Video Transcript
A planet’s mass is 0.48 times the mass of Earth. The acceleration due to gravity at the surface of that planet is 0.12 times that on Earth’s surface. Given that the radius of Earth is 6.34 times 10 to the power of six meters, calculate the radius of the other planet.
In this problem, what we’re looking at is a relationship between the mass, the radius, and the acceleration due to gravity at the surface of different planets. So because we have this information, what we’re going to look at is Newton’s law of universal gravitation. And what we have in this scenario is a formula that we can use. And that is that 𝑔 is equal to capital 𝐺 multiplied by 𝑚 over 𝑟 squared, where 𝑔 is the acceleration due to gravity on the surface of the planet, capital 𝐺 is the universal gravitation constant, 𝑚 is the mass, and 𝑟 is the radius.
Well, before we use this formula, what we’re going to do is write down the information we’ve been given. So first of all, we know that a planet’s mass is 0.48 times the mass of the Earth. So we can say that 𝑚 sub 𝑃 is equal to 0.48𝑚 sub 𝐸. Then what we’re told is that the acceleration due to gravity at the surface of the planet is 0.1 times that on the Earth’s surface. So we can say that 𝑔 sub 𝑃 is equal to 0.12𝑔 sub 𝐸. And then we know that the radius of the Earth is equal to 6.34 times 10 to the power of six, and that’s in meters, which is what we need when we’re looking at this kind of problem. So what we’re trying to find though is the radius of the planet.
So now that we have all the information that we’ve been given, let’s see if we can actually work out the answer to the problem using the formula that we have. So the first thing we’re going to do is in fact look in terms of the Earth. So we can say that 𝑔 sub 𝐸 is equal to capital 𝐺 multiplied by 𝑚 sub 𝐸 over 6.34 times 10 to the power of six all squared.
It’s worth noting here that you might think, well, can’t we just put in what we know the acceleration due to gravity at the surface of the Earth to be? However, in the question, it doesn’t state whether this is 9.81 or 9.8 or in fact any other degree of accuracy. So what we’re going to do is actually leave things in terms of 𝑔 sub 𝐸 and 𝑚 sub 𝐸.
So we’re now going to rearrange the formula, so change the subject of the formula, to make capital 𝐺, the universal gravitation constant, the subject of the formula. So what we’re gonna do is multiply through by 6.34 times 10 to the power of six all squared and divide through by 𝑚 𝐸. And this gives us 6.34 times 10 to the power of six all squared 𝑔 sub 𝐸 over 𝑚 sub 𝐸 is equal to our capital 𝐺, the universal gravitation constant. Okay, great, now we have this in terms of 𝑔 sub 𝐸 and 𝑚 sub 𝐸. We’re gonna move on to the planet.
Now, when we’re looking at this other planet, what we’re gonna remember is the first two facts that we were given from the question. So now we can write out our equation. So we’ve got 0.12𝑔 sub 𝐸 is equal to 6.34 times 10 to the power of six all squared 𝑔 sub 𝐸 over 𝑚 sub 𝐸 multiplied by 0.48𝑚 sub 𝐸 all over 𝑟 sub 𝑃 squared. And we get that because we know that our universal gravitation constant, capital 𝐺, is 6.34 times 10 to the power of six all squared multiplied by 𝑔 sub 𝐸 over 𝑚 sub 𝐸, and then the other two relationships we’ve just mentioned as well.
So now if we multiply through by 𝑟 sub 𝑃 squared, we get 0.12 multiplied by 𝑟 sub 𝑃 squared multiplied by 𝑔 sub 𝐸 equals 6.34 times 10 to the power of six all squared multiplied by 𝑔 sub 𝐸 over 𝑚 sub 𝐸 multiplied by 0.48𝑚 sub 𝐸. So next, on the right-hand side of the equation, we can divide through by 𝑚 sub 𝐸. And then we can divide through both sides of the equation by 𝑔 sub 𝐸. So we’ve now canceled out our 𝑔 sub 𝐸’s and 𝑚 sub 𝐸’s. So what we are left with is 0.12 multiplied by 𝑟 sub 𝑃 squared is equal to 6.34 multiplied by 10 to the power of six all squared multiplied by 0.48. So now if we divide through by 0.12, we’re gonna get 𝑟 sub 𝑃 squared is equal to 6.34 multiplied by 10 to the power of six all squared multiplied by 0.48 divided by 0.12.
So therefore, we can say that 𝑟 sub 𝑃 squared is equal to 1.607824 times 10 to the power of 14. Then, taking the square root of both sides of the equation, we get 𝑟 sub 𝑃 is equal to 12680000. So therefore, we can say that the radius of the other planet is going to be 1.268 times 10 to the power of seven meters.
