Question

A projectile is projected with velocity $k{v}_e$ in vertically upward direction from the ground into the space ($v_e$ is escape velocity and $k<1$). If air resistance is considered to be negligible, then the maximum height from the center of earth till which it can go will be (R= radius of earth).

• A. $\frac{R}{k^2+1}$
• B. $\frac{R}{k^2-1}$
• C. $\frac{R}{1-k^2}$
• D. $\frac{R}{k+1}$

Answer 1

The correct option is C $\frac{R}{1-k^2}$

Potential energy at surface $U_i=\frac{-GMm}{R}$

Potential energy at height h from the center is final Potential energy $U_f=\frac{-GMm}{h}$

Where $M$ and $m$ are mass of the earth and the projectile respectively.

Difference in potential energy is $U_f-U_i=\frac{GMm}{R}(1-R/h)$

$\text{this difference in potential energy is nothing but the kinetic energy supplied at the time of projection}$

We know that escape velocity is $\sqrt[2]{22GM/R}$ so projection velocity $v=k\times \sqrt[2]{22GM/R}$ so the kinetic energy $K.E.=m{v}^2/2=m{k}^{2}GM/R$

equating the $K.E.$ and the change in potential energy

we get $h=\frac{R}{1-k^2}$

Option C is correct.