Two Jupiter-size planets are released from rest 1.0 x 10¹¹ m apart. What are their speeds as they crash together?

Video transcript

Hey, everyone. So this problem is dealing with gravitational force. Let's see what it's asking us. We have two twin exoplanets separated by a distance of two tanks 10 to the eighth meters that are released from rest and approach approach each other under mutual gravitational attraction until they collide together. We're asked to spine the speeds at the collision. If the mass and the radius are given as mass is equal to two times 10 to the 25 kg and the radius of each exoplanet is five times 10 to the five m. Our multiple choice answers are a 1.8 times 10 to the three m per second. B 2.6 times 10 to the three m per second. C 3.6 times 10 to the four m per second or D 3.7 times 10 to the five m per second. So the first thing we can do with this problem is recognize that there are no outside forces acting on these bodies. The only force acting is this mutual gravitational attraction or the gravitational force. And so we have a conservation of energy where our initial potential energy plus our initial kinetic energy is equal to sorry A K E is equal to our final potential energy plus our final Connecticut. And in turn, our potential energy 42 um bodies with a gravitational force is given by negative G M one M two divided by D. And our kinetic energy is given by one half and square. And so if we log this these two equations into our conservation of energy equation, we can start to isolate this speed variable V that we are ultimately solving for in this problem. Now, the problem tells us that the planets are released from rest. So our initial kinetic energy term is zero. So this equation then becomes negative G and the masses are the same. So M one and M two both equal M. So it will be G M squared divided by D equals negative G M squared divided by. Now, this D in the uh general equation is the distance between the two centers of, of masses. And so when they collide, they are only um separated by their, their radius or the radii. Uh so that would be two R and then plus our kinetic energy terms for both stars, they're both exoplanets will be one half M V squared. So one half and B squared plus one half and B squared. And so we can simplify this equation, we will, and this term to the left hand side of the equation and simplify the kinetic energy terms when we factor out G M squared. On the left hand side, you get G M squared multiplied by the quantity negative one divided by D plus one divided by two R is equal to M B squared. And so one of those masses cancel and then we can take the square route to isolate that velocity term. So V equals the square root of G M multiplied by negative one divided by D plus one divided by two R. And when we look at each of these terms, we have what we need to plug in and solve for this problem. So we can recall G or gravitational constant is 6.67 times 10 to the negative Newton meters squared kilogram squared. And that's multiplied by the mass which was given as two times to the 25 kg and then multiplied by one divided by negative one divided by D. That's the original distance that these planets were separated from each other. And that was given as two times 10 to the eighth meters plus one divided by two, multiplied by the radius of each which was given as five times 10 to the five m. When we plug that in to our calculator, we get a speed of 3.6 times 10 to the four m per second. And that is the final answer for this problem. We look at our multiple choice answers and it aligns with answer choice C so that's all we have for this one, we'll see you in the next video.