Answer
(a) less than
(b) $4.91m/s^2$
Work Step by Step
We know that the gravitational acceleration on a planet of mass $M_p$ and radius$R_p$ is given as
$g_p=\frac{G\rho_p(\frac{4}{3}\pi R_p^3)}{R_p^2}$
$\implies g_p=\frac{4}{3}\pi \rho_pGR_p$
As $R_p=\frac{R_E}{2}$ and $\rho_p=\rho_E$
$\implies g_p=\frac{4}{3}\pi \rho_E G\frac{R_E}{2}$
This simplifies to:
$g_p=\frac{g_E}{2}$.
Thus, the gravitational acceleration is less than that of the Earth.
(b) We can find the gravitational acceleration on the planet as follows:
$g_p=\frac{g_E}{2}$
$\implies g_p=\frac{9.81m/s^2}{2}$
$\implies g_p=4.91m/s^2$
