Chapter 10, Problem 47QAP

To determine

(a)

The gravitational potential energy of the Moon-Pacific Ocean system when the Pacific is facing away from the Moon.

Answer to Problem 47QAP

Gravitational potential energy of the Moon-Pacific Ocean system is $-9.1\times {10}^{24}J$

Explanation of Solution

Given:

Density of the water of Ocean = $\rho =1030kg/m^3$

Volume of water of the Ocean = $V=7.0\times {10}^{8}k{m}^3=7.0\times {10}^8{({10}^{3}m)}^3=7.0\times {10}^{17}{m}^3$

Radius of earth = $R_{earth}=6.4\times {10}^{6}m$

The distance between moon and center of the earth = $r=3.84\times {10}^{8}m$

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

Formula used:

Mass of the object is defined as,

  $\begin{array}{l}\text{Mass = volume}\times \text{density}\\ i.e,m=\rho \times V\end{array}$

Gravitational potential energy is defined as,

  $U=-\frac{GMm}{r}$

Calculation:

Mass of the Ocean = $m=\rho \times V=1030kg/m^3\times 7.0\times {10}^{17}{m}^3$

  $m=7.210\times {10}^{20}kg$

In this situation, Pacific Ocean is facing away from the moon. So, the distance between from the center of moon to the Ocean,

  $r\text{'}=r+R_{earth}$

So, gravitational potential energy,

  $\begin{array}{l}U\text{'}=-\frac{GMm}{r\text{'}}\\ U\text{'}=-\frac{GMm}{(r+R_{earth})}\\ U\text{'}=-\frac{6.67\times {10}^{-11}N{m}^2/k{g}^2\times 7.35\times {10}^{22}kg\times 7.21\times {10}^{20}kg}{(3.84\times {10}^{8}m+6.4\times {10}^{6}m)}\\ U\text{'}=-9.1\times {10}^{24}J\end{array}$

Conclusion:

Gravitational potential energy of Moon-Pacific system is $-9.1\times {10}^{24}J$.

To determine

(b)

The gravitational potential energy when Pacific Ocean faces towards the moon

Answer to Problem 47QAP

The gravitational potential energy of the Pacific Ocean-Moon system, when Pacific Ocean faces towards the Moon, is $-9.4\times {10}^{24}J$.

Explanation of Solution

Given:

Density of the water of Ocean = $\rho =1030kg/m^3$

Volume of water of the Ocean = $V=7.0\times {10}^{8}k{m}^3=7.0\times {10}^8{({10}^{3}m)}^3=7.0\times {10}^{17}{m}^3$

Radius of earth = $R_{earth}=6.4\times {10}^{6}m$

The distance between moon and center of the earth = $r=3.84\times {10}^{8}m$

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

Calculation:

The distance between center of the Moon and Ocean = $r\text{'}\text{'}=r-R_{earth}$

Then, gravitational potential energy,

  $\begin{array}{l}U\text{'}\text{'}=-\frac{GMm}{r\text{'}\text{'}}\\ U\text{'}\text{'}=-\frac{GMm}{(r-R_{earth})}\\ U\text{'}\text{'}=-\frac{6.67\times {10}^{-11}N{m}^2/k{g}^2\times 7.35\times {10}^{22}kg\times 7.21\times {10}^{20}kg}{(3.84\times {10}^{8}m-6.4\times {10}^{6}m)}\\ U\text{'}\text{'}=-9.4\times {10}^{24}J\end{array}$

Conclusion:

Thus, gravitational potential energy of the Moon-Ocean system, when Ocean is facing towards the Moon, is $-9.4\times {10}^{24}J$.

To determine

(c)

The maximum speed of Ocean due to change in gravitational potential energy

Answer to Problem 47QAP

The maximum speed of the Ocean is $29m/s$.

Explanation of Solution

Given:

Mass of the Ocean = $m=\rho \times V=1030kg/m^3\times 7.0\times {10}^{17}{m}^3$

  $m=7.210\times {10}^{20}kg$

  $U\text{'}=-9.1\times {10}^{24}J$

  $U\text{'}\text{'}=-9.4\times {10}^{24}J$

Formula used:

By conservation of mechanical energy,

Change in kinetic energy = - change in gravitational potential energy

  $K.E_f-K.E_i=-(U\text{'}\text{'}-U\text{'})$

Calculation:

Initial kinetic energy = $K.E_i=0$

Final kinetic energy = $K.E_f=\frac{1}{2}m{v}^2;v\text{ is the speed of Ocean}\text{.}$

Now, $K.E_f-K.E_i=-(U\text{'}\text{'}-U\text{'})$

  $\begin{array}{l}\frac{1}{2}m{v}^2-0=-(-9.4\times {10}^{24}J-(-9.1\times {10}^{24}J))\\ \frac{1}{2}(7.21\times {10}^{20}kg)v^2=0.3\times {10}^{24}J\\ v=\sqrt{\frac{2\times 0.3\times {10}^{24}J}{7.21\times {10}^{20}kg}}=28.85m/s\approx 29m/s\end{array}$

Conclusion:

Thus, maximum speed of the Ocean is $29m/s$.