Chapter 7, Problem 99A

(a)

To determine

Whether the tides on the Earth are caused by the pull of the Moon.

The force in terms of mass, that the moon and the sun exert on mass m of water on the earth.

Answer to Problem 99A

The forces that the moon and the sun exert on mass m of water on the earth are:

  $F_{sun}=\left(5.89\times {10}^{-3}m\right)\text{ N}$

  $F_{moon}=\left(3.22\times {10}^{-5}m\right)\text{ N}$

Explanation of Solution

Given info:

Mass of sun, $M_s=1.99\times {10}^{30}\text{ kg}$

Gravitational constant, $\text{G}=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Distance of earth from the sun, ${\text{r}}_s=1.5\times {10}^{11}\text{ m}$

Mass of the moon, ${\text{M}}_m=7.35\times {10}^{22}\text{ kg}$

Distance of earth from the moon, ${\text{r}}_m=13.9\times {10}^8\text{ m}$

Formula used:

The gravitation force of sun on mis

  $F_{sun}=\frac{G{M}_{s}m}{R_s{}^2}$

The gravitation force of moon on m is

  $F_{moon}=\frac{G{M}_{m}m}{R_m{}^2}$

Calculation:

Tides are natural phenomenon in which the level of water on earth increases and decreases due to the gravitational forces of the sun and the moon.

$\begin{array}{c}{F}_{sun}=\frac{G{M}_{s}m}{R_s{}^2}\\ =\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{ kg}\right)m}{{\left(1.5\times {10}^{11}\text{ m}\right)}^2}\\ =\left(5.89\times {10}^{-3}m\right)\text{ N}\end{array}$

  $\begin{array}{c}{F}_{moon}=\frac{G{M}_{m}m}{R_m{}^2}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(7.35\times {10}^{22}\text{ kg}\right)m}{{\left(13.9\times {10}^8\text{ m}\right)}^2}\\ =\left(3.22\times {10}^{-5}m\right)\text{ N}\end{array}$

Conclusion:

The forces that the moon and the sun exert on mass m of water on earth is

  $F_{sun}=\left(5.89\times {10}^{-3}m\right)\text{ N}$

  $F_{moon}=\left(3.22\times {10}^{-5}m\right)\text{ N}$

(b)

To determine

Whether the sun or moon has greater pull on the earth.

Answer to Problem 99A

The gravitation force of sun is more than gravitational force of moon

  $F_{sun}>F_{moon}$

Explanation of Solution

Given:

Mass of sun, $M_s=1.99\times {10}^{30}\text{ kg}$

Gravitational constant, $\text{G}=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Distance of earth from the sun, ${\text{r}}_s=1.5\times {10}^{11}\text{ m}$

Mass of the moon, ${\text{M}}_m=7.35\times {10}^{22}\text{ kg}$

Distance of earth from the moon, ${\text{r}}_m=13.9\times {10}^8\text{ m}$

Formula used:

The gravitation force of sun on m is

  $F_{sun}=\frac{G{M}_{s}m}{R_s{}^2}$

The gravitation force of moon on m is

  $F_{moon}=\frac{G{M}_{m}m}{R_m{}^2}$

Calculation:

  $\begin{array}{c}{F}_{sun}=\frac{G{M}_{s}m}{R_s{}^2}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{ kg}\right)m}{{\left(1.5\times {10}^{11}\text{ m}\right)}^2}\\ =\left(5.89\times {10}^{-3}m\right)\text{ N}\end{array}$

  $\begin{array}{c}{F}_{moon}=\frac{G{M}_{m}m}{R_m{}^2}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(7.35\times {10}^{22}\text{ kg}\right)m}{{\left(3.9\times {10}^8\text{ m}\right)}^2}\\ =\left(3.22\times {10}^{-5}m\right)\text{ N}\end{array}$

  $F_{sun}=\left(5.89\times {10}^{-3}m\right)\text{ N}$

  $F_{moon}=\left(3.22\times {10}^{-5}m\right)\text{ N}$

  $F_{sun}>F_{moon}$

Conclusion:

The gravitation force of sun is more than gravitational force of moon

  $F_{sun}>F_{moon}$

(c)

To determine

The difference in the force exerted by the moon on water at the near surface and water at the far surface of Earth.

Answer to Problem 99A

The difference between gravitational forces due to moon on water at near and far surface is:

  $F_{near}-F_{far}\text{=}\left(\text{0}{\text{.22×10}}^{\text{-5}}\text{m}\right)\text{ N}$

Explanation of Solution

Given info:

Mass of sun, $M_s=1.99\times {10}^{30}\text{ kg}$

Gravitational constant, $\text{G}=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Distance of earth from the sun, ${\text{r}}_s=1.5\times {10}^{11}\text{ m}$

Mass of the moon, ${\text{M}}_m=7.35\times {10}^{22}\text{ kg}$

Distance of earth from the moon, ${\text{r}}_m=13.9\times {10}^8\text{ m}$

Formula used:

The gravitation force of sun on m is

  $F_{sun}=\frac{G{M}_{s}m}{R_s{}^2}$

The gravitation force of moon on m is

  $F_{moon}=\frac{G{M}_{m}m}{R_m{}^2}$

Calculation:

Near distance

  $\begin{array}{c}{R}_{near}=3.9\times {10}^8-0.063\times {10}^8\\ =3.837\times {10}^8\text{ m}\end{array}$

Far distance

  $\begin{array}{c}{R}_{far}=3.9\times {10}^8+0.063\times {10}^8\\ =3.963\times {10}^8\text{ m}\end{array}$

Near gravitation force

  $\begin{array}{c}{F}_{near}=\frac{G{M}_{m}m}{R_{near}{}^2}=\frac{\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}\right)\left(7.35\times {10}^{22}\text{ kg}\right)m}{{\left(3.837\times {10}^8\text{ m}\right)}^2}\\ =\left(3.44\times {10}^{-5}m\right)\text{ N}\end{array}$

Far gravitation force

  $\begin{array}{c}{F}_{far}=\frac{G{M}_{m}m}{R_{far}{}^2}=\frac{\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}\right)\left(7.35\times {10}^{22}\text{ kg}\right)m}{{\left(3.963\times {10}^8\text{ m}\right)}^2}\\ =\left(3.22\times {10}^{-5}m\right)\text{ N}\end{array}$

So, the difference between gravitational forces due to moon on water

  $\begin{array}{c}{F}_{near}-F_{far}=\left(\text{3}{\text{.44×10}}^{\text{-5}}\text{m}\right)\text{ N-}\left(\text{3}{\text{.22×10}}^{\text{-5}}\text{m}\right)\text{ N}\\ F_{near}-F_{far}\text{=}\left(\text{0}{\text{.22×10}}^{\text{-5}}\text{m}\right)\text{ N}\end{array}$

Conclusion:

The difference between gravitational forces due to moon on water

  $F_{near}-F_{far}\text{=}\left(\text{0}{\text{.22×10}}^{\text{-5}}\text{m}\right)\text{ N}$

(d)

To determine

The difference in force exerted by the sun on water at near and far surface of earth.

Answer to Problem 99A

The difference between gravitational forces due to sun on water

  $F_{near}-F_{far}=\left(.101\times {10}^{-5}m\right)\text{ N}$

Explanation of Solution

Given info:

Mass of sun $M_s=1.99\times {10}^{30}\text{ kg}$

Gravitational constant $\text{G=}6.67\times {10}^{-11}{\text{ m}}^{3}k{g}^{-1}{m}^{-2}$

Distance from the sun ${\text{r}}_s=1.5\times {10}^{11}\text{ m}$

Mass of the moon ${\text{M}}_m=7.35\times {10}^{22}\text{ kg}$

Distance from the moon ${\text{r}}_m=3.9\times {10}^8\text{ m}$

Formula used:

The gravitation force of sun on m is

  $F_{sun}=\frac{G{M}_{s}m}{R_s{}^2}$

Calculation:

Near distance

  $\begin{array}{c}{R}_{near}=1.5\times {10}^{11}-0.000063\times {10}^{11}\\ =1.499937\times {10}^{11}\text{ m}\end{array}$

Far distance

  $\begin{array}{c}{R}_{far}=1.5\times {10}^{11}+0.000063\times {10}^{11}\\ =1.500063\times {10}^{11}\text{ m}\end{array}$

Near gravitation force

  $\begin{array}{c}{F}_{near}=\frac{GMm}{R_{near}{}^2}=\frac{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{1}{\text{.99×10}}^{\text{30}}\text{kg}\right)m}{{\left(1.499937\times {10}^{11}\text{ m}\right)}^2}\\ =\left(5.89975\times {10}^{-3}m\right)\text{ N}\end{array}$

Far gravitation force

  $\begin{array}{c}{F}_{far}=\frac{GMm}{R_{far}{}^2}=\frac{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{1}{\text{.99×10}}^{\text{30}}\text{kg}\right)m}{{\left(\text{1}{\text{.500063×10}}^{\text{11}}\text{m}\right)}^2}\\ =\left(5.89874\times {10}^{-3}m\right)\text{ N}\end{array}$

So, the difference between gravitational forces due to sun on water

  $\begin{array}{c}{F}_{near}-F_{far}=\left(5.89975\times {10}^{-3}m\right)\text{ N}-\left(5.89874\times {10}^{-3}m\right)\text{ N}\\ \text{=}\left(0.101\times {10}^{-5}m\right)\text{ N}\end{array}$

Conclusion:

The difference between gravitational forces due to sun on water

  $F_{near}-F_{far}=\left(.101\times {10}^{-5}m\right)\text{ N}$

(e)

To determine

The celestial body which has greater difference in pull from one side of earth to other side.

Explanation of Solution

Introduction:

Tides are natural phenomenon in which the level of water on earth increases and decreases due to the gravitational forces of the sun and the moon.

From the above gravitational force difference we can find

Moon on water

  $F_{near}-F_{far}\text{=}\left(\text{0}{\text{.22×10}}^{\text{-5}}\text{m}\right)\text{ N}$

Sun on water

  $F_{near}-F_{far}=\left(.101\times {10}^{-5}m\right)\text{ N}$

Therefore the difference in pull from one side of earth to the other is greater for the moon than the sun.

Conclusion:

The difference in pull from one side of earth to the other is greater for the moon as compared for due tothe sun.

(f)

To determine

The gravitational effect of moon and the sun on the tides.

Explanation of Solution

Introduction:

Tides are natural phenomenon in which the level of water on earth increases and decreases due to the gravitational forces of the sun and the moon.

The gravitational effect of the sun on the earth is more than the moon because of the massive size of the sun compared to moon. Therefore it is misleading to say that the tides result from the pull of the Moon. However, as the moon is nearer to the earth than the sun,its gravitational gradient is greater than that of the sun, hence the effect of the moon’s pull is more dominant. Gravitational gradient means that the difference in effect by the moon’s gravity with change in distance from the near to the far side is greater than it is for the sun.

Conclusion:

The gravitational effect of the sun on the earth is more than the moon because of the massive size of the sun compared to moon. Therefore it is misleading to say that the tides result from the pull of the Moon. However, as the moon is nearer to the earth than the sun its gravitational gradient is greater than that of the sun, hence the effect of the moon’s pull is more dominant.