Have you ever wondered how fast a cannonball would have to be fired into the sky for it to leave the world behind and fly into space? Well, the speed needed for this to happen is called the escape velocity.
Escape velocity is a concept used for any field that exerts an attractive force on an object. Simply put, it's the speed required for an object to escape the influence of the attractive field starting from a certain spatial point. In this case, we're specifically talking about the gravitational field created by the Earth.
To put it simply, a rocket must travel at a speed greater than the escape velocity of the Earth to travel into space. So, if you want to explore the final frontier, you better make sure your rocket is ready to achieve escape velocity!
The formula for the escape velocity is derived from Newton's law of gravitation, which states that the force of gravity between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The formula for the escape velocity is given by:
v_esc = \sqrt{\frac{2GM}{r}}
where G is the gravitational constant (with an approximate value of 6.67 \times 10^{-11} m^3/kg \cdot s^2), M is the mass of Earth (approximately 5.97 \times 1024 kg), and r is the radial distance between them measured in m (whose presence indicates spherical symmetry).
Imagine an object like a ball or a space rocket on the surface of the Earth. If it's not moving, it has energy But push, its velocity will determine the amount of kinetic energy it has (and its total energy).
Now, let's think about a point infinitely far away from Earth. Since the radial distance is in the denominator of the potential energy formula, as the distance gets larger, the potential energy approaches zero. This means that a stationary object at an infinite distance from Earth has zero energy. But if the object has a certain speed, it will have kinetic energy and, therefore, total energy.
According to the principle of conservation of energy, the total energy of an isolated system remains constant. Since we're only looking at the Earth and the object, the total energy of the object on Earth must be the same as the total energy of the object at an infinite distance.
With this in mind, we can classify the possible trajectories of an object:
That's correct! The escape velocity is the minimum velocity required for an object to escape the gravitational pull of a planet or other celestial body. For the Earth, the escape velocity is approximately 11.2km/s, which is independent of the mass of the object. This means that any object, regardless of its mass, must achieve a velocity of at least 11.2km/s to escape the Earth's gravitational pull and reach an infinite radial distance. It's a key concept in space exploration and understanding the dynamics of celestial bodies.
Yes, that's correct! To calculate the escape velocity for Mars, we can use the formula:
v = √[(2GM)/r]
where G is the gravitational constant, M is the mass of Mars, and r is the radius of Mars. Plugging in the given values, we get:
v = √[(2 * 6.67e-11 * 6.39e23)/3.34e6]
v = √(12.65e13)
v = 5.05e3 m/s
So the escape velocity for Mars is approximately 5.05 km/s. This means that any object on the surface of Mars must achieve a velocity of at least 5.05 km/s to escape the gravitational pull of the planet and reach infinite radial distance.
If an object is fast enough to make it into the atmosphere but not fast enough to escape, it is trapped under the influence of the Earth’s gravitational field and will follow (closed) orbits around Earth. Closed orbits for the interaction determined by Newton’s gravitational law are usually elliptical. Ellipses can take many forms that are determined by specific parameters. For a unique combination of these parameters, we get a circle, which is simply a very special case of an ellipse with singular properties. While studying circular orbits is a huge simplification, it helps us understand the concept of escape velocity.
There is a relationship between the orbital speed v0 of the object and the radial distance r from the centre of the Earth. We can derive this by using the equation for centripetal force, as this is provided by the gravitational attraction to the object in orbit: This implies that depending on the orbit’s altitude, bodies will orbit at a certain speed. The orbital speed should not be confused with escape velocity, as they differ by a factor of √2.This means any satellite with any stable orbit around the Earth would have to increase its velocity by 41.42% to escape Earth’s gravity.
If you look at the image below see differentories an object based on its initial velocity. In A, the object falls back towards Earth. In B, the object enters a stationary orbit with an orbital velocity. In C, the object’s velocity is too high for a stationary orbit but too low to leave the Earth’s gravitational field. Therefore, it will fall into an orbit at a higher altitude. In D, the object leaves the Earth’s gravitational field with the escape velocity.
The orbital speed and escape velocity are two different concepts that should not be confused with each other. The escape velocity is the minimum velocity required for an object to escape the gravitational field of a planet or a celestial body, while the orbital speed is the velocity required for an object to maintain a stable orbit around that body. And as you mentioned, if an object is fast enough to make it into the atmosphere but not fast enough to escape it, it will follow (closed) orbits around the Earth.
What is escape velocity?
The escape velocity is the speed needed for the object to escape the influence of the attractive field starting from a certain spatial point.
How do you calculate the escape velocity?
The escape velocity can be calculated by equating the initial kinetic energy needed with the gravitational potential lost and solving the equation for the initial velocity.
What is an example of escape velocity?
An example of escape velocity is the velocity needed to escape the Earth’s gravitational field. The escape velocity of a body on the surface of the Earth is independent of its mass.
