Gravitational Pull of The Earth-Escape Velocity

JEE Escape Velocity

By definition, escape velocity is the minimum velocity an object must acquire to escape the gravitational pull of the earth without falling back. The concept of escape velocity has huge significance in the field of space science. So let us delve into this a little deeper.

We know, that if we throw a body upwards, it falls back after rising for a while by the gravitational pull of the earth. Gravity decelerates the body progressively at the rate of g m/s² (acceleration due to gravity) until its velocity becomes 0 at the highest point. After exhausting its initial upward velocity, the body begins to now accelerate at the rate g but downwards i.e. towards the surface of the earth and falls back.

However, if we were to impart sufficient velocity to the body such that the body crosses the gravitational field of the earth without exhausting its velocity, it would be able to escape into space. A good way to calculate the escape velocity is to use the principle of conservation of energy.

If the earth’s gravitational field extends upto a height, h above the earth’s surface, body’s potential energy at that height would be.

P.E. = GMm/(R + h)

where G is the universal Gravitational constant

M is the mass of the earth

m is the mass of the body

and R is the radius of the earth.

Since, h<<R, we can simplify the potential energy to following expression.

P.E. = GMm/R

Let’s say the escape velocity of the body is v. The body must have greater energy than its gravitational binding energy.

? ½(m v²) = GMm/R

or v = ?(2GM/R)

As you can see, this velocity is independent of the mass of the body and therefore, same for all bodies. On putting the values of G, M m, R and V in the expression, we get

v = 11.1 Km/s


Practise This Question

Which of the following corrborales the fact the genetic code is degenerate?