Question

Expression for the escape velocity of a body from the surface of the earth

Answer 1

Answer:

To find:

Escape velocity of an object from the earth surface .

Concept:

Escape velocity is that velocity with which a particle when projected, goes out of the gravitational pull of a planet

(i.e. infinity )

And we have to find the minimum velocity to do so.

So on reaching infinity, both kinetic and Potential energy will be zero.

Calculation:

So we will be applying Conservation Of Mechanical Energy theorem.

Let escape Velocity be v , mass of object be m , mass of Earth be M ,and radius be r.

$\therefore \: ke1 + pe1 = ke2 + pe2$

$= > \dfrac{1}{2} m {v}^{2} + ( - \dfrac{gmM}{ r } ) = 0 + 0$

$= > \dfrac{1}{2} m {v}^{2} = \dfrac{gmM}{r}$

$= > {v}^{2} = \dfrac{2gM}{r}$

$= > v = \sqrt{ \dfrac{2gM}{r} }$

So from this Equation, we can clearly understand that escape velocity is not dependent on the mass of the object.

So final answer :

$\boxed{ \red{ v \: esc.= \sqrt{ \dfrac{2gM}{r} }}}$

Answer 2

$\huge{\sf{Escape \ Velocity}}$

Velocity required for a macroscopic particle to escape the gravity of earth is known as Escape Velocity

According to the law of conservation of mechanical energy,

$\sf Initial \ KE = Final \ PE$

$\longmapsto \: \sf \: \dfrac{1}{2} M {v_e}^{2} = Mg(R_E + h) \\ \\ \longmapsto \: \sf{ {v_e}^{2} = 2g(R_e + h) } \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{v_e = \sqrt{2g(R_E + h)} }}}$

For smaller heights,

Radius of Earth » H

$\longmapsto \boxed{ \boxed{\sf{ \: v_e = \sqrt{2gR_E} }}}$

In terms of gravitational constant and Mass of Earth,

$\longmapsto \boxed{\boxed{\sf{v_e = \sqrt{\dfrac{2G M_E}{R_E}}}}}$