Question

For a satellite very close to the earth surface $h<<R_e$ $i.e.r=R_e$
$v_0=\sqrt{\frac{GM}{R_e}}+\sqrt{g{R}_e}=8km/sec$ prove this.

Answer 1

At very close to earth-

$r=R_e+h\approx R_e$

Now, to move on a circular path the centripetal force must be equal to gravitational force.

$⟹\frac{GMm}{R_e^2}=\frac{m{v}_o^2}{R_e}$

$⟹v_o=\sqrt{\frac{GM}{R_e}}$

And we know that $g=\frac{GM}{R_e^2}$

$⟹v_o=\sqrt{\frac{GM}{R_e}}=\sqrt{\frac{GM{R}_e}{R_e^2}}$

$⟹v_o=\sqrt{g{R}_e}$

$R-e=6400km=64\times {10}^{5}m$

On putting values -

$v_o=\sqrt{10\times 64\times \times {10}^5}=8\times {10}^{3}m/s$

$⟹v_o=8km/s$