 Let's solve a question on gravitational potential energy. Here we have a system which consists of 3 particles each of past 5 kgs which are located at the corners of an equilateral triangle with sides of 30 m and we can see 3 particles over here. We need to figure out the potential energy of the system and second part is assume the particles are released simultaneously, describe the subsequent motion of each. Will any collisions take place? Alright, before I get into this, why don't you pause the video and first attempt this one on your own. Okay, hopefully you have given this a shot. Now potential energy of a system, let's say if we had two particles, if we had any two particles like this, let's say this is of mass m1 and this is of mass m2. Now the gravitational potential energy of this system, this is given by u which is equal to minus g m1 into m2 divided by the distance between their centers and if let's say that is r, this is divided by r, here the system has 3 particles. So the total potential energy of the system will be the sum of potential energy of each of the pairs that is the sum of the potential energy of mass 1 and 2 plus the potential energy of mass 2 and 3 plus the potential energy of mass 3 and 1. So the u, the u of the system, this is equal to potential energy of the first mass and the second mass plus the potential energy of the second and third mass and the potential energy of the third and the first mass. So let's calculate that, let me remove this part over here. So for the first and the second particle, the potential energy, this is minus g and the masses are the same. So I'm writing m2 because m1 and m2 would just become m2, masses are the same, m2 divided by r which is still 30 but let's just write r and we will add this 3 times. Even u, the potential energy of the second and third particle, even that would be minus gm2 by r because the distance is still the same, this is an equilateral triangle and even the masses are the same. So really what we are doing is we are multiplying this with 3, we are adding it 3 times and now when we put in the numbers this becomes minus 3 into g into m2 which is 25 divided by 30. After putting in the value of g and remember that g is 6.67 into 10 to the power minus 11 meter cube divided by kilogram second square. Okay so after putting in this number, the potential energy is 16.6 into 10 to the power minus 11 joules. And now the second part is assume that the particles are released simultaneously, describe the subsequent motion. So let's think about what will happen, when these particles are released there is already a gravitational force of attraction among all of these 3 particles, there is a gravitational force in this direction, the particle 1 is experiencing a force in this direction due to particle 2, 2 is experiencing a force in this direction due to 1, but 1 is also experiencing a force in this direction due to 3 and the same 3 is also experiencing a force in this direction due to 1. If you think about 2 and 3 they are also exerting gravitational forces on top of each other and as a result of which look at all the forces that are acting on any one particle, there are 2 forces acting. So if you think about the resultant, there will be a resultant which will be in this direction for particle 1, in this direction for particle 2 and in this direction for particle 3. Turns out when you release them simultaneously, that is when you release them together, they all start moving towards the center of the triangle and after a point of time this is where they will land up, they will all come towards each other because of the gravitational force of attraction among all of them. So yes there will be a collision and they will collide together right at the center of the equilateral triangle. You can try more questions from this exercise in the lesson and if you are watching on YouTube, do check out the exercise link which is added in the description.