Q:

what is actual defination of escape velocity?Is there any example of it?

- Angel Pudasaini (age 16)

Ktm, basundhara

- Angel Pudasaini (age 16)

Ktm, basundhara

A:

Escape velocity is defined as the speed at which an objects kinetic energy is equal to its gravitational potential energy, for speeds greater than this, an object will escape the orbit of the body which is exerting the gravitational force.

to illustrate the idea, if we fire a bullet from a gun, with no additional propulsion on board it begins with kinetic energy

KE=^{1}/_{2 }mv^{2}

Where m is the mass of the bullet, and v is its velocity upon leaving the barrel.

the bullet also has gravitational potential energy from Newton's Law of Gravitation

U=-GMm/r^{}

Where G is Newtons gravitational constant, M is the mass of the earth (or whatever gravitational body is in question), m is still the mass of the bullet, and r is the distance from the center of the earth/other body, in this case just the radius of the earth if we're firing from the surface.

So we set these equal and solve for v

v=(2GM/r)^{(1/2)}So, theoretically, if the bullet is fired faster than this, it will escape the earth's gravitational pull and go off into space. It's interesting to note that by this equation, the mass of the bullet doesn't matter, nor does the angle at which it is fired as long as it doesn't hit anything on the way out. So for any nicely aerodynamic* object ejected from the surface of the earth, without additional propulsion, the escape velocity is a little over 11 kilometers/second.

*It's important to note that this formula does not account for things like drag due to air resistance, which could cause the escape velocity to raise rather significantly depending on the aerodynamics of the object in question. For the case of a small, heavy bullet though this is probably pretty small and this formula should be fairly accurate.

There is also an additional factor, if viewed from the surface of the earth, due to the rotation of the planet. This causes the apparent escape velocity to vary based on whether the object is fired eastward or westward, and how near or far from the equator. This difference near the equator can be about 10.735 km/s relative to Earth for a bullet fired east, and to the west requires an initial velocity of about 11.665 km/s relative to Earth.

Thanks for the question,

Mike Boehme

to illustrate the idea, if we fire a bullet from a gun, with no additional propulsion on board it begins with kinetic energy

KE=

Where m is the mass of the bullet, and v is its velocity upon leaving the barrel.

the bullet also has gravitational potential energy from Newton's Law of Gravitation

U=-GMm/r

Where G is Newtons gravitational constant, M is the mass of the earth (or whatever gravitational body is in question), m is still the mass of the bullet, and r is the distance from the center of the earth/other body, in this case just the radius of the earth if we're firing from the surface.

So we set these equal and solve for v

v=(2GM/r)

*It's important to note that this formula does not account for things like drag due to air resistance, which could cause the escape velocity to raise rather significantly depending on the aerodynamics of the object in question. For the case of a small, heavy bullet though this is probably pretty small and this formula should be fairly accurate.

There is also an additional factor, if viewed from the surface of the earth, due to the rotation of the planet. This causes the apparent escape velocity to vary based on whether the object is fired eastward or westward, and how near or far from the equator. This difference near the equator can be about 10.735 km/s relative to Earth for a bullet fired east, and to the west requires an initial velocity of about 11.665 km/s relative to Earth.

Thanks for the question,

Mike Boehme

*(published on 02/06/2012)*