The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.


Escape speed is that after attaining which a body moves just out of the earth’s Gravitational influence

As body will be out of earth’s influence when its total energy is Zero or positive and we know the Total energy of a body is the sum of kinetic energy and potential energy


T = K + U


Where T is the total energy, U is potential energy and K is kinetic energy kinetic energy of a body which is always positive and depends upon the speed of the body, potential energy is negative and decreases as we move away from orbiting the planet and at an infinite distance from the planet


Kinetic Energy is given as



Where K is the kinetic energy of a body of mass m, moving with velocity v


The potential energy of a body above the surface of the earth is given as


U = -GMm/R


Where U is the gravitational Potential Energy of body of mass m at a distance R from the centre of the earth and M is the mass of earth G is universal gravitational Constant


If Body have to move out of Earth’s influence its total energy should be positive i.e.


T ≥ 0


Or


Let m be the mass of the projectile and Ve be the escape speed from surface of earth. So kinetic energy of a body at escape speed Ke is


Ke = 1/2 m Ve2


when distance of body from centre of earth will be equal to radius of earth, Potential energy of particle at surface of earth is


U = -GMm/r


i.e. for a body to just escape total energy should be zero i.e.



i.e. Ke = 1/2 m Ve2 = -(-GMm/r)


or U = -Ke


so potential energy of a body at surface of earth is equal to negative of kinetic energy at escape velocity


We are given initial speed Vi is three times escape speed i.e.


Vi = 3Ve


Let m be the mass of the projectile, then its initial kinetic energy will be


Ki = 1/2 mVi2 = 1/2 m(3Ve)2


= 9(1/2 m Ve2) = 9 Ke


So initial kinetic energy of projectile is 9 times kinetic energy at escape speed


Now initial potential energy of projectile when it is at surface of earth will be


Ui = -GMm/R


We know it will be equal to negative of the kinetic energy of the same particle at escape velocity


Ui = -Ke


So total initial energy of projectile will be


Ti = Ki + Ui


i.e. Ti = 9 Ke + (-Ke) = 8 Ke


so total initial energy of projectile will be 8 times its kinetic energy at escape velocity


now finally when projectile will be at an infinite distance, its potential energy will be zero


Uf = 0


And let us assume particle of mass m has gained a velocity Vf, so the final kinetic energy of projectile will be


Kf = 1/2 mVf2


So total final energy will be


Tf = Kf + Uf


i.e. Tf = 1/2 mVf2


using the law of conservation of energy we know total initial energy must be equal to total final energy so we have


Ti = Tf


Or we can say


8(Ke) = 1/2 mVf2


8 x (1/2 mVe2) =  mVf2


On solving we get the relation between final speed Vf and Escape speed velocity Vi as


Vf2 = 8Ve2


Or


Now we know escape velocity is


Ve = 11.2 km/s = 11.2 × 103 m/s


the final velocity of projectile Vf far away from earth will be



i.e. Vf = 31.68 × 103 m/s = 31.68 km/s


so final velocity of Projectile at distance far away from earth is 31.68 km/s

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