NEET 2024 — Physics solution

NEET2024Physics

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ from the surface of the earth is:

A
$\frac{G m M}{2 R}$
B
$\frac{G m M}{3 R ^{2}}$
C
$\frac{5 G m M}{6 R}$
D
$\frac{2 G m M}{3 R}$

Solution

Orbital radius from earth's centre: $r = 3 R$ .

Total energy at orbit: $E_{or bi t} = - \frac{GM m}{2 r} = - \frac{GM m}{6 R}$ .

Energy at surface: $E_{s u r f} = - \frac{GM m}{R}$ .

Required: $Δ E = E_{or bi t} - E_{s u r f} = - \frac{GM m}{6 R} + \frac{GM m}{R} = \frac{5 GM m}{6 R}$ .

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