Chapter 21, Problem 1SP

(a)

To determine

The number of seconds in a year.

Answer to Problem 1SP

The number of seconds in a year is $3.15\times {10}^7\text{s}$.

Explanation of Solution

One year has $365\text{days}$, one day has $24\text{h}$, $1\text{h}$ has $60\min$ and $1\min$ has $60\text{s}$. The number of seconds in a year is obtained as,

$\begin{array}{c}\text{Number of seconds in }1\text{y}=\left(1\text{y}\right)\left(\frac{365\text{days}}{1\text{y}}\right)\left(\frac{24\text{h}}{1\text{day}}\right)\left(\frac{60\min }{1\text{h}}\right)\left(\frac{60\text{s}}{1\min }\right)\\ =3.15\times {10}^7\text{s}\end{array}$

Conclusion:

Therefore, the number of seconds in a year is $3.15\times {10}^7\text{s}$.

(b)

To determine

The number of meters in $1\text{ly}$.

Answer to Problem 1SP

The number of meters in $1\text{ly}$ is $9.46\times {10}^{15}\text{m}$.

Explanation of Solution

One light-year ($1\text{ly}$) is the distance travelled by light ray in one year. The light travels with a speed of $3\times {10}^8\text{m/s}$ through vacuum. So, the number of meters in $1\text{ly}$ is the distance travelled by the light ray in meters, during the time of $1\text{y}$.

Write the expression for the distance travelled by the light ray in a given time.

$d=ct$

Here,

$d$ is the distance travelled by light ray

$c$ is the speed of light

$t$ is the time

Substitute $3\times {10}^8\text{m/s}$ for $c$ and $1\text{y}$ for $t$ to find $d$.

$\begin{array}{c}d=\left(3\times {10}^8\text{m/s}\right)\left(1\text{y}\right)\\ =\left(3\times {10}^8\text{m/s}\right)\left(1\text{y}\times \frac{365\text{days}}{1\text{y}}\times \frac{24\text{h}}{1\text{day}}\times \frac{60\min }{1\text{h}}\times \frac{60\text{s}}{1\min }\right)\\ =9.46\times {10}^{15}\text{m}\end{array}$

Since there distance travelled by light ray in one year is obtained as $9.46\times {10}^{15}\text{m}$, the number of meters in $1\text{ly}$ is equal to the same.

Conclusion:

Therefore, the number of meters in $1\text{ly}$ is $9.46\times {10}^{15}\text{m}$.

(c)

To determine

The distance to the nearest star to the sun in meters.

Answer to Problem 1SP

The distance to the nearest star to the sun in meters is $3.78\times {10}^{16}\text{m}$.

Explanation of Solution

Given info: The distance to the nearest star to sun is about $4\text{ly}$.

The distance in meters, equivalent to one light-year is $9.46\times {10}^{15}\text{m}$. Therefore, the distance ($d$) equivalent to $4\text{ly}$ is obtained as,

$\begin{array}{c}d=4\left(9.46\times {10}^{15}\text{m}\right)\\ =3.78\times {10}^{16}\text{m}\end{array}$

Hence, the distance between the sun and the nearest star is $3.78\times {10}^{16}\text{m}$.

Conclusion:

Therefore, the distance to the nearest star to the sun in meters is $3.78\times {10}^{16}\text{m}$.

(d)

To determine

The time taken to reach the nearest star if one travels with the speed one-tenth of the speed of light.

Answer to Problem 1SP

The time taken to reach the nearest star if one travels with the speed one-tenth of the speed of light is $40\text{y}$.

Explanation of Solution

Given info: The distance between the sun and the nearest star is $3.78\times {10}^{16}\text{m}$ and the speed of the traveler is $c/10$.

Write the expression for the time of travel in terms of distance and speed.

$t=\frac{d}{v}$

Here,

$t$ is the time of travel

$d$ is the distance

$v$ is the speed

Substitute $3.78\times {10}^{16}\text{m}$ for $d$ and $c/10$ for $v$ with $c=3\times {10}^8\text{m/s}$ to find the time taken to reach the nearest star if one travels with the speed one-tenth of the speed of light.

$\begin{array}{c}t=\frac{3.78\times {10}^{16}\text{m}}{\left(\frac{c}{10}\right)}\\ =\frac{3.78\times {10}^{16}\text{m}}{\left(\frac{3\times {10}^8\text{m/s}}{10}\right)}\\ =1.26\times {10}^9\text{s}\times \frac{1\text{h}}{3600\text{s}}\times \frac{1\text{day}}{24\text{h}}\times \frac{1\text{y}}{365\text{days}}\\ =40\text{y}\end{array}$

Conclusion:

Therefore, the time taken to reach the nearest star if one travels with the speed one-tenth of the speed of light is $40\text{y}$.