INSTRUCTOR: The mass of asteroid 1 is 750,000 kilograms. And the mass of asteroid is 130,000 kilograms. Assume asteroid 1 is located at the origin. And asteroid is located at the point 15 comma negative 5 comma 10, measured in units of 10 to the eighth power kilometers. Given that the universal gravitational constant is equal to 6.67384 times 10 to the negative 11 meters cubed times kilograms to the negative 1 times seconds to the negative 2, find the gravitational force vector that asteroid 1 exerts on asteroid 2.

The equation we need for this is F equals G times m1 times m2 divided by the magnitude of r squared, multiplied by-- or end fraction multiplied by r hat. Where r hat is the unit vector in the direction of r. Now, we can fill in a lot of this information already, so that F is equal to 6.67384 times 10 to the negative 11 times 750,000 times 130,000.

All divided by whatever the magnitude of r squared is. Then multiply that outside of our fraction by r hat. Why don't we go ahead to the side and find the magnitude of r squared?

First, let's just note that r is the vector 15 times 10 to the 8 comma negative 5 times 10 to the 8 comma 10 times 10 to the 8. So that the magnitude of this squared is equal to 15 times 10 to the 8 quantity squared. Plus parentheses negative 5 times 10 to the 8, close parentheses squared.

Plus 10 times 10 to the eighth parentheses squared. That is equal to 3.5 times 10 to the 18. Substituting that into our expression, we now have a very large constant, multiplied by the unit vector in the direction of r.

Let's go ahead and find that unit vector, r hat. Now, because we already know what the vector is, and we now know the magnitude of that, it's a square root of the magnitude squared, we can say that our vector is open bracket, 15 times 10 to the 8, over the square root of 3.5 times 10 to the 18 end square root comma negative 5 times 10 to the 8 divided by the square root of 3.5 times 10 to the 18 end square root comma 10 times 10 to 8 divided by the square root of 3.5 times 10 to the 18. And square root, and vector.

If we multiply the constant found earlier, the G times m1 times m2 divided by the magnitude of r squared, by this unit vector in the direction of r, we find that the force vector is equal to angle bracket 1.491 times 10 to the negative 18, comma, negative 4.969 times 10 to the negative 19 comma 9.938 times 10 to the negative 19. And our vector, and that is force in Newtons.