00:01
Question number 43 is asking us about the voyager 1 spacecraft.
00:05
How far away it is in terms of kilometers, how long its signals take to reach earth from its current position, and how does its distance compare with that of the nearest star that isn't the sun.
00:23
So voyager 1 was launched in 1977, and it flew by the planet's jupiter and saturn, in 1979 and 1980, respectively.
00:34
It then left the solar system on the path that i'm illustrating here.
00:39
And now it's the furthest man -made object in space.
00:44
So the problem is asking about the distance between earth and voyager 1 in kilometers, given that the distance in astronomical units is 125.
00:56
Or at least it was, at the time of this book's publishing.
01:00
Voyager 1's moving mighty fast through space, and so there's a good chance this figure may have changed recently.
01:08
One astronomical unit is the distance between earth and the sun, which is 150 million kilometers, or as i'm going to write it in scientific notation, because when we multiply it out, the number will be huge.
01:24
1 .5 times 10 to the 8th kilometers.
01:29
And when we multiply through our units cancel, and 125 times 150 million gives us a distance of about 1 .88 times 10 to the 10th kilometers, or about 18 .8 billion kilometers.
01:55
And the next question, they ask, how long does it take voyager's signals to reach earth from 125 astronomical units away? well, voyager's signals would travel to us at the speed of light, which we refer to as c.
02:13
And speed of light is a constant in a vacuum that is three times 10 to the 8 meters per second, or, or, converting to kilometers per second, that is, three times 10 to the fifth kilometers per second.
02:35
So the way we find out how long voyager's signals take to reach earth from the 125 astronomical unit distance is to divide that distance we found earlier, the 1 .88 times 10 to the 10th kilometers.
02:55
Divide that by our 3 times 10 to the 5th kilometers for second.
03:04
Now, to solve our problem, the units cancel, and seconds would be brought up to the numerator, and we would find up with a time period of 6 .27 times 10 to the 4 seconds...