A) \[2R\]
B) \[4R\]
C) \[\frac{1}{4}R\]
D) \[\frac{1}{2}R\]
Correct Answer: D
Solution :
Key Idea: Acceleration due to gravity is the acceleration in vertically downward direction acting on every object on earth. Its value is given by Newton's second law which is \[g=\frac{F}{m}\] The acceleration due to gravity on an object of mass \[m\] \[g=\frac{F}{m}\] but from Newton's law of gravitation \[F=\frac{GMm}{{{R}^{2}}}\] where M is the mass of the earth and R the radius of earth. \[g=\frac{GMm/{{R}^{2}}}{m}=\frac{GM}{{{R}^{2}}}\] Given: \[{{\rho }_{planet}}=2{{\rho }_{earth}}\] Also, \[{{g}_{planet}}={{g}_{earth}}\]\[\frac{G{{M}_{p}}}{R_{p}^{2}}=\frac{G{{M}_{g}}}{R_{e}^{2}}\] or \[\frac{G\times \frac{4}{3}\pi R_{p}^{3}{{\rho }_{p}}}{R_{p}^{2}}=\frac{G\times \frac{4}{3}\pi R_{e}^{3}{{\rho }_{e}}}{R_{e}^{2}}\] or \[{{R}_{p}}{{\rho }_{p}}={{R}_{e}}{{\rho }_{e}}\] or \[{{R}_{p}}\times =2{{\rho }_{e}}=R{{\,}_{e}}{{\rho }_{e}}\] or\[{{R}_{p}}=\frac{{{R}_{e}}}{2}=\frac{R}{2}\]
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