The ratio of the kinetic energy required to be given to a satellite so that it escapes the gravitational field of Earth to the kinetic energy required to put the satellite in a circular orbit just above the free surface of Earth is

Question

The ratio of the kinetic energy required to be given to a satellite so that it escapes the gravitational field of Earth to the kinetic energy required to put the satellite in a circular orbit just above the free surface of Earth is (A) 1 (B) 2 (C) 3 (D) 9

Tardigrade Admin · Accepted Answer

ve=√(2 G M/R) ⇒ ke=(1/2) m ve2=(G M m/R) vo=√(G M/R+h) here h=0, ko=(1/2) m vo2=(G M m/2 R) ⇒ (ke/ko)=2

Q. The ratio of the kinetic energy required to be given to a satellite so that it escapes the gravitational field of Earth to the kinetic energy required to put the satellite in a circular orbit just above the free surface of Earth is

1

2

3

9

$v_{e} = \frac{2 GM}{R}$
$\Rightarrow k_{e} = \frac{1}{2} m v_{e}^{2} = \frac{GM m}{R}$
$v_{o} = \frac{GM}{R + h}$
here $h = 0, k_{o} = \frac{1}{2} m v_{o}^{2} = \frac{GM m}{2 R}$
$\Rightarrow \frac{k _{e}}{k _{o}} = 2$

Q. The ratio of the kinetic energy required to be given to a satellite so that it escapes the gravitational field of Earth to the kinetic energy required to put the satellite in a circular orbit just above the free surface of Earth is

1

2

3

9

ve​=R2GM​​ ⇒ke​=21​mve2​=RGMm​ vo​=R+hGM​​ here h=0,ko​=21​mvo2​=2RGMm​ ⇒ko​ke​​=2

$v_{e} = \frac{2 GM}{R}$
$\Rightarrow k_{e} = \frac{1}{2} m v_{e}^{2} = \frac{GM m}{R}$
$v_{o} = \frac{GM}{R + h}$
here $h = 0, k_{o} = \frac{1}{2} m v_{o}^{2} = \frac{GM m}{2 R}$
$\Rightarrow \frac{k _{e}}{k _{o}} = 2$