Question

The initial velocity $v_1$ required to project a body vertically upward from the surface of the earth to just reach a height of $10R$, where$R$ is the radius of the earth, may be described in terms of escape velocity$v_e$ such that $v_i=\sqrt{\left(\frac{x}{y}\right)}{v}_e$. The value of x will be_______.

Answer 1

Step 1: Draw the required diagram:

Step 2: Given data :

Note that from the figure,

The initial velocity $=v_i$

$h=10R$

Radius of the earth$=R$

Escape velocity $v_e$

Mass of earth $m_e$

Step 3: Formula used:

The potential energy at height h from the earth's surface is-

$P.E=-\frac{G{m}_{e}m}{\left(R+h\right)}$

Kinetic energy is defined as-

$K.E=\frac{1}{2}m{v}^2$

Step 4: Calculate the value of $x$

Total height

$$\begin{gathered} h=10R+R \\ h=11R \end{gathered}$$

Total energy at the earth's surface will be equal to total energy at height h-

The kinetic energy at height h will be-

$\frac{1}{2}m{v_i}^2=0$

From energy conservation-

${\left(K.E\right)}_{earthsurface}+{\left(P.E\right)}_{earthsurface}={\left(K.E\right)}_h+{\left(P.E\right)}_h$

$$\begin{gathered} \Rightarrow -\frac{G{m}_{e}m}{R}+\frac{1}{2}m{v_i}^2=-\frac{G{m}_{e}m}{11R}+0\left[R+10R=11R\right] \\ \Rightarrow \frac{1}{2}m{v_i}^2=\left(\frac{10}{11}\right)\frac{G{m}_{e}m}{R} \\ \Rightarrow v_i=\sqrt{\left(\frac{20}{11}\right)\frac{G{m}_e}{R}}\left(Becauseescapevelocity{v}_e=\sqrt{\frac{G{m}_e}{R}}\right) \\ \Rightarrow v_i=\sqrt{\left(\frac{20}{11}\right)}{v}_e \end{gathered}$$

Now compare the given $v_i=\sqrt{\left(\frac{x}{y}\right)}{v}_e$

Therefore the value of $x=20$.