A space station is at a height equal to the radius of the Earth. If 'vE' is the escape velocity on the surface of the Earth, the same on the space station is ....... times vE.

Question

A space station is at a height equal to the radius of the Earth. If 'vE' is the escape velocity on the surface of the Earth, the same on the space station is ....... times vE. (A) (1/2) (B) (1/4) (C) (1/√2) (D) (1/√3)

Tardigrade Admin · Accepted Answer

us = (GMm/2R) K.E. = P.E. (1/2)mv2s = (GMm/2R) v2s = (GM/R) vs = √gR [∵ GM=gR2] But ve=√2 g R=√2 √g R ve=√2 vs vs=(ve/√2)

Q. A space station is at a height equal to the radius of the Earth. If $^{'} v_{E}^{'}$ is the escape velocity on the surface of the Earth, the same on the space station is ....... times $v_{E}$ .

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{3}$

$u_{s} = \frac{GM m}{2 R}$
K.E. = P.E.
$\frac{1}{2} m v_{s}^{2} = \frac{GM m}{2 R}$
$v_{s}^{2} = \frac{GM}{R}$
$v_{s} = g R$
$[∵ GM = g R^{2}]$
But $v_{e} = 2 g R = 2 g R$
$v_{e} = 2 v_{s}$
$v_{s} = \frac{v _{e}}{2}$

Q. A space station is at a height equal to the radius of the Earth. If ′vE′​ is the escape velocity on the surface of the Earth, the same on the space station is ....... times vE​.

21​

41​

2​1​

3​1​

us​=2RGMm​ K.E. = P.E. 21​mvs2​=2RGMm​ vs2​=RGM​ vs​=gR​ [∵GM=gR2] But ve​=2gR​=2​g​R ve​=2​vs​ vs​=2​ve​​

Q. A space station is at a height equal to the radius of the Earth. If $^{'} v_{E}^{'}$ is the escape velocity on the surface of the Earth, the same on the space station is ....... times $v_{E}$ .

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{3}$

$u_{s} = \frac{GM m}{2 R}$
K.E. = P.E.
$\frac{1}{2} m v_{s}^{2} = \frac{GM m}{2 R}$
$v_{s}^{2} = \frac{GM}{R}$
$v_{s} = g R$
$[∵ GM = g R^{2}]$
But $v_{e} = 2 g R = 2 g R$
$v_{e} = 2 v_{s}$
$v_{s} = \frac{v _{e}}{2}$