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NEET-XII-Physics
07: Circular Motion
- #1Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon = 3.85 × 105 km and the time taken by the moon to complete one revolution around the earth = 27.3 days.Ans : Distance between the Earth and the Moon:
`` r=3.85\times {10}^{5}\,\mathrm{\,km \,}=3.85\times {10}^{8}\,\mathrm{\,m \,}``
Time taken by the Moon to revolve around the Earth:
`` T=27.3\,\mathrm{\,days \,}``
`` =24\times 3600\times 27.3\,\mathrm{\,s \,}=2.36\times {10}^{6}\,\mathrm{\,s \,}``
`` ``
`` \,\mathrm{\,Velocity \,}\,\mathrm{\,of \,}\,\mathrm{\,the \,}\,\mathrm{\,Moon \,}:``
`` v=\frac{2\pi r}{T}``
`` =\frac{2\times 3.14\times 3.85\times {10}^{8}}{2.36\times {10}^{6}}=1025.42\,\mathrm{\,m \,}/\,\mathrm{\,s \,}``
`` \,\mathrm{\,Acceleration \,}\,\mathrm{\,of \,}\,\mathrm{\,the \,}\,\mathrm{\,Moon \,}:``
`` a=\frac{{v}^{2}}{r}=\frac{(1025.42{)}^{2}}{2.36\times {10}^{6}}=0.00273\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2}``
`` \Rightarrow a=2.73\times {10}^{-3}\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2}``
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