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Explain the variations of acceleration due to gravity inside and outside the earth and draw the graph.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`implies`In `g(r) = 4/3 pi Grrho, 4/3 piGrho`Gp is <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> . <br/>`:. g(r) prop r ` <br/>Means, the gravitational <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> (g) at a point inside the earth is directly proportional to the distance of that point from the centre of the earth. <br/> `implies` And `g(r) = (<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/r^2` where `r gt gt R_E`so ,where `g(r) prop 1/r^2` where ` r gt gt R_E`. Hence starting from the centre of the earth g(r) increases in directly proportion as r increases and then <a href="https://interviewquestions.tuteehub.com/tag/outside-1143075" style="font-weight:bold;" target="_blank" title="Click to know more about OUTSIDE">OUTSIDE</a> the surface g(r) decreases as inverse square of distance.<br/> `implies` The variations in gravitational acceleration with below the surface of earth and above the height from the surface is shown as in figure.<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C08_E01_017_Q01.png" width="80%"/></body></html>


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