question_answer 57)

                  (a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth's rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? What is it at latitude \[\theta \]? How does these acceleration compare with \[g=9.8\text{ }m/{{s}^{2}}\]? (b) Earth also moves in circular orbit around sun once every year with on orbital radius of\[1.5\times {{10}^{11}}m\]. What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with \[g=9.8\,m/{{s}^{2}}\] ?

Answer:

                  Acceleration = centripetal acceleration                 \[a=\,\frac{{{\upsilon }^{2}}}{R}\]                 But \[\upsilon =\omega R\]                 \[\therefore \] \[\,a\,=\,{{\omega }^{2}}R=\frac{4{{\pi }^{2}}R}{{{T}^{2}}}\,\]                 \[=\,\frac{4\,\times \,{{(3.14)}^{2}}\times \,6.4\,\times \,{{10}^{6}}}{{{(24\times \,3600)}^{2}}}\]                 \[=3.38\times {{10}^{2}}m{{s}^{2}}\]                 Hence, \[\frac{a}{g}\,=\,\frac{3.38\,\times \,{{10}^{-2}}}{9.8}\,=\,3.45\,\times \,{{10}^{-3}}\]                 At latitude \[\theta ,\,a\,=g\,\left( 1-\frac{R{{\omega }^{2}}\,{{\cos }^{2}}\theta }{g} \right)\]