While visiting Planet Physics, you toss a rock straight up at 11 m/s and catch it 2.5 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet's radius every 230 min. What are the (a) mass and (b) radius of Planet Physics?

Video transcript

Hey, everyone. So this problem is working with gravitational forces. Let's see what it's asking us while exploring my mysterious planet Syra. You throw a ball vertically upwards with the velocity of 14 m per second and catch it after 3.2 seconds. Meanwhile, your spaceship is maintaining a stable orbit around the planet at an altitude equal to planet Xaus radius, taking 245 minutes for a full revolution, determine one the mass and two, the radius of this enigmatic planets. Arax. Our multiple choice answers here are a one, 2.21 times 10 to the 28 kg, 24 times 10 to the second meters. B 13.39 times 10 to the kg to five point or sorry, five times 10 to the fifth meters. C 14.71 times 10 to the kg to six times 10 to the six m or D 15.42 times 10 to the kg to eight times 10 to the four m. OK. So the first thing we're going to do here is actually recall from our kinematics equations that V F equals V I plus a multiply by T where A is the acceleration or the gravity? Um Zara, our T is equal to half of the total time. So you throw the ball up so it goes up and then comes down. So the time that we're looking at here is going to be half of that, it's gonna be one half, multi 553.2 seconds. So that equals 1.6 seconds. And then our V F is zero because you've caught the ball and the eye is 14 m per second. That's the speed at which you initially throw the ball. So when we saw for our celebration on planet SIA, which we can determine from this um ball throwing event, we have V F minus the I divided by T we plug in our known values. So we have zero m per second minus 14 m per second, divided by 1.6 seconds. And we plug that into our calculators and we get negative 8. m per second. So the next thing we can recall is that the, from our gravitational force equations, the magnitude of the gravitational force, the gravitational field, excuse me, the magnitude of the gravitational field is given by G equals uppercase G M over our square. And here our G gravity is going to be this acceleration that we solve for. In the first part of the problem. We can also recall from Kepler's third. Yeah, that T squared equals four pi squared over G M multiplied by R Q. And so when we look at the values or the information that's given to us in the problem, we know that t our time that the spaceship is taking to make a full revolution around the planet is mess. We want to get that into standard units. So we'll just use our 60 seconds per minute conversion factor and that is 1. multiplied by or 10, 10 of the fourth seconds. And then the radius of the orbit of our spaceship is equal to two times the radius of the planet. Because the radius of the orbit is the radius of the planet plus the altitude of the spaceship, which is equal to the radius of the planet. So the radius of our overall orbit is equal to two multiplied by the radius of our planet Syra. And so the, and the time that we're looking at here is this time of our work. So when we go to Seoul for our radius or R sub Z, we are actually looking at here this radius where we have T O, the time of our orbit and R O our the radius of our orbit. So we can rearrange this equation to be T squared equals squared multiplied by G M or sorry. Four pi were divided by G M and then R O cubed, we can substitute N for two R Z cubed which when we simplify, that is going to be four pi square divided by G M multiplied by eight R as we swear. And we can multiply it through about four and eight. So we'll get 32 pi squared R Z multiplied by R Z E squared divided by G M. So we can pull out this rie squared and the um upper case geo gravitational constant. And, and recognize that that is the magnitude of our gravitational field. So this equation for our time of revolution greatly simplifies to 32 pi squared R Z divided by cheap our G. We so, so sorry, excuse me, we are trying to solve for this R sub Z the radius of I don't plan its name. Yes. And so we can arrange, we can rearrange this equation to isolate that variable. And we will have T squared multiplied by G divided by 32 5 squared. It's funny and our known values so T is 1.47 times 10 to the four seconds that quantity squared multiplied by G which we solve for in the first part, 8.7 seconds meters, 8.75 m per second squared, all divided by 32 pi squared. We plug that into our calculator and we get six times 10 with six m. And so that is our radius. So when we go up to our multiple choice answers the only one that has the correct radius is actually answer choice C but let's solve for our mass just to make sure that we're on the right track. So we can go back up to that um magnitude of gravitational field equation. And so for our maps, so I'll rewrite that here, G gravity is equal to upper case G multiplied by M divided by R Z squared. We can recall that G is our gravitational constant 6.67 times 10 to the negative 11. And that is in meters cubed per kilogram second squared. And then we can rearrange this equation to solve for our mass. So we'll have G multiplied by a radius R Z squared divided by our gravitational constant. So again, this gravity on this special planet we have determined is 8.75 meters per second squared. Our radius we just saw 46 times 10 to the six liters. That quantity is gonna be squared all divided by our gravitational constant 6.67 times 10 to the negative 11 meters cubed per kilogram second square. We plug that into our calculator and we get 4. times 10 to the 24 kg. So that is the answer for the mass of this plant. And that does align with answer choice C so C is the correct answer for this problem. That's all we have for this one. We'll see you in the next video