Humans have always been explorers. From ancient nomads to Columbus, our hunger to find new places and make new discoveries can never seem to be filled. With our planet fully explored, we have now turned out sights to the stars, and are now beginning to explore space. Before we could leave the planet, however, we had to face one of the greatest forces in the galaxy: gravity. Thankfully, math and science were on the astronauts’ side: they were able to apply the laws of physics to safely leave the planet. Through a series of experiments and discoveries, they discovered the speed and trajectory necessary to propel their rockets away from Earth’s gravitational pull. They discovered Earth’s escape velocity. Like many others, I have always been interested in space, planets, and the forces that bind the universe together. One day I want to be an astronaut. Before I can make it to Mars, however, I have to be able to leave Earth. For this reason, I wanted to learn more about escape velocity. In this exploration, I hope to understand escape velocity, understand how to derive the equation, and be able to fill in any variables without leaving the planet.Mathematicians from the earliest recorded civilizations labored to apply their skill to what they saw beyond the sky. However, it took until the eighteenth century before the laws of the universe were described and given accurate formulas. Isaac Newton found ways to understand forces and how they act on one another. Gravity is one of his primary findings. Newton’s formulas describing gravity were so accurate, many are still used today so it is no surprise that escaping gravity uses many of his equations. Escape Velocity can be defined as the speed needed to leave a body’s gravity with no speed left over at the end. It is the minimum speed needed, and therefore the most efficient. For rocket scientists, it is very important to be efficient because thrust, and thereby fuel, has a growing cycle. The more fuel a rocket has, the more it will weigh. The more a rocket weighs, the more fuel it needs. This is why it is important to waste as little thrust as possible. When I first began researching Escape Velocity, I was easily able to find the equation.V?=2GM/r(Where V? is Escape Velocity, G is the gravitational constant, M is the mass of the escaped body, and r is the radius of the escaped body.)I then asked myself how this equation was derived, something many of the internet sites I reviewed failed to explain. To answer this question, I had to turn back and analyze some of Newton’s other discoveries. More specifically, I had to understand the law of conservation of energy. Energy cannot be created or destroyed, so there must always be the same amount of energy in a system before an action as there is after. There are two types of energy, kinetic and potential. Kinetic energy comes from movement. If something is not moving, it does not have any kinetic energy. Potential energy is the possibility of an object’s future movement in relation to other objects. For example, an object a few feet in the air has the potential to fall. The law of conservation allows for energy to flow from kinetic to potential without any loss. The equation for this law is: U? + K? = Uf + Kf(Where U is potential energy, K is kinetic energy, ? is initial, and f is final.)I wanted to test this law for myself in a simple way, so I designed an experiment that allowed me to compare a recorded speed of an object with the speed predicted by Newton’s equation. I held a block up 2 meters and, using a camera to record the block’s position over time, measure the speed of the block just before it hit the ground. Based on the video recording, I found that the block was moving at 6.24 m/s. I then used the law of conservation of energy to see how closely the formula would match the recorded speed. While holding the block, it is not moving and therefore has no kinetic energy. Just as the block is hitting the ground, all of its potential energy is expended. With these facts in mind, I know that K?and Uf are zero. This helps to simplify the equation to a point that we can expand the potential and kinetic equations.U?= Kfmgh = 12mv²(Where m is mass in kilograms, g is the acceleration of gravity in meters per second squared, h is height in meters, and v is velocity in meters per second.)From here, we can rearrange the equation to solve for v, the velocity right before the block hits the ground. It was interesting to me that the mass of the block did not matter in the final equation. This makes sense though, as all objects fall at the same rate. I plugged in all the variables and found that the velocity should have been 6.26 m/s.v = 2gh = 2(9.8)(2) = 39.2 = 6.261(m*s-2)*m = m*s-1 While working through this equation, I remembered that velocity is a vector, so it can not have a negative magnitude. In future problems like this, I will not need to use a plus-minus symbol after a square root. In all, though, the equation worked well. Any difference between the math and the recorded speed was likely because of a limitation of the camera. Satisfied that the equation did work in a real-world example, I turned my attention to applying the law of conservation of energy to find escape velocity. There is a separate potential energy for gravity, so I added it to the base equation. Ug + U? + K?= Ug + Uf + KfThen I thought about what the equation should be describing. Before the action, there should be no potential from height, as the rocket would be on the ground. There would be energy from gravity and movement. Afterwards, there should be no movement at all, because that is the point of escape velocity.Ug + K0= 0K0= -UgThis was a nice and simple equation, so I expanded the equations and rearranged them to find v, the escape velocity. Velocity is a vector, so it will have a positive magnitude.12mv² = -(-GMmr)(Where m is the smaller mass in kilograms, v is escape velocity, G is the gravitational constant, M is the larger mass, and r is the distance between the center of the two masses.)v² = 2GMr V? = 2GMr I was pleased to find that the escape velocity I had derived was the same as the one that could be found on the internet. I was also surprised to see that the mass of the rocket does not matter. Any object with the right speed could escape gravity. The rocket’s mass is only needed for calculating how much speed an amount of fuel can give.Now I was faced with a new problem. There are two variables and a constant in this equation. How did the early scientists calculate the mass and radius of Earth? What is the gravitational constant? I wanted to be able to answer these questions for myself, without specialized equipment that I would not understand. I started with the mass of the Earth. We can actually calculate the mass using the moon and knowledge about circles. The moon is kept in orbit by centrifugal force, the same force that keeps people in their seats in an upside-down roller coaster loop. In a roller coaster, this force comes from friction, but the moon is not touching anything else to cause friction. For the moon, centrifugal force is caused by gravity. I set these equations as equal to one another and solved for the mass of Earth.Fgravity = FcentrifugalGMmr2 = mv2rMearth = v2rG As with many other of the equations I have been arranging, the mass of the smaller object does not matter. This means that scientists can measure the mass of any planet just by sending a satellite to orbit the planet. I think that this will be important in exploring our galaxy, as the mass of a planet can tell scientists a lot about what it is made of. The velocity of the moon is angular, meaning that it travels in a circle. We can find the velocity by dividing the distance the moon travels by the time it takes to make a loop. The distance is really just the circumference of a circle with a radius the distance between the Earth and the Moon. We can then plug this back into the original equation and simplify.vmoon = 2rTMearth = (2rT)2rGMearth = 42r3T2G(Where r is the distance in meters between the Earth and the moon, T is the time in seconds that the moon takes to complete an orbit, and G is the gravitational constant.)The moon travels around the Earth once every 29.53 days, translating to 2,551,392 seconds. The only other information I needed for this was the distance between the Earth and the moon. This can be found by holding up a coin and lining it up with the moon at night. = tan-1(xy)tan() = zww = zcos() After completing this experiment for myself, I found that the Moon is approximately _____ meters away. This is a bit off from the NASA official distance of 384,400,000 meters, but still close enough.Now that I had all the information, I could plug in the time of a rotation, the gravitational constant, and the distance between the Earth and the moon. We can see that the Earth is roughly 5.83*1024kg. By setting up the equation with just the units, many of the exponents cancel out, leaving kilograms as the final unit.Mearth = 4(2)(384,400,003)(2,551,0002)(6.67408*10-11) = 5.83*1024 (m3)(s2)(m3kg-1s-2) = kg On a NASA fact sheet, the Earth is said to have a mass of 5.97 * 1024. By using the equation for percent error, I know that my measurements were too low by 2.345%.% Error = Calculated – ActualActual*100(5.83*1024)-(5.98*1024)(5.98*1024)*100 = -2.345% Delighted that my equations had given me an acceptable margin of error, I began to research how early astronomers tried to calculate the radius of Earth. The first recorded attempt was around 100 BCE, by a Greek scientist named Eratosthenes. Although his measurements were off by a noticeable amount, the idea behind his measurements is still useful. He used trigonometry to calculate the angle of a sector of the globe. He knew the distance between two cities, Alexandria and Sienna. Eratosthenes put a stick with a length of one stade, a Greek measurement, in Alexandria. He then measured the shadow of the stick at the exact time that the sun was immediately above Sienna. This time was decided by looking into a well in Sienna. If one could see their shadow in the water at the bottom of the well, it was assumed the sun was right overhead. Once Eratosthenes had calculated the angle of the sector, he could find the radius of Earth. Unfortunately, Eratosthenes got many of his numbers wrong, so he got an inaccurate radius. Modern scientists have rerun his experiment with the help of clocks, cameras, and better measuring tools. They get much more accurate measurements. According to NASA, Earth has a radius of 6,378,000 meters. I tested this number by using two equations: one that calculates the force of gravity using the mass and radius of the earth, and one that finds the same force using acceleration from gravity.Fgravity = GMmr2 = gm (Where G is the gravitational constant, M is the mass of the Earth, m is the mass of another object, r is the distance between the two objects or the radius of Earth, and g is the acceleration from gravity.) Surprisingly, I had all the variables besides the radius already figured out. I found the mass of Earth earlier in the exploration and proved the acceleration of gravity in my first experiment. I could rearrange the two equations to solve for radius and plug in my numbers.GMmgm = r2r = GMg = (6.67408*10-11)(5.83*1024)9.8 = 6,301,000 meters (m3kg-1s-2)(kg)m*s-2 = m When I used the mass I had calculated, I got a radius of 6,301,000 meters. This has a percent error of 1.21% when compared to the NASA approved radius. However, if I use the NASA calculated mass, I get a radius of 6,376,000 meters. This has a percent error of 0.0314%. No matter which measurement I end up using, it does find the radius of Earth in an acceptable margin of error. The last number I needed to find was the Gravitational Constant. This constant was first described by Newton, but it took over a century to measure it. Henry Cavendish calculated the attracting pull of two masses using complicated machinery. The constant he measured has remained nearly the same as modern scientists redo the experiment. I cannot measure the Gravitational Constant. It is important for me to remember that I cannot do everything by myself, so I am trusting the numbers found by other scientists to plug in for the Gravitational Constant. Finally, I had all the information needed to plug into my equation for escape velocity. There is definitely a difference if I use my numbers instead of NASA’s. I did the equation twice to show this, first with my information and second with official data.V? = 2GMr = 2(6.67408*10-11)(5.83*1024)(6.301*106) =11,113 m*s-1 = 11.1 km*s-1 V? = 2GMr = 2(6.67408*10-11)(5.98*1024)(6.378*106) = 11,187 m*s-1 = 11.2 km*s-1 An official NASA fact sheet says that the escape velocity of Earth is 11.2 km*s-1, thereby proving that the math behind escape velocity works. Even using my data, which anyone could calculate using math, only had a percent error of 0.662%. Is this velocity of any use to scientists? What it means is that any rocket with the intent to leave the planet must be moving at 11.2 kilometers every second from the surface of Earth. Mass does not move that way, however. You can not simply pick a speed and expect an object to begin moving at it. Objects can accelerate and decelerate to hit a wanted speed, but nothing can begin moving at a high speed. For this reason, it is silly to think that a rocket could move at 11.2 kilometers a second from the ground. Instead, many rockets use several stages to get into the correct speed at a different altitude. The each stage will release fuel from a different storage area, continuing to accelerate the rocket. When the fuel tank runs out, it detaches itself from the main rocket, reducing the weight. Acceleration is difficult to obtain with gravity because they are both vectors. Often, they are forces pulling in the exact opposite directions, gravity pulling down and acceleration pushing up. This means that in order to go up, the thrust of a rocket must have a magnitude of over 9.8 meters per second squared. In order to reach a high speed, it will have to accelerate much more than that.The Earth’s atmosphere is, on average, 480 kilometers thick. Let’s say that we want to have reached escape velocity by 100 kilometers above the surface. There is an equation that I rearranged to help me find the acceleration needed to reach escape velocity by that altitude. vx2 = v0x2 + 2ax(x – x0) (Where vx is the wanted velocity in the x direction, v0x is the starting velocity in the x direction, and ax is the acceleration in the x direction.)ax = vx22(x) = (athrust – agravity)Since there is no speed or altitude at the beginning of launch, I could cancel out v0x and x0. Then I replaced the velocity with the escape velocity equation, modifying it to have the added altitude. Finally, I plugged in the values for the Mass of Earth, the radius, and the added altitude. vx = 2GM(r+x) = 2(6.67408*10-11)(5.83*1024)((6.301*106)+100000)= 11036 m*s-1athrust = vx22(x)+agravity = (11036)22(100000)+ 9.8 = 619 m*s-2(m*s-1)2m + m*s-2 = m*s-2 A rocket in these conditions would need to gain 619 meters per second of speed every second. This is a huge acceleration just to get to the speed needed to leave the orbit. Most of the rockets NASA sends out, however, do not need to reach escape velocity at first. Instead, some fall back to Earth and others are pulled into orbit. Even some of now distant satellites orbited Earth for a while, because the escape velocity is much lower that far from the earth’s gravitational pull. Scientists may not use escape velocity directly very often, but they do need it for many other equations. Overall, the escape velocity equation has many variables that seem complicated at first. However, the majority of them can be found using one’s eyes, the laws of physics, and a bit of math. It is an amazing feeling to plug in my own variables and get the right velocity. It does not take fancy equipment to derive complicated measurements.