### Video Transcript

When two masses are a distance π sub one apart, the gravitational force between them is πΉ sub one. If the distance between them changes so that they become π sub two apart, the gravitational force πΉ sub two between them becomes a half πΉ sub one. Find the ratio π sub one to π sub two.

So, when we take a look at the question, we can see it concerns gravitational force. Therefore, we know what weβre going to be looking at is Newtonβs law of universal gravitation. And what we have is a formula to help us. And this formula states that πΉ sub πΊ is equal to capital πΊ multiplied by π sub one multiplied by π sub two over π squared, where πΉ sub πΊ is the gravitational force, capital πΊ is the universal gravitational constant, π is the mass, and π is the separation or distance between our two bodies.

So now that we have this, what we can do is use our formula and the information that weβve been given to create a couple of equations. Well, first of all, what we want to do is create an equation in terms of π sub one. So, to begin with, what we have is πΉ sub one is equal to capital πΊ multiplied by π sub one multiplied by π sub two over π sub one squared.

Well, as weβd already stated previously, what weβre looking to do is make π sub one the subject of this. And to do that, what weβre going to do is actually swap the πΉ sub one and the π sub one squared. What this actually means mathematically is multiplying both sides of the equation by π sub one squared and then dividing by πΉ sub one. And when we do that, what we get is π sub one squared is equal to capital πΊ multiplied by π sub one multiplied by π sub two over πΉ sub one. Well, in fact, we still donβt have π sub one as the subject because we have π sub one squared. So, therefore, what we want to do is take the square root of both sides. And when we do that, weβre left with π sub one is equal to the square root of capital πΊ multiplied by π sub one multiplied by π sub two over πΉ sub one.

Well, now, we have this in terms of π sub one. What we can do is take a look at π sub two. What we can do is start in the same way as we have done previously. And we can say that πΉ sub two is equal to capital πΊ multiplied by π sub one multiplied by π sub two over π sub two squared. And then, once again, we can rearrange. And what weβre gonna have is π sub two all squared is equal to capital πΊ multiplied by π sub one multiplied by π sub two over πΉ sub two.

And itβs at this point we can use some more information that weβve been given in the question about the relationship between πΉ sub two and πΉ sub one. And that is that πΉ sub two is equal to a half πΉ sub one. And in fact, what we can do is use this to give us π sub two squared in terms of πΉ sub one because what we can do is substitute πΉ sub two for a half πΉ sub one.

So, now, what we have is π sub two squared is equal to capital πΊ multiplied by π sub one multiplied by π sub two over a half πΉ sub one. So, now, what we can do to remove the fraction from the denominator is multiply both sides of the equation by a half, which is gonna give us π sub two squared over two is equal to capital πΊ multiplied by π sub one multiplied by π sub two over πΉ sub one. So then, if we take the square root of both sides, what weβre gonna have is π sub two over root two is equal to the square root of capital πΊ multiplied by π sub one multiplied by π sub two over πΉ sub one.

Now, what we could do at this point to make π sub two the subject is multiply through by root two. However, weβre not going to do this. And the reason weβre not going to do this is because we can see that the right-hand sides are in fact the same because weβve got the square root of capital πΊ multiplied by π sub one multiplied by π sub two over πΉ sub one. So therefore, this means that we can equate the left-hand sides. So, doing this gives us π sub one is equal to π sub two over root two. So then, what we can do is multiply through by root two. And when we do this, what we have is root two π sub one is equal to π sub two. So, what this means is that one π sub two is equal to root two π sub ones.

So, therefore, we can say that π sub two is root two times bigger than π sub one. So, with this information, what we can do is solve the problem because what weβre asked to find is the ratio π sub one to π sub two. And that ratio is one to root two.