The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that \({v_i} = \sqrt\frac{x}{y} \times {v_e}\). The value of x will be ______.

CONCEPT:

The law of conservation of energy is defined as the sum of the initial kinetic energy and potential energy is equal to the final kinetic energy and potential energy and it is written as;

K_i + U_i = K_f + U_f

Here we have K_o as the initial kinetic energy, U_o as the initial potential energy, K_i as the final kinetic energy, and U_i as the final potential energy.

CALCULATIONS:

Given: The radius of the earth = R

The initial kinetic energy is written as;

K_i = \(\frac{1}{2} mv_i^2\)

and potential energy is written as;

U_i = \(-\frac{Gm_em}{R}\)

The final kinetic energy is written as;

K_f = 0

and potential energy is written as;

U_f = \(-\frac{Gm_em}{(10 R +R)}\)

⇒ Uf = \(-\frac{Gm_em}{11R}\)

By using the law of conservation of energy we have;

Ki + Ui = Kf + U_f

⇒ \(\frac{1}{2} mv_i^2\) \(-\frac{Gm_em}{R}\) = 0 \(-\frac{Gm_em}{11R}\)

⇒ \(\frac{1}{2} mv_i^2\) = \(\frac{10Gm_em}{11R}\)

⇒ \(v_i^2\) = \(\frac{20Gm_e}{11R}\)

⇒ \(v_i\) = \(\sqrt {\frac{20Gm_e}{11R}}\)     ----(1)

As we know that the escape velocity is;

\(v_e = \sqrt{\frac{2Gm_e}{R}}\)

Now, on putting this value in equation (1) we have;

v_i = \(\sqrt{\frac{10}{11}}\) v_e

Hence, x is equal to 10.