question_answer 62)

                  Earth's orbit is an ellipse with eccentricity 0.0167. Thus, earth's distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant through the year. Assume that earth's spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year?

Answer:

                  According to the law of conservation of angular momentum                                 \[{{I}_{1}}{{\omega }_{1}}\,={{I}_{2}}{{\omega }_{2}}\]                 or \[\frac{2}{5}\,\,MR_{p}^{2}\,{{\omega }_{p}}=\,\frac{2}{5}\,MR_{a}^{2}\,{{\omega }_{a}}\]                 or \[R_{p}^{2}{{\omega }_{p}}=\,R_{a}^{2}{{\omega }_{a}}\] or \[\frac{{{\omega }_{p}}}{{{\omega }_{a}}}\,\,=\,{{\left( \frac{{{R}_{a}}}{{{R}_{p}}} \right)}^{2}}\]                ? (i)                 Now \[{{R}_{p}}=a\,(1-e),\] where \[a=\] semi major axis and \[{{R}_{a}}\,=a\,(1+\,e)\]                 \[\therefore \] eqn. (i) becomes \[\frac{{{\omega }_{p}}}{{{\omega }_{a}}}=\,\frac{{{(1+\,e)}^{2}}}{{{(1-\,e)}^{2}}}\]                 \[=\,\frac{(1.0167){{\,}^{2}}}{{{(0.9833)}^{2}}}\,\,=1.0691\]                 Now \[{{\omega }_{p}}\,=\,\frac{2\pi }{{{T}_{p}}}\] and \[{{\omega }_{a}}\,=\,\frac{2\pi }{{{T}_{a}}}\]                 \[\therefore \] \[\frac{{{\omega }_{p}}}{{{\omega }_{p}}}=\frac{{{T}_{a}}}{{{T}_{p}}}\,=\,\frac{1}{1.0693\,}\,=\,0.9354\]                        ... (ii)                 Also \[\frac{{{T}_{a}}+\,{{T}_{p}}}{2}=24\]\[\therefore \] \[{{T}_{a}}\,+{{T}_{p}}=48\]         ... (iii)                 From eqns. (ii) and (iii), \[{{T}_{p}}=24.8\,h\] and \[{{T}_{a}}=23.20\,h\]