 What goes up must eventually come down. Or is it? In 1865 Schuylverne proposed in his book the Laterale Alune from the earth to the moon that if we would shoot a projectile from a cannon with sufficient initial speed, it would actually go all the way to the moon and never fall down. In chapter 3 of the book he even gives us the speed. According to the book we are needing a speed of 12,000 yards per second. This is in metric unit about 1.1 times 10 to the 4 meters per second. Now the question I want to be solving in this video is Is this number actually accurate? Would an object if I shoot it upwards with 1.1 times 10 to the 4 meters per second as Schuylverne predict not come back down, but actually go all the way to the moon or beyond. What I want to calculate is what would be the speed of an object needed to leave the earth and not be pulled back by gravity. So I'm going to do this by looking at conservation of energy. So we have energy final is energy initial plus worked on by any external forces. So in my case once I'm very very far up at the highest point of my trajectory I will have potentially potential energy and kinetic energy and initially when I start from the surface of the planet I have my potential energy initial plus my kinetic energy initial plus the work done by any unbalanced force. In this case if I'm at my highest point I assume if I just make it to my highest point I will actually come to a stop. So at the highest point I will have no kinetic energy while at the initial point I have some potential energy from gravity and I have some kinetic energy from gravity. Now we have one problem for the potential energy of gravity. We cannot use Mgh. Why? Because G the further I go away from the planet the lower G will be. So definitely the change in potential energy is not Mgh. But I can use the old trick that the change in potential energy is minus the work done by gravity. So in case of so instead of considering the potential energies I'm going to be considering the work done by gravity and gravity being the only force that actually acts on my object. If I do my free body diagram here of my projectiles, I'm going to be shooting up. I'm going to have gravity pulling back down to the earth and we're traveling in upward direction. It's called this DX. So if I simplify this equation, what I'm going to get is that the kinetic energy initial so the one-half mv initial square is equal to minus the work done by gravity. So how do we calculate the work done by gravity? It's the force times displacement. In the case that the force is changing we will have to do the integration. So fg dot dx. Now I'm well aware that most of you have not taken calculus 2 yet, so you don't know what the what I'm actually doing here with the integration, but once you did take calculus 2 you can come back here and look if you now understand what I'm doing. For the moment just follow what I'm doing and we're going to see what the number is that I come up with. Also, I hope this gives you a good idea of why it is important to take calculus 2 and learn about integration. This must be equal to minus I'm going to be integrating from where I start from the surface of the planet so our initial and I want to go to infinity and the force of gravity here we're using Newton's universal law of gravity. So we have capital, gravitational constant, times mass of the projectile, times mass of the planet Earth over the distance squared and now the x, and if I look at my free body diagram the x and fg is pointing in opposite directions, so I have a cosine of 180 So I'm going to put a minus here so the minus actually removes the minus, so I can simplify as g is a constant, mass is a constant, mass integral from initial position to infinity from x to the power of minus 2 dx So what does this give me? This gives me g times mass times mass times x to the power of minus 1 over minus 1 plus integration constant evaluated from our initial to infinity which gives me g m 1 over infinity minus 1 over infinity plus 1 over our initial. Now let me rewrite this so we had one half and v squared is equal to capital G mass of the projectile, mass of the planet over r So the mass of the projectile interestingly falls out and what I get is v is equal to square root of 2 times capital G times the mass of the planet. I'm leaving from times the distance from the center of that planet I'm leaving for Now something that's interesting to see here is that if I would be leaving from a planet with less mass, I would need less initial Velocity to do so. So to go on an interstellar mission instead of leaving from the surface of the earth It could be actually interesting to leave from the surface of the moon or eventually from the surface of the Mars Even better than leaving from the surface if we're Leaving from a bigger distance from the center of the object for example from the orbit from a space station We would also need much less velocity So to go to Mars most likely we will not leave with a rocket directly from earth We will first transport everything into the earth orbit and then leave from the orbit around the earth Or even better from the orbit around the moon to go To Mars and to go on interstellar mission if we could make a pit stop at the Mars First transport everything to the orbit of Mars and then for our interstellar mission leave from the orbit of Mars We would need much less energy to do so. But now back to Silver's numbers of 12,000 yards per second or 1.1 times 10 to minus 4 meters seconds Well, what happens if now I plug in my numbers G as 6.6 7 times 10 to minus 11 Newton square meters per kilogram square the mass of the earth as 5.97 times 10 to the 24 kilograms and The radius of the earth assuming we are leaving just from t-level Then we would have a radius of 6.37 times 10 to the 6 meters Now what I will get is the following 1.1 times 10 to the 4 meters per second So 1.1 times 10 to the 4 meters per second is the speed needed So if you shoot the projectile at this speed at the surface of the earth, it will never come back down It will actually go to Infinity that's how we call this velocity the escape velocity Now back to the number of chelverne Look at this. It is absolutely exact. So in 1865 a French author correctly predicts the escape velocity from earth Just think about it 1865 there was no space program yet. There were no rockets yet So he was really ahead of his tongue Actually the speed that he gave to shoot the gun to leave the earth is is enough to go to infinity He only wants to go to the moon. So to go to the moon. You actually need less than that Why because somewhere between earth and moon? so if you have earth here and You have the moon here Somewhere in between the gravity of the moon Will actually be equal than the gravity of the earth So you don't even need to shoot the projectile to infinity All you need to shoot it is to that point where the gravity of the moon is Equal to the gravity of the earth and then we'll fall down onto the moon automatically. So The speed that should when we're was giving that is required to leave earth Is Sufficient to not just go to the moon. It would have been sufficient to go all the way to infinity