# A Scientific Guide to Understanding Escape Velocity

## What is Escape Velocity?

When an object reaches **Escape Velocity** it will eventually travel an infinite distance away from the planet, star, or moon that is exerting a gravitational pull on it, and *never* fall back. Escape velocity is, therefore, the velocity at which an object must have in order to escape the gravitational pull of a planet or object. That is of course if nothing else happens, such as an astronaut deciding to reverse his course. The velocity that is needed to escape to an infinite distance is an "Instantaneous Velocity." This means it does not need to be maintained rather the object will gradually slow down but will never stop.

Escape velocity is the velocity at which an object must have in order to escape the gravitational pull of a planet or object.

## What Determines Escape Velocity?

The escape velocity needed depends, of course, on *what* we are escaping from. In other words, the **mass** of that body (e.g. planets, moons, the sun, and satellites). It also depends on how **far** (distance) we are away from that body initially. Armed with just those two facts, we can calculate this figure for any "body" in the Universe using a specific type of working procedure.

## Calculating Escape Velocity

How, exactly, is escape velocity calculated? Imagine, for example, that we could drop an object from an infinite height above the Earth to a distance, **"r"**, from the center of Earth. Now, imagine that there was a gigantic trampoline at that position in space. The object would hit that trampoline and rebound with the **same** velocity, the escape velocity needed to return to the point an infinite distance away (if the trampoline were perfect). Obviously, this experiment would take an infinitely long time and would be quite costly.

Instead, let us conduct a mathematical experiment. In the example we have just looked at, the Gravitational or Potential Energy (energy an object holds due to its place within a gravitational field) of the object we dropped turned into Kinetic Energy—the energy of motion. The value of this kinetic energy will give us the final velocity, or the escape velocity. However, it is not quite that simple. Why? Because we need to keep in mind that the acceleration is not constant. The gravitational force **increases** as the object gets nearer to the Earth. We need to, therefore, use calculus to arrive at a formula to calculate the escape velocity.

## Integrating Gravity

The increasing speed and kinetic energy of a falling object are due to the **size** of the force acting on it and on the **distance** it falls. An object falling from ten feet, for example, has *twice* the kinetic energy it would have falling from five feet. There is virtually no difference in the strength of gravity over these small distances. However, gravitational force weakens dramatically over thousands of miles and eventually becomes zero at infinity. To illustrate how gravity plays a part in all this, let's take for example the popular fidget spinner. On Earth, the velocity of the fidget spinner is largely influenced by the Earth's gravitational pull. This means, no matter how much effort you put into spinning the fidget, eventually the spinner will come to a stop. This is because of Earth's

To illustrate how gravity plays a part in all this, let's take for example the popular fidget spinner. On Earth, the velocity of a fidget spinner toy is largely influenced by the Earth's gravitational pull. This means, no matter how much effort you put into spinning the fidget, eventually the spinner will come to a stop. This is because of Earth's gravitational force that is put upon it. Now, if we were to perform the same action in zero gravity the same fidget spinner could spin an infinite number of times as there is no gravitational being put against it.

To better understand how to calculate escape velocity, we have to separate the fall of the object into small segments where the gravity does not change. Then we multiply the small distance by the gravity at that point and add all these segments together. This mathematical process is called **integration**. To use integration, we need the formula for the gravitational force between two objects.

*Gravitational Force*,**F = GMm/R²**(**G**is the gravitational constant;**M**is the mass of the Sun, planet or moon;**m**is the mass of the object;**R**is the distance between the center of both masses).*Kinetic Energy*= sum of all (force x small change in distance).

The following equations appear complicated, but they leave us with a simple result. The interval of integration is between infinity and *r*. We take the constants (the things that do not change) outside of the integration.

## Integration

mv²/2 = GMm ∫ (1/R²) dR = GMm(1/r) – GMm(1/∞).

As 1/∞ = 0 we can write:

mv²/2 = GMm(1/r) – 0.

mv²/2 = Gmm/r.

v²/2 = GM/r .

v² = 2GM/r.

Escape Velocity, **v**** = ****√(2GM/r).**

Notice that the mass of the object does not appear in the final equation. Objects of different sizes and masses have the **same** acceleration in the same gravitational field. This was shown dramatically when an Apollo astronaut dropped a feather and a hammer together on the Moon, and they landed simultaneously. However, the amount of fuel needed to reach escape velocity would be proportional to the mass of the object. *Georgia State University* has developed a working formula to help calculate Escape Velocity. If the formula above seems complicated, give it a try.

## Escaping Earth

Let us find what this velocity is for, say, a satellite at a height of 100 miles or 160 kilometers. That is the radius of the Earth + (160 x 10³) meters = (6371 x 10³) + (160 x 10³) = 6531 x 10³ meters. (Remember mass and distance figures must be in kilograms and meters because **G **is a metric constant).

- Escape Velocity, v = √[2 x 6.672 x 10‾ ¹¹ x 597.8 x 10²² x (1/6531) x 10‾³].
- v = √(0.0122 x 10¹°).
- v = 11 x 10³ meters per second or 11 kilometers per second.

That is 6.875 miles per second or 24,750 miles per hour. Most sources give Earth's escape velocity at 11.2 kilometers per second, but this is calculated from ground level and would result in immediate destruction in Earth's atmosphere for any spacecraft. Another interesting fact comes from this equation, and it is that an object’s escape velocity also equals its circular orbital velocity multiplied by √2.

## Stronger Forces

The only problem with the concept of escape velocity is that it assumes the object and the body it is escaping from are all that exist in the Universe. If a spacecraft reached the Earth's escape velocity, it would still be under the influence of a much larger body. In the solar system, the Earth is dwarfed by the Sun. So, to escape our solar system, a spacecraft would need to reach the escape velocity of the Sun at our position in space. We can find that velocity by multiplying Earth's circular orbital velocity by the square root of 2. We arrive at a final velocity of 94,747 miles per hour needed to escape our solar system.

**© 2017 The Daily Syndicate**

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