Chapter 9, Problem 39QAP

(a)

To determine

The amount of mass that would Comet Halley loses in a year.

Answer to Problem 39QAP

The amount of mass that would Comet Halley loses in a year is $9.46\times {10}^{11}\text{kg}$ .

Explanation of Solution

Write the expression for mass lost in a year.

$\Delta M=\Delta mt$        (I)

Here, $\Delta m$ is mass lost by comet per second, $t$ is number of seconds in a year and $\Delta M$ is total mass lost in a year.

Conclusion:

Calculate the mass lost per second by comet.

$\begin{array}{c}\Delta m=20,000\text{kg}/\text{s}+10,000\text{kg}/\text{s}\\ =30,000\text{kg}/\text{s}\end{array}$

Calculate the number of seconds in a year.

$\begin{array}{c}t=\left(1\text{y}\right)\left(\frac{365\text{days}}{1\text{y}}\right)\left(\frac{24\text{h}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ =31536000\text{s}\end{array}$

Substitute $30,000\text{kg}/\text{s}$ for $\Delta m$ and $31536000\text{s}$ for $t$ in equation (1).

$\begin{array}{c}\Delta M=\left(30,000\text{kg}/\text{s}\right)\left(31536000\text{s}\right)\\ =9.46\times {10}^{11}\text{kg}\end{array}$

Thus, the amount of mass that would Comet Halley loses in a year is $9.46\times {10}^{11}\text{kg}$ .

(b)

To determine

The fraction of mass lost to the total mass.

Answer to Problem 39QAP

The fraction of mass lost to the total mass is $0.43\%$ .

Explanation of Solution

Write the expression for percentage of mass lost.

$P=\left(\frac{\Delta M}{M}\times 100\right)\%$        (II)

Here, $M$ is the total mass of Halley.

Conclusion:

Substitute $9.46\times {10}^{11}\text{kg}$ for $\Delta M$ and $2.20\times {10}^{14}\text{kg}$ for $M$ in equation (II).

$\begin{array}{c}P=\left(\frac{9.46\times {10}^{11}\text{kg}}{2.20\times {10}^{14}\text{kg}}\times 100\right)\%\\ =0.43\%\end{array}$

Thus, the fraction of mass lost to the total mass is $0.43\%$ .

(c)

To determine

The number of trips Comet Halley would make before complete destruction.

Answer to Problem 39QAP

The number of trips Comet Halley would make before complete destruction is $3$ .

Explanation of Solution

The time taken by Halley for one trip is $76\text{years}$ .

Write the expression for number of trips.

$n=\frac{M}{T\times \Delta M}$        (III)

Here, $T$ is the time period of Halley and $n$ is the number of trips.

Conclusion:

Substitute $76\text{years}$ for $T$ , $9.46\times {10}^{11}\text{kg}/\text{year}$ for $\Delta M$ and $2.20\times {10}^{14}\text{kg}$ for $M$ in equation (III).

$\begin{array}{c}n=\frac{2.20\times {10}^{14}\text{kg}}{\left(76\text{years}\right)\left(9.46\times {10}^{11}\text{kg}/\text{year}\right)}\\ =3.0599\\ \approx 3\end{array}$

Thus, the number of trips Comet Halley would make before complete destruction is $3$ .

(d)

To determine

The time taken by Comet Halley for complete destruction.

Answer to Problem 39QAP

The time taken by Comet Halley for complete destruction is $232.5524\text{years}$ .

Explanation of Solution

Conclusion:

The time taken by Halley for one trip is $76\text{years}$ .

The number of trips that Comet Halley would make before complete destruction is $3.0599$ .

Calculate the number of years Comet Halley would survive.

$\begin{array}{c}t=n\times T\\ =\left(3.0599\right)\left(76\text{years}\right)\\ =232.5524\text{years}\end{array}$

Thus, the time taken by Comet Halley for complete destruction is $232.5524\text{years}$ .