Variation Of Gravitational Acceleration 'g'

<div dir="ltr" style="text-align: left;" trbidi="on"> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px; text-align: center;"> <img height="171" src="http://metrouk2.files.wordpress.com/2013/11/gravity3.jpg" width="320" /></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px; text-align: center;"> <br /></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <b><span style="font-family: Times, Times New Roman, serif; font-size: large;">At poles or longitude:</span></b></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> </div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;">So,</span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;">And</span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;"><br /></span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;">Dividing the equations we get :</span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;"><br /></span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;"><br /></span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;">When ‘ω’ increases g’ becomes more and more less.</span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <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> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;"><br /></span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;">                                                               g= GM/R^2 </span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <span style="font-family: Times, Times New Roman, serif; font-size: large;"><br /></span></div> <div style="background-color: white; line-height: 29px; margin: 0px; padding: 0px;"> <div style="text-align: center;"> <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> </div>

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...

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