Explanation:
The density of earth [tex]\rho_E[/tex] is given by
[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]
and in terms of this density, we can write the acceleration due to gravity on earth as
[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]
Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by
[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]
[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]
We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as
[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]
[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]
[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]
