Question

1. the voyager 1 spacecraft was launched in 1977 to explore the outer planets. it left the solar system in august of 2012, and is the farthest man - made probe from earth. voyager 1 travels at about 17 km/s and was about 20.6 billion km from the earth in 2017. voyager 1 will stop transmitting data back to earth in 2025 due to power losses on the craft. if you assume that it continued to travel at a constant 17 km/s for the eight years from 2017 - 2025, how far from earth will it be when it stops data transmission? you may assume an exact eight more years of transmission.
2. using the data in problem 7, how long will it take voyager 1 to travel one light - year? one light - year is the distance light travels in one earth year. you may use 3×10^8m/s for the speed of light and 365.2422 days of 86400 seconds each for one earth year.
3. if you drive from huntsville to nasas johnson space center in houston (96 miles) at a constant speed of 120 km/h, tour the museum and grounds for two hours, then drive back to huntsville at 110 km/h, what is the average speed (not velocity, but speed) of your entire trip? you can use the simplification that your position did not change during your two hours at the museum, and that 1 km = 0.621371 miles.
4. the international space station travels at about 7.66 km/s. its total path for one trip around the earth is about 42600 km. how many times does it go around the earth in one day?

Explanation:

7.

Step1: Calculate the distance traveled from 2017 - 2025

The time interval from 2017 - 2025 is $t = 8$ years. First convert years to seconds. One year has $365\times24\times3600$ seconds. So $t=8\times365\times24\times3600\ s$. The speed of Voyager 1 is $v = 17\ km/s=17000\ m/s$. The distance traveled $d_1=v\times t$.
\[

$$\begin{align*} t&=8\times365\times24\times3600\ s\\ &=8\times31536000\ s\\ & = 252288000\ s \end{align*}$$

\]
\[d_1 = 17000\times252288000\ m=4.288896\times 10^{11}\ m = 4.288896\times10^{8}\ km\]
In 2017, it was $d_0 = 20.6\times10^{9}\ km$ from Earth. The distance $D$ from Earth when it stops transmitting in 2025 is $D=d_0 + d_1$.
\[D=(20.6\times 10^{9}+4.288896\times 10^{8})\ km\approx21.0288896\times 10^{9}\ km\approx21.03\times 10^{9}\ km\]

Step1: Calculate the distance of one - light year

The speed of light $c = 3\times10^{8}\ m/s$, and the number of seconds in one Earth - year $t_y=365.2422\times24\times3600\ s = 31556952\ s$. The distance of one - light year $d_{ly}=c\times t_y$.
\[d_{ly}=3\times 10^{8}\times31556952\ m\approx9.4670856\times 10^{15}\ m\]
The speed of Voyager 1 is $v = 17000\ m/s$. Let the time taken to travel one - light year be $T$. Using the formula $d = vT$, we have $T=\frac{d_{ly}}{v}$.
\[T=\frac{9.4670856\times 10^{15}}{17000}\ s\]
\[T\approx5.568874\times10^{11}\ s\]
Convert seconds to years: $T_y=\frac{T}{31556952}\ years$
\[T_y=\frac{5.568874\times 10^{11}}{31556952}\ years\approx17647.1\ years\]

Step1: Convert the distance from miles to km

The distance from Huntsville to Houston is $d = 96$ miles. Since $1\ km=0.621371$ miles, then $d=\frac{96}{0.621371}\ km\approx154.5\ km$.

Step2: Calculate the time taken to drive to Houston

The speed to Houston $v_1 = 120\ km/h$. Using the formula $t=\frac{d}{v}$, the time taken to drive to Houston $t_1=\frac{154.5}{120}\ h\approx1.29\ h$.

Step3: Calculate the time taken to drive back

The speed on the return trip $v_2 = 110\ km/h$. The time taken to drive back $t_2=\frac{154.5}{110}\ h\approx1.405\ h$.
The total time of the trip $t_{total}=t_1 + t_2+2\ h$ (the 2 - hour stay at the museum).
\[t_{total}=(1.29 + 1.405+2)\ h=4.695\ h\]
The total distance traveled $D = 2\times154.5\ km = 309\ km$.
The average speed $v_{avg}=\frac{D}{t_{total}}$.
\[v_{avg}=\frac{309}{4.695}\ km/h\approx65.81\ km/h\]

Answer:

$21.03$ billion km

8.