A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, its velocity must be increased

Question

A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, its velocity must be increased (A)  100 %  (B)  41.4 %  (C)  50 %  (D)  59.6 %

Tardigrade Admin · Accepted Answer

Orbital velocit vo=√gRe Escape velocity ve=√2gRe where Re is radius of earth. % increase =(ve-vo/vo)×100 % increase =(√2gRe-√gRe/√gRe)×100 =(√2-1)×100 =(1.414-1)×100=41.4 %

Q. A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, its velocity must be increased

$100%$

$41.4%$

$50%$

$59.6%$

Orbital velocit $v_{o} = g R_{e}$
Escape velocity $v_{e} = 2 g R_{e}$
where $R_{e}$ is radius of earth.
$%$ increase $= \frac{v _{e} - v _{o}}{v _{o}} \times 100$
$%$ increase $= \frac{2 g R _{e} - g R _{e}}{g R _{e}} \times 100$
$= (2 - 1) \times 100$
$= (1.414 - 1) \times 100 = 41.4%$

Q. A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, its velocity must be increased

100%

41.4%

50%

59.6%

Orbital velocit vo​=gRe​​ Escape velocity ve​=2gRe​​ where Re​ is radius of earth. % increase =vo​ve​−vo​​×100 % increase =gRe​​2gRe​​−gRe​​​×100 =(2​−1)×100 =(1.414−1)×100=41.4%

$100%$

$41.4%$

$50%$

$59.6%$

Orbital velocit $v_{o} = g R_{e}$
Escape velocity $v_{e} = 2 g R_{e}$
where $R_{e}$ is radius of earth.
$%$ increase $= \frac{v _{e} - v _{o}}{v _{o}} \times 100$
$%$ increase $= \frac{2 g R _{e} - g R _{e}}{g R _{e}} \times 100$
$= (2 - 1) \times 100$
$= (1.414 - 1) \times 100 = 41.4%$