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