A body is projected vertically up with a speed v0 = √gR from earth’s…

Let the body of mass m be projected from point 1 and attains a height h at point 2.

Conservation of energy of the body between 1 and 2 yields, U₁ + K₁ = U₂ + K₂

$\Rightarrow\left(-\frac{GMm}{R}\right)+\left(\frac{1}{2} mv_1^2\right)=\left(-\frac{GMm}{R+h}\right)+\left(\frac{1}{2} mv_2^2\right)$

Putting $v_2=0$ and $v_1=v_0$, we obtain,

$\frac{1}{2} mv_0^2=G M m\left(\frac{1}{R}-\frac{1}{R+h}\right)$

$\Rightarrow \frac{v_0^2}{2}=\frac{GMh}{R(R+h)}$

$\Rightarrow \frac{v_0^2}{2}=\left(\frac{GM}{R^2}\right)\left(\frac{Rh}{R+h}\right)$

$\Rightarrow \frac{v_0^2}{2}=\frac{gRh}{R+h} \quad\left(∵ \frac{G M}{R^2}=g\right)$

$\frac{R+h}{h}=\frac{2 gR}{v_0^2}$

$\Rightarrow \frac{R}{h}+1=\frac{2 gR}{v_0^2}$

$\Rightarrow h=\frac{R}{\frac{2 gR}{v_0^2}-1}$

Putting v₀ = $\sqrt{gR}$ , we obtain,

⇒ h = R = 6400 km.