1.

A sky lab of mass 2xx10^3 kg is first launched from the surface of earth in a circular orbit of radius 2R (from the centre of earth ) and then it is shifted from this circular orbit to another circular orbit of radius 3R . Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab form first orbit to the second orbit . Given , R = 6400 Km and g = 10m//s^2

Answer» <html><body><p></p>Solution :The energy of the sky lab on the <a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> of <a href="https://interviewquestions.tuteehub.com/tag/earth-13129" style="font-weight:bold;" target="_blank" title="Click to know more about EARTH">EARTH</a> <br/> `E_(s)=KE +PE=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>+(-(GMm)/R)=-(GMm)/R`<br/> And the energy of the sky lab in an orbit of radius r <br/> `E=1/2mv_(0)^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)+[-(GMm)/r]=(-GMm)/(2r)[ " as" v_(0) = sqrt((GM)/r)]`<br/> (a) So the energy required to place the lab from the surface of earth to the orbit of radius 2R, <br/> `E_(1)-E_(s)=-(GMm)/(2(2R))-[-(GMm)/(R)]=3/4(GMm)/R` <br/> i.e `DeltaE=3/4m/RxxgR^(2)=3/4mgR"" ["asg" =(GM)/R^2]`<br/> i.e. `DeltaE=3/4(2xx10^(3)xx10xx6.4xx10^6)=` <br/> `3/4(12.8xx10^(10))=9.6xx10^(10)J` <br/> (b) As for <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a> orbit `r = 3R , E_(11)=-(GMm)/(2(3R))=-(GMm)/(6R)` <br/> `:. E_(11)-E_(1)=-(GMm)/(6R)-(-(GMm)/(4R))=1/12 (GMm)/R` <br/> But as `g=(Gm//R^2), ` i.e. `GM=gR^2` or <br/> `DeltaE=1/2mgR=1/12(12.8xx10^(10))=1.1xx10^(10)J`</body></html>


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