Concept Builders - A Complete Guide To Entrance Test Examination: Physics

How Albert Einstein Started His Journey towards E=mc^2 [ The life of Founder of Relativity ]

How Albert Einstein Started His Journey towards E=mc^2 [ The life of Founder of Relativity ]

Variation Of Gravitational Acceleration 'g'

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<b><span style="font-family: Times, Times New Roman, serif; font-size: large;">At poles or longitude:</span></b></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                          It is ‘constant’ and ‘maximum’ i.e (9.8 m/s.s)<br /><br /><b>At equator or latitude:</b></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                          It ‘decreases’ with increasing distance from center of Earth.<br /><br /><b>At some altitude (height):</b>    </span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                               It decreases exponentially , and is inversely proportional to the square of height H above the surface of earth. The formula for g at the surface of earth is given by:</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                            <span style="color: #990000;"> g = GM/R<sup style="margin: 0px; padding: 0px;">2</sup></span></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">Here ‘g’ is the acceleration due to gravity, ‘G’ is the gravitational constant, ‘M’ is the mass of Earth and ‘R’ is the mean radius of Earth.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">Let ‘g<sub style="margin: 0px; padding: 0px;">o</sub>’ be the value of acceleration due to gravity at radius ‘R’ from center of Earth. ‘g<sub style="margin: 0px; padding: 0px;">H</sub>’ be the value of acceleration due to gravity at a height ‘H’ from Earth’s surface.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">So,</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                          <span style="color: #cc0000;">  g<sub style="margin: 0px; padding: 0px;">o</sub>= GM/R<sup style="margin: 0px; padding: 0px;">2</sup> </span></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">And</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                       <span style="color: #cc0000;"> g<sub style="margin: 0px; padding: 0px;">H </sub>= GM / [(R+H)] <sup style="margin: 0px; padding: 0px;">2</sup></span> </span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">Dividing the equations we get :</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                      <span style="color: #cc0000;">  g<sub style="margin: 0px; padding: 0px;">H </sub>= g<sub style="margin: 0px; padding: 0px;">o</sub> (R / (R+H))<sup style="margin: 0px; padding: 0px;">2</sup></span></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><b><span class="Apple-tab-span" style="margin: 0px; padding: 0px; white-space: pre;"> </span>In Depth:</b><br />                        It decreases linearly, and is inversely proportional to the depth in Earth. Let ‘g<sub style="margin: 0px; padding: 0px;">o</sub>' be the value of gravitational acceleration on Earth’s surface from center of Earth and the radius be ‘R’ and 'g<sub style="margin: 0px; padding: 0px;">x</sub>' be the value of gravitational acceleration at a depth of ‘(R – x)’ , So, a formula can be derived which  gives</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                        <span style="color: #cc0000;"> g<sub style="margin: 0px; padding: 0px;">x</sub> = g<sub style="margin: 0px; padding: 0px;">o</sub> (1- x/R  )         </span> </span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">When x = R i.e. center of Earth g<sub style="margin: 0px; padding: 0px;">x</sub> = 0. So at center of Earth acceleration due to gravity is zero.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">Its value at sun is <b><span style="color: #38761d;">274 m/s<sup style="margin: 0px; padding: 0px;">2</sup></span></b> . At moon it is <b><span style="color: #38761d;">1.6 m/s<sup style="margin: 0px; padding: 0px;">2</sup></span></b></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><b><span style="margin: 0px; padding: 0px;">Some important concepts:-</span></b></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><span style="color: purple;"><span class="Apple-tab-span" style="margin: 0px; padding: 0px; white-space: pre;"> </span>Just imagine the Earth starts rotating very fast. What will happen to the person standing on it?</span> Obviously, he will fly off tangentially to the surface of the Earth. It means the gravitational force on that person has decreased. In this case the value of g will be ‘<span style="color: #38761d;">MINIMUM</span> ‘.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">The variation of gravitational acceleration with rotation of earth is given by </span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                       <span style="color: #cc0000;"> g<sup style="margin: 0px; padding: 0px;">' </sup>= g - R<sub style="margin: 0px; padding: 0px;">e</sub>ω^2 cos^2θ</span></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">When ‘ω’ increases g’ becomes more and more less.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><span style="color: purple;"><span class="Apple-tab-span" style="margin: 0px; padding: 0px; white-space: pre;"> </span>Now what happens if the Earth stops suddenly? </span><br />The value of g will become maximum or infinity in that case.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">In this case ‘ω’ becomes zero so does the factor <span style="color: #38761d;">R<sub style="margin: 0px; padding: 0px;">e</sub>ω^2cos^2θ</span> so g<sup style="margin: 0px; padding: 0px;">' </sup>= g, which is maximum value.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><span class="Apple-tab-span" style="margin: 0px; padding: 0px; white-space: pre;"> </span>As we know gravitational force of attraction is the force that attracts a body towards the center of Earth. Now if a body is placed at the center of the Earth what will be the gravitational force of attraction in this case. It will be zero as the body is already placed at the center of the Earth. (already discussed in variation of ‘g’ with depth.)</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><span class="Apple-tab-span" style="margin: 0px; padding: 0px; white-space: pre;"> </span>Now, if a body starts moving from equators to poles value of g will increase continuously increases till it reaches its maximum value at poles. And if a body starts moving from poles to equators, value of g will decrease till it reaches its minimum value at equators. </span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">This can also be explained on the basis of this formula</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                     <span style="color: #cc0000;">    g<sup style="margin: 0px; padding: 0px;">' </sup>= g - R<sub style="margin: 0px; padding: 0px;">e</sub>ω^2 cos^2θ</span></span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">==> At poles the value of θ is 90 which make the factor <span style="color: #38761d;">R<sub style="margin: 0px; padding: 0px;">e</sub>ω^2cos^2θ </span>equal to zero so gravitation acceleration is maximum i.e g’ = g.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">==> At equator the value of θ is 0 which gives the factor <span style="color: #38761d;">R<sub style="margin: 0px; padding: 0px;">e</sub>ω^2cos^2θ</span> its maximum value so g<sup style="margin: 0px; padding: 0px;">' </sup>< g.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">==> Now, as we move from equator to poles the value of θ increases and cosine value gets lowers making g' greater and vice verse.</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">This effect can also be simply under by Newton’s law of gravitation which gives</span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                               g= GM/R^2 </span></div>
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<span style="font-family: Times, Times New Roman, serif; font-size: large;"><img src="http://www.open.edu/openlearnworks/pluginfile.php/66047/mod_page/content/1/S103_3_003i.jpg" /></span></div>
<span style="font-family: Times, Times New Roman, serif; font-size: large;">At poles the radius is smaller so gravitational acceleration is large and at equator radius is large so value of acceleration due to gravity gets smaller.</span></div>
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Variation Of Gravitational Acceleration 'g'

At poles or longitude: It is ‘constant’ and ‘maximum’ i.e (9.8 m/s.s) At equator or latitu...

Physics Chapterwise Important Formulas

<div dir="ltr" style="text-align: left;" trbidi="on"><div class="separator" style="clear: both; text-align: center;"></div><div style="text-align: left;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiw2cq9nyLhMBk0s7qXVs2n04riFqD6xJKPTQE8oTIuQnBhhVbnDgGtBTLzmyuNGoj0P1Vtk_oXP2eSADeytt28paIzw_1n_hEJD0LUOr1dbRBBNEN92ZUqOmSFSpMiysuW19tQIzwXbnM/s1600/images.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="117" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiw2cq9nyLhMBk0s7qXVs2n04riFqD6xJKPTQE8oTIuQnBhhVbnDgGtBTLzmyuNGoj0P1Vtk_oXP2eSADeytt28paIzw_1n_hEJD0LUOr1dbRBBNEN92ZUqOmSFSpMiysuW19tQIzwXbnM/s320/images.png" width="320" /></a></div><h2 style="text-align: left;"><span style="color: red; font-family: Times, Times New Roman, serif; font-size: large;">Assalam O Alaikum Guys !!</span></h2></div><div style="text-align: left;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEikjvbBVq-vDLiTFUlj9q-ibghmOoknGOjMqlm-YrponMVTWgwjZGBCOn-YM9bQnhsthWXUayp_eCE-4bXFKracSGq_phAXeHaOcjOCyDnNujhnlMJ8B5rQsIAPuPM0nuobx7QVOywE5x8/s1600/images.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><br />
</a></div><span style="font-family: Times, Times New Roman, serif; font-size: large;">                   I hope you''ll be fine. As we know, physics is based on mathematical relations and formulas. Here, I have summarized all important formulas of physics which we can use in solving physics problem.  If you don't have enough time to repeat your whole books, you can just read these formulas and Insha'Allah! this will help you.</span></div><div class="separator" style="clear: both; text-align: center;"></div><br />
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Physics Chapterwise Important Formulas

Assalam O Alaikum Guys !! I hope you''ll be fine. As we know, physics is based on mathematical relations and for...