Acceleration due to Gravity

Acceleration due to gravity $g = \frac{GM}{R^2}$ For the planet: $M' = \frac{M}{10}$, $R' = \frac{R}{2}$ $g' = \frac{G(M/10)}{(R/2)^2} = \frac{4GM}{10R^2} = \frac{4}{10}g$ $g' = \frac{4}{10} \times 9.8 = 3.92\ ms^{-2}$

$g = \frac{GM_p}{R^2}$, and $R = \frac{D_p}{2}$
$g = \frac{GM_p}{(D_p/2)^2} = \frac{4GM_p}{D_p^2}$
It does not depend on the mass m of the particle.

$g = \dfrac{GM}{R^2}$, and the mass is $M = \dfrac{4}{3}\pi R^3\rho$
$g = \dfrac{G \cdot \frac{4}{3}\pi R^3\rho}{R^2} = \dfrac{4}{3}\pi G\rho R$
So for the same density, $g \propto R$.
The new planet has a radius 3 times that of the earth.
$\dfrac{g'}{g} = \dfrac{3R}{R} = 3$
$\Rightarrow g' = 3g$

Acceleration due to gravity at the surface: $g = \dfrac{GM}{R^2}$
Mass in terms of density: $M = \dfrac{4}{3}\pi R^3 \rho$
$g = \dfrac{G \cdot \frac{4}{3}\pi R^3\rho}{R^2} = \dfrac{4}{3}\pi G\rho R$
So $g \propto \rho R$
According to the question $g_{planet} = g_{earth}$:
$\dfrac{4}{3}\pi G\rho_{planet}R_{planet} = \dfrac{4}{3}\pi G\rho_{earth}R_{earth}$
With $\rho_{planet} = 2\rho_{earth}$: $2\rho_{earth}R_{planet} = \rho_{earth}R$
$\Rightarrow R_{planet} = \dfrac{R}{2}$

The man jumps with the same take-off speed u on both planets, so the kinetic energy he gives himself is the same.
At the highest point this kinetic energy becomes potential energy: $\dfrac{1}{2}mu^2 = mgh$, so $gh$ = constant, i.e. $h \propto \dfrac{1}{g}$.
Height of jump on planet B $= \dfrac{g_A}{g_B} \times$ height of jump on planet A
$h_B = 9 \times 2 = 18$ m
So the man jumps 18 m on planet B.