A geostationary satellite is revolving around | vedclass

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(B) The orbital velocity of a satellite is given by $v_0$ and the escape velocity is given by $v_e = \sqrt{2} v_0 \approx 1.414 v_0$.To escape the gra

Vedclass · Accepted Answer

$41.4$

Vedclass · Answer

(B) The orbital velocity of a satellite is given by $v_0$ and the escape velocity is given by $v_e = \sqrt{2} v_0 \approx 1.414 v_0$.To escape the gravitational field,the satellite's velocity must be increased from $v_0$ to $v_e$.The required increase in velocity is $\Delta v = v_e - v_0 = 1.414 v_0 - v_0 = 0.414 v_0$.The fractional increase is $\frac{\Delta v}{v_0} = \frac{0.414 v_0}{v_0} = 0.414$.Therefore,the percentage increase required is $0.414 	imes 100\% = 41.4\%$.

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

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