A small moon orbits its planet in a circular orbit at a speed of 7.5 km/s. It takes 28 hours to complete one full orbit. What is the mass of the planet?

Video transcript

Hey, everyone. So this problem is dealing with Kepler's law. Let's see what it's asking us. We have a satellite moving around its host plant qunar along a circular path at a given speed of 9. kilometers per second. It tells us that it takes 34 hours to complete a single orbit. And we're asked to calculate the mass of the parent body. This Quinta on our multiple choice answers are a 3. times 10 to the 25 kg. B 2.26 times 10 to the 28 kg C 3.27 times 10 to the 24 kg for D 2. times 10 to the 26 kg. So the first part or the first step to this problem is we're calling Kepler's Third law, which is given by the equation T squared is equal to four pi squared are cubed divided by G M where R is the radius of the circular path. And so we can also recall our uniform circular motion equation for velocity is given by B equals two pi R over T. So when we rearrange, the second equation for radius in terms of velocity. Because we were given velocity, we get radius is equal to velocity multiplied by T the period of orbit divided by two pi. So we can plug this into our Cutler's third law equation. First, we're going to um isolate or isolate the variable that we're asked for mass. And so when we rearrange the equation to get mass alone, on the left hand side, we get mass as equal to four pi squared divided by G T squared multiplied by R Qed. Now we've solved for R in terms of velocity. So we're going to plug it in. So that's V T divided by two pi and that whole term cut. So that looks like when we solve this out, four pi squared multiplied by V cube T cubed divided by G T squared multiplied by two cubed I cubed. And so we can simplify this greatly. So four is so pi squared, we can um cancel with pi cube. So just pi to the first, same with the four that becomes to the first T squared cancels with T cubes, that's just one T left. And so we are left with B Q T divided by two pi G. And now we have everything we need to solve this problem from the problem. We're given velocity is equal to 9. kilometers per second. So we're gonna keep that in standard units 9.2 times 10 to a third meters per second RT is our period of orbit. And that was given as 32 hours. Again, we're gonna multiply that by 3600 seconds per hour to keep us in standard units. And that comes out to 1. times 10 to the fifth seconds. And then G is our gravitational constant. We can recall that that 6.67 times 10 to the negative 11 m cubed per kilogram second squared. And so when we plug all of that in to our mass equation, we've got 9.2 times 10 to the third meters per second. And that quantity cubed multiplied by 1.152 times 10 to the fifth seconds. All of that divided by two pi multiplied by 6.67 times 10 to the negative m cubed per kilogram second squared, plug that all into our calculator. And we are left with 2. times 10 to the 26 kg. When we look at our multiple choice answers for this problem that aligns with answer choice D. So that's all we have for this one. We'll see you in the next video.