Question

Qᴜᴇꜱᴛɪᴏɴ
$\rule{190pt}{2pt}$
What is Escape Velocity ? Obtain expression for it from maximum height attained by a body projected vertically upwards and show that the ratio of an artificial satellite moving close to earth is
$\sqrt{2} :1$
$\rule{190pt}{2pt}$

​

Answer 1

★ Concept

If we project a body upward then after reaching a certain height, the body returns due to gravitational attraction of the Earth. If we go on increasing the initial velocity of the body, then it returns after reaching greater and greater height Ultimately, at a certain velocity of projection the body will go out the gravitational field of the earth and will never return to the earth. This initial velocity of the body is called the 'escape velocity'. Also, the kinetic energy given to the body to protect it with the escape velocity is called the 'escape energy' .

Proceeding to the second part of the question first understand essential parameters.

Now let's assume gravitational potential energy zero at infinity then for a mass 'm' placed on the earth's surface is given by

$\sf \bf {U }= - \dfrac{GM_e m}{R_e}$

where M and R with respect to e is the mass and radius of the Earth. Therefore in order to take a body from the earth's surface to infinity, the work required is GMe m/Re.

Note

It is evident that if we throw a body of mass 'm' with such velocity that it's kinetic energy is GMe m/Re, then it will go in space outside the gravitational field of the earth.

$\sf \bf{ \red{ Escape \: Energy}= + \red{ \dfrac{GM_e m}{R_e}}}$

Now , proceeding to the third part of the question

Before starting the above derivation for Escape velocity below are provided essential points :-

• The escape velocity is independent of the mass (m) of the body.
• If velocity of projection of the body (v) from the surface of planet is greater than the escape velocity (Ve) of that planet then the body will escape out of gravitational field of that planet.

$\footnotesize \sf \dfrac{1}{2} m {v}^{2} + {\huge (} - \dfrac{GM_e m}{R_e} {\huge)}= \dfrac{1}{2} m {v'}^{2} + 0$

$\sf \footnotesize \dfrac{ {v}^{2} }{2} = \dfrac{ {v'}^{2} }{2} + \dfrac{ {v_e}^{2} }{2}$

$\sf {v'}^{2} = {v}^{2} - {v_e}^{2}$

Relation between Orbital Velocity and Escape Velocity

$\sf \: \dfrac{v_o}{v_e} = \dfrac{ \sqrt{gR_e} }{ \sqrt{2gR_e} } = \dfrac{1}{ \sqrt{2} }$

$\underline{ \boxed{ \sf v_e = \sqrt{2} \: v_o}}$

$\bf v_e : v_o = \sqrt{2} : 1$

That is if the orbital velocity of a satellite revolving close to the earth happens to increase to √2 times, the satellite would escape.

$\rule{190pt}{2pt}$

More to know

If a body is projected from earth/planet with velocity 'v' such that v < ve (escape velocity) then after reaching a certain height either the body will move in an orbit around the earth/planet or it will fall back to the planet.

Answer 2

Answer:

hello neart how are you??