(II) Three meteorites in outer space lie along a straight line...

Question

(II) Three meteorites in outer space lie along a straight line. At a certain instant in time, the meteorite on the right is 53 m from the center meteorite and the lefthand meteorite is 21 m from the center meteorite. If the mass of the righthand meteorite is 1300 kg and the net gravitational force on the center meteorite is zero, what is the mass of the lefthand meteorite?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Scientists are closely observing a group of three satellites situated in the void of space. They are arranged in a straight line and their thrusters are not working. There is nothing else in their close vicinity. The leftmost satellite is at a distance of 23 m from the one at the center. While the rightmost one is 43 m away from the one in the middle, given that the leftmost satellite has a mass of 1700 kg. And the net gravitational force on the satellite in the middle is zero, calculate the mass of the rightmost satellite. So that's our end goal. Ultimately, we're trying to figure out what the mass of the rightmost satellite is provided the given information in the prom itself. So final, our final answer is we're trying to figure out what the mass of the rightmost satellite is awesome. So now that we have an idea of what we're solving for let's look at our multiple choice answers and read them off to see what our final answer might be. And let us note that they're all in the same units of kilograms. A is 3.1 multiplied by 10 to the power of three B is 5.9 multiplied by 10 to the power of three C is 1.1 multiplied by 10 to the power of four and D is 2.1 multiplied by 10 to the power of four. So first off, let us draw a diagram to help us better visualize this problem. So I've gone ahead and drawn a figure for us to use to help us visualize this problem. I have used the dean, I was, I've used to denote the leftmost satellite or the mass of leftmost satellite. So on the far left capital M subscript capital L is the mass of our leftmost satellite. The mass of our middle satellite is capital M subscript lowercase M and then the mass of our rightmost satellite is capital M subscript, capital R. Awesome. And then from the S our middle satellite to our leftmost satellite, we have some radius R, subscript capital R and some force F capital R. And then similarly on our left side, for our leftmost satellite, we have a radius of capital R subscript capital L and then our force L which is capital F, subscript capital L and the leftmost force is pointing to the left whereas the rightmost force is pointing to the right and they're all pointing away from the middle satellite. Awesome. So now that we've defined all of our variables, we now need to note that since the net force on the satellite in the middle is zero, it will be pulled with equal forces in both directions by the other two satellites. So the middle satellite is gonna be being pulled by the, the left and the right satellite respectfully. Awesome. So moving right along here. So therefore, we can recall and use the following equations for the forces on the left and right satellites. So once again, we're gonna take fl which is the force of the leftmost satellite. And as we should recall, we can say that this is equal to capital G where capital G is our gravitational constant multiplied by ML multiplied by mm divided by RL square. And we can also write, I'll do a little and symbol and FR is similarly is equal to capital G multiplied by Mr multiplied by mm divided by RR squared. Awesome. So now we need to note and let's write this little note in blue that FL is equal to Fr. Thus, we can write the following that capital G multiplied by Mr multiplied by mm divided by RR squared is equal to capital G multiplied by ML multiplied by mm divided by RL square. So now if we need to start rearranging this equation to isolate and solve for Mr because we're trying to solve for the mass of the rightmost satellite, which is our final answer. So to do that, we'll use a little bit of algebra to do some rearranging. And we will find that Mr is equal to RR squared divided by RL squared multiplied by ML. So now at this point, we can plug in all of our known variables to solve for our final answer. So let us note that RR is 43 m squared divided by RL, which RL is equal to 23 m. And we need to make sure that our RL value is squared. And note that this is given to us in the prong itself multiplied by the mass of our leftmost satellite, which is given to us as 1700 kilograms. So when we plug that into our calculator, we will find that Mr is equal to 5.9 multiplied by 10 to the power of three kilograms when we round to one decimal place. And we also write it in scientific notation and that's it we saw for our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter B 5.9 multiplied by 10 to the power of 3 kg. And that's it. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

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