00:01
For this problem, we are told that a planet of mass 4 times 10 to the power of 24 kilograms is at location negative 6 times 10 to the power of 11, 3 times 10 to the power of 11, 0 meters.
00:11
We're told that a star of mass 6 times 10 to the power of 30 kilograms is that the location 5 times 10 to the power of 11, negative 3 times 10 to the power of 11, 0 meters.
00:21
We are asked, what is the force exerted on the planet by the star? so to begin, we need to find the vector from the planet to the star.
00:30
Our r -hat vector, which will be given by taking the location of the star and subtracting the location of the planet.
00:39
So this would be 1 times 10 to the power of 11 meters times, let's see here, it would be times negative 6 -30 minus.
00:53
Oh, excuse me, no, i have that the wrong way around.
00:56
That's the location of the planet.
00:59
So this is from the planet to the star.
01:03
That's our r hat vector here so this would be 5 negative 3 0 minus negative 6 3 0 which will be equal to 1 times 10 to the power of 11 meters times the vector oops you fix it there so we have 5 minus negative 6 which is then 5 plus 6 and so that's 11 negative 3 minus 3 which is negative 6, 0...