A) 2R
B) 4R
C) \[\frac{1}{4}R\]
D) \[\frac{1}{2}R\]
Correct Answer: D
Solution :
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: \[{{\rho }_{planet}}=20{{\rho }_{earth}}\] Also, \[{{g}_{planet}}={{g}_{earth}}\] \[\frac{GM}{R_{p}^{2}}=\frac{G{{M}_{e}}}{R_{e}^{2}}\] 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 \[{{R}_{p}}\times 2{{\rho }_{p}}={{R}_{e}}{{\rho }_{e}}\] or \[{{R}_{p}}=\frac{{{R}_{e}}}{2}=\frac{R}{2}\] Aliter: \[g=\frac{4}{3}\pi \rho GR\] \[\Rightarrow \]\[\frac{{{R}_{p}}}{{{R}_{e}}}=\left( \frac{{{g}_{p}}}{{{g}_{e}}} \right)\left( \frac{{{\rho }_{e}}}{{{\rho }_{p}}} \right)=(1)\times \left( \frac{1}{2} \right)\] \[\Rightarrow \]\[{{R}_{p}}=\frac{{{R}_{e}}}{2}=\frac{R}{2}\]
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