CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    According to Keplers law of planetary motion if T represents time period and r is orbital radius, then for two planets these are related as:

    A)  \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{3}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\]

    B)         \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{\frac{3}{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\]

    C)  \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\]

    D)         \[\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{\frac{3}{2}}}\]

    E)  \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\]

    Correct Answer: D

    Solution :

    \[{{T}^{2}}\propto {{r}^{3}}\] \[\Rightarrow \]               \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\]


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