What exactly is escape velocity? In this article we will explore what this concept is, how to calculate this velocity for an object and also look at how this velocity varies for different celestial objects.
Defining the Term
The escape velocity of the Earth is defined as the velocity required by a projectile to escape the gravitational field of Earth. This means that the velocity of an object, such as a cannon ball or a rocket ship, has to be equal to or greater than the value of this velocity for the Earth in order for it to escape the pull of gravity exerted by the Earth.
For an object to attain this velocity, the kinetic energy of the object must equal the gravitational potential energy of the object at the surface of the Earth. Isaac Newton had designed a thought experiment in which he pictured a cannon on to top a high tower. A cannon ball would be launched in different scenarios with varying results. If the cannon ball was not launched with enough velocity it would fall back to the ground. If the cannon ball is shot with sufficient velocity to miss the ground, then it would make an entire revolution of the Earth and strike the cannon from behind. In a third scenario, the cannon ball is launched with just enough velocity to make an elliptical orbit of the Earth, yet remain in the gravitational field of the Earth, Finally, in the fourth scenario the cannon ball is launched with velocity that is sufficient enough to escape the gravity well of Earth.
It should be noted that this velocity of an object is not dependant on the mass of the object. Another interesting thing to note is that Increasing or decreasing the radius of the Earth (or any celestial body of interest) will cause a change in this value as well. This is because the radius of the Earth effects its escape velocity as it is directly related to the gravitational pull of the Earth. Next, let us take a look at how this velocity for the Earth is calculated.
Calculating This Velocity for the Earth
The formula to calculate this velocity for an object on Earth is:
v = √(2GM⁄ R)
- v is the escape velocity
G is the gravitational constant (6.67 x 10-20 km3/kg-s2)
M is the mass of the Earth (5.974 x 1024 kg)
- R is the radius from the center of the Earth (~6371 km)
In order to calculate the escape velocity of the Earth, all we have to do is to plug in the above numbers in the equation and we arrive at a number equal to 11.2 km/s (about 7 mi/sec). This is the speed at which an object must travel at to escape the gravitational pull of the Earth.
Escape Velocity for Various Celestial Objects
Different planets and moons in our Solar System have different values of this velocity. For instance, the escape velocity of the Moon is only 2.4 km/s, whereas that of Jupiter is 59.5 km/s. The Sun has the highest gravitational field in our Solar System and therefore this velocity is very high for the Sun, nearly equal to 617.5 km/s. This means that an object on the surface of the Sun would have to be travelling 617.5 km/s to escape the gravitational pull of the Sun. Objects with greater gravity wells, such as black holes, have these velocities nearing the speed of light, which means that a ray of light would have to travel faster than its normal speed of nearly 300,000 km/s to escape the gravitational pull of a black hole. No object can escape a black hole, including light, due to its enormous gravity.