A) \[4.9\,m/{{s}^{2}}\]
B) \[0.49\,m/{{s}^{2}}\]
C) \[0.98\,m/{{s}^{2}}\]
D) \[9.8\,m/{{s}^{2}}\]
Correct Answer: B
Solution :
From law of gravitation the force of attraction acting on moon due to earth is \[F=G\frac{{{M}_{e}}m}{R_{e}^{2}}\] ... (i) where G is gravitational constant and Mg is mass of earth. From Newtons second law, \[F=mg\] ? (ii) From Eqs. (i) and (ii), we get \[g=\frac{G{{M}_{e}}}{{{R}_{e}}^{2}}\] \[\Rightarrow \] \[\frac{{{g}_{e}}}{{{g}_{m}}}=\frac{{{M}_{e}}}{{{M}_{m}}}{{\left( \frac{{{R}_{m}}}{{{R}_{e}}} \right)}^{2}}\] Given, \[{{M}_{e}}=80\,{{M}_{m}},\,\,{{R}_{e}}=2\,{{R}_{m}}\] \[\therefore \] \[\frac{{{g}_{e}}}{{{g}_{m}}}=\frac{80\,\,{{M}_{m}}}{{{M}_{m}}}{{\left( \frac{{{R}_{m}}}{2{{R}_{m}}} \right)}^{2}}\] \[=80\times \frac{1}{4}=20\] \[\Rightarrow \] \[{{g}_{m}}=\frac{{{g}_{e}}}{20}=\frac{9.8}{20}=0.49\,m/{{s}^{2}}\]You need to login to perform this action.
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