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 2R from Earth’s surface is

Question

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 2R from Earth’s surface is (A)  (5GMm/6R) (B)  (2GMm/3R) (C)  (GMm/2R) (D)  (GMm/3R)

Tardigrade Admin · Accepted Answer

- (GMm/R)+K.E.=(-GMm/3R)+ (1/2)m ((GM/3R)) K.E = (GMm/R)- (GMm/3R)+(GMm/6R) ∴ K.E. = (5GMm/6R)

Q. 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 2R from Earth’s surface is

$\frac{5 GM m}{6 R}$

$\frac{2 GM m}{3 R}$

$\frac{GM m}{2 R}$

$\frac{GM m}{3 R}$

$- \frac{GM m}{R} + K . E . = \frac{- GM m}{3 R} + \frac{1}{2} m$ $(\frac{GM}{3 R})$
K.E = $\frac{GM m}{R} - \frac{GM m}{3 R} + \frac{GM m}{6 R}$
$∴ K . E . = \frac{5 GM m}{6 R}$

Q. 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 2R from Earth’s surface is

6R5GMm​

3R2GMm​

2RGMm​

3RGMm​

−RGMm​+K.E.=3R−GMm​+21​m (3RGM​) K.E = RGMm​−3RGMm​+6RGMm​ ∴K.E.=6R5GMm​

$\frac{5 GM m}{6 R}$

$\frac{2 GM m}{3 R}$

$\frac{GM m}{2 R}$

$\frac{GM m}{3 R}$

$- \frac{GM m}{R} + K . E . = \frac{- GM m}{3 R} + \frac{1}{2} m$ $(\frac{GM}{3 R})$
K.E = $\frac{GM m}{R} - \frac{GM m}{3 R} + \frac{GM m}{6 R}$
$∴ K . E . = \frac{5 GM m}{6 R}$