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 V_e be the escape speed from surface of earth. So kinetic energy of a body at escape speed K_e is

K_e = 1/2 m V_e²

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. K_e = 1/2 m V_e² = -(-GMm/r)

or U = -K_e

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 V_i is three times escape speed i.e.

V_i = 3Ve

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

K_i = 1/2 mV_i² = 1/2 m(3V_e)²

= 9(1/2 m V_e²) = 9 K_e

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

U_i = -GMm/R

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

U_i = -K_e

So total initial energy of projectile will be

T_i = K_i + U_i

i.e. T_i = 9 K_e + (-K_e) = 8 K_e

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

U_f = 0

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

K_f = 1/2 mV_f²

So total final energy will be

T_f = K_f + U_f

i.e. T_f = 1/2 mV_f²

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

T_i = T_f

Or we can say

8(K_e) = 1/2 mV_f²

8 x (1/2 mV_e²) =  mV_f²

On solving we get the relation between final speed V_f and Escape speed velocity V_i as

V_f² = 8V_e²

Or

Now we know escape velocity is

V_e = 11.2 km/s = 11.2 × 10³ m/s

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

i.e. V_f = 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