A satellite is revolving round the earth in a circular orbit with a velocity of 8km/s. at a height where acceleration due to gravity is 8m//s^(2). How high is the satellite from the earth ? (Take R = 6000 km)

A satellite is revolving round the earth in a circular orbit with a ve

1.	A satellite is revolving round the earth in a circular orbit with a velocity of 8km/s. at a height where acceleration due to gravity is 8m//s^(2). How high is the satellite from the earth ? (Take R = 6000 km)
Answer» <html><body><p></p>Solution :As centripetal acceleration equals to acceleration due to gravity at the height, then <br/> `therefore a = (<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(r) = g_(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)rArr (V^(2))/(r) = <a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> = (64 xx 10^(6))/(R+h) = 8` <br/> `rArr R + h = 8 xx 10^(6)` <br/> `h = 8 xx 10^(6) - 6 xx 10^(6) = 2 xx 10^(6)m` <br/> = 2000 km</body></html>