A) \[g'=3\,g\]
B) \[g'=\frac{g}{9}\]
C) \[g'=9\,g\]
D) \[g'=27\,g\]
Correct Answer: A
Solution :
The acceleration due to gravity on the new planet can be found using the relation \[g=\frac{GM}{{{R}^{2}}}.....(i)\] but \[M=\frac{4}{3}\,\pi {{R}^{3}}\rho ,\,\rho \] being density Thus, Eq. (i) becomes \[\therefore g=\frac{G\times \frac{4}{3}\,\pi \,{{R}^{3}}\rho }{{{R}^{2}}}\] \[=G\times \frac{4}{3}\pi \,\,R\,\rho \] \[\Rightarrow \,R\,\,\propto \,\,R\] \[\therefore \frac{g'}{g}=\frac{R'}{R}\] \[\Rightarrow \frac{g'}{g}=\frac{3\,R}{R}=3\] \[\Rightarrow g'=3g\]
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