravitational force between two masses ( M and m ) separated by distance r is:

  $F=\frac{GMm}{r^2}$

G is the gravitational constant.

Proof:

Express the force exerted by the sun on a body of water of mass, $m$ ,

  $F_s=\frac{G{M}_{s}m}{r_s^2}.\mathrm{..}(1)$

Express the force exerted by the moon on a body of water of mass $m$ ,

  $F_m=\frac{G{M}_{m}m}{r_m^2}.\mathrm{..}(2)$

Divide equation (1) by (2)

  $\frac{F_s}{F_m}=\frac{M_s{r}_m^2}{M_m{r}_s^2}$

Substitute the values and solve:

  $\begin{array}{l}\frac{F_s}{F_m}=\frac{\left(1.99\times {10}^{30}\text{kg}\right){\left(3.84\times {10}^8\text{m}\right)}^2}{\left(7.36\times {10}^{22}\text{kg}\right){\left(1.50\times {10}^{11}\text{m}\right)}^2}\\ \frac{F_s}{F_m}=177\end{array}$

(b)

To determine

ToShow: $dF/F=\left(-2dr\right)/r$

Explanation of Solution

Given information :

Mass of the Sun $=M_s$

Mass of the Moon $=M_m$

Distance of the particle from Earth to Sun $=r_s$

Distance of the particle from Earth to Moon $=r_m$

Formula used :

Gravitational force between two masses ( M and m ) separated by distance r is:

  $F=\frac{GMm}{r^2}$

G is the gravitational constant.

  $$

Calculation:

  $F=\frac{G{m}_1{m}_2}{r^2}$

Differentiate the expression with respect to r :

  $\begin{array}{l}\frac{dF}{dr}=-\frac{2G{m}_1{m}_2}{r^3}\\ \frac{dF}{dr}=-2\frac{F}{r}\end{array}$

Re-arrange the expression:

  $\frac{dF}{F}=-2\frac{dr}{r}$

(c)

To determine

To Show:For a small difference in distance compared to the average distance, the ratio of the differential gravitational force exerted by the Sun to the differential gravitational force exerted by the moon on Earth’s oceans is given by

  $\raisebox{1ex}{$\Delta F_s$}\!\left/ \!\raisebox{-1ex}{$\Delta F_m$}\right.\approx \raisebox{1ex}{$\left(M_s{r}_m^3\right)$}\!\left/ \!\raisebox{-1ex}{$\left(M_m{r}_s^3\right)$}\right.$ and to calculate the ratio.

Explanation of Solution

Given information :

Mass of the Sun $=M_s$

Mass of the Moon $=M_m$

Distance of the particle from Earth to Sun $=r_s$

Distance of the particle from Earth to Moon $=r_m$

Formula used :

From part b, $\Delta F=-2\frac{F}{r}\Delta r$

Calculations:

The change in force $\Delta F$ for a small change in distance $\Delta r$ .

  $\Delta F=-2\frac{F}{r}\Delta r$

  $\begin{array}{l}\Delta F_s=-2\frac{\raisebox{1ex}{$Gm{M}_s$}\!\left/ \!\raisebox{-1ex}{$r_s^2$}\right.}{r_s}\Delta r_s\\ \Delta F_s=-2\frac{Gm{M}_s}{r_s^3}\Delta r_s\end{array}$

  $\Delta F_m=-2\frac{Gm{M}_m}{r_m^2}\Delta r_m$

  $\begin{array}{l}\frac{\Delta F_s}{\Delta F_m}=\frac{\raisebox{1ex}{$M_s$}\!\left/ \!\raisebox{-1ex}{$r_s^3$}\right.\Delta r_s}{\raisebox{1ex}{$M_m$}\!\left/ \!\raisebox{-1ex}{$r_m^3$}\right.\Delta r_m}\\ \frac{\Delta F_s}{\Delta F_m}=\frac{M_s{r}_m^3\Delta r_s}{M_m{r}_s^3\Delta r_m}\end{array}$

Because the difference in the distance is small, $\Delta r_s$ and $\Delta r_m$ can be neglected.

  $\frac{\Delta F_s}{\Delta F_m}\approx \frac{M_s{r}_m^3}{M_m{r}_s^3}$

Substitute the numerical values

  $\begin{array}{l}\frac{\Delta F_s}{\Delta F_m}=\frac{\left(1.99\times {10}^{30}\text{kg}\right){\left(\text{3}{\text{.84×10}}^{\text{8}}\text{m}\right)}^3}{\left(7.36\times {10}^{22}\text{kg}\right){\left(\text{1}{\text{.50×10}}^{11}\text{m}\right)}^3}\\ \frac{\Delta F_s}{\Delta F_m}=0.454\end{array}$