A) 1/2 times
B) \[\sqrt{2}\] times
C) \[{{2}^{2/3}}\]times
D) \[{{2}^{3/2}}\] times
Correct Answer: D
Solution :
According to Keplers law \[{{T}^{2}}\propto {{R}^{3}}\] (Here: R is orbital radius and T is time period) Now according to question when orbital radius is doubled, then period will be \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{R}{2R} \right)}^{3/2}}\] \[{{T}_{2}}={{2}^{3/2}}{{T}_{1}}\] (Here: \[{{R}_{1}}=R,\,\,{{R}_{2}}=2R\]) Hence, the time period will become \[{{2}^{3/2}}\] times.
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