Question Video: The Relation between the Gravitational Force and the Distance between Two Bodies Mathematics

When two masses are a distance 𝑑₁ apart, the gravitational force between them is 𝐹₁. If the distance between them changes so that they become 𝑑₂ apart, the gravitational force 𝐹₂ between them becomes 1/2 𝐹₁. Find the ratio 𝑑₁ : 𝑑₂.

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Video Transcript

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

So, when we take a look at the question, we can see it concerns gravitational force. Therefore, we know what we’re going to be looking at is Newton’s law of universal gravitation. And what we have is a formula to help us. And this formula states that 𝐹 sub 𝐺 is equal to capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over π‘Ÿ squared, where 𝐹 sub 𝐺 is the gravitational force, capital 𝐺 is the universal gravitational constant, π‘š is the mass, and π‘Ÿ 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 we’ve been given to create a couple of equations. Well, first of all, what we want to do is create an equation in terms of 𝑑 sub one. So, to begin with, what we have is 𝐹 sub one is equal to capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over 𝑑 sub one squared.

Well, as we’d already stated previously, what we’re looking to do is make 𝑑 sub one the subject of this. And to do that, what we’re going to do is actually swap the 𝐹 sub one and the 𝑑 sub one squared. What this actually means mathematically is multiplying both sides of the equation by 𝑑 sub one squared and then dividing by 𝐹 sub one. And when we do that, what we get is 𝑑 sub one squared is equal to capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over 𝐹 sub one. Well, in fact, we still don’t have 𝑑 sub one as the subject because we have 𝑑 sub one squared. So, therefore, what we want to do is take the square root of both sides. And when we do that, we’re left with 𝑑 sub one is equal to the square root of capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over 𝐹 sub one.

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

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

So, now, what we have is 𝑑 sub two squared is equal to capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over a half 𝐹 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 𝑑 sub two squared over two is equal to capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over 𝐹 sub one. So then, if we take the square root of both sides, what we’re gonna have is 𝑑 sub two over root two is equal to the square root of capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over 𝐹 sub one.

Now, what we could do at this point to make 𝑑 sub two the subject is multiply through by root two. However, we’re not going to do this. And the reason we’re not going to do this is because we can see that the right-hand sides are in fact the same because we’ve got the square root of capital 𝐺 multiplied by π‘š sub one multiplied by π‘š sub two over 𝐹 sub one. So therefore, this means that we can equate the left-hand sides. So, doing this gives us 𝑑 sub one is equal to 𝑑 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 𝑑 sub one is equal to 𝑑 sub two. So, what this means is that one 𝑑 sub two is equal to root two 𝑑 sub ones.

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

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