A) 2 R
B) 4 R
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. \[\therefore \] \[g=\frac{GMm/{{R}^{2}}}{m}=\frac{GM}{{{R}^{2}}}\] Given: \[{{\rho }_{planet}}=2{{\rho }_{earth}}\] Also, \[{{g}_{planet}}={{g}_{earth}}\] \[\Rightarrow \] \[\frac{G{{M}_{P}}}{R_{P}^{2}}=\frac{G{{M}_{e}}}{R_{e}^{2}}\] 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 \[{{R}_{P}}=\frac{{{R}_{e}}}{2}=\frac{R}{2}\]You need to login to perform this action.
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