Gravitation Part 1

Let the mass of planet be $m_{p}$ and that of earth be $m$ .Given, $m_{p}=2 m$Density of planet $(\rho_{p})=$ Density ofearth $(\rho)$As we know that,Weight of object on earth is W $w=m g \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)$where, $g$ is acceleration due to gravity.and $\;\; g=\;\frac{G m}{R^{2}}$where, $\;\; G=$ universal gravitationalconstant, $R \;\; =\;\text{ radius of earth }\;$ $\;\text{ and }\; \;\; m_{e} \;\; =\;\text{ mass of earth. }\;$ $\Rightarrow g \;\; =\;\frac{G \rho \frac{4}{3} \pi R^{3}}{R^{2}}$ $\Rightarrow g \;\; =G \rho \;\frac{4}{3} \pi R$Putting this value in Eq. (i), we get $w=m G \rho \;\frac{4}{3} \pi R \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)$As densities of planets earth is same.i.e., $\;\; \rho_{p}=\rho$ $\Rightarrow \;\frac{m_{p}}{\frac{4}{3} \pi R_{p}^{3}} \;\; =\;\frac{m}{\frac{4}{3} \pi R^{3}}$ $\Rightarrow \;\; R_{p} \;\; =\left(\;\frac{m_{p}}{m} R^{3}\right)^{\frac{1}{3}}$ $\;\; =\left(\;\frac{2m}{m} R^{3}\right)^{\frac{1}{3}}$ $\Rightarrow \;\; R_{p} \;\; =(2)^{\frac{1}{3}} R$Weight on planet, $w_{p}=\;\frac{4}{3}(2)^{\frac{1}{3}} \pi m G \rho R \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(iii)$Dividing Eq. (iii) by Eq. (ii), we get $\;\frac{w_{p}}{w} \;\; =\;\frac{ \;\frac{4}{3}(2)^{\frac{1}{3}} \pi m G \rho R }{ m \rho G \;\frac{4}{3} \pi R }$ $\Rightarrow \;\; w_{p} \;\; =2^{\frac{1}{3}} w$