The acceleration due to gravity (g) on a planet is given by the formula:

$ g = G \frac{M}{R^2} $

where:

• $G$ is the universal gravitational constant,
• $M$ is the mass of the planet, and
• $R$ is the radius of the planet.

In this question, we know that:

• The mass of the planet $ M_p $ is $\frac{1}{10}$ of the mass of the earth $ M_e $, so $ M_p = \frac{M_e}{10} $.
• The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius $ R_p $ is also half the radius of the earth $ R_e $. Therefore, $ R_p = \frac{R_e}{2} $.

Using the formula for acceleration due to gravity and substituting the above information:

$$ g_p = G \frac{M_p}{R_p^2} = G \frac{\frac{M_e}{10}}{\left(\frac{R_e}{2}\right)^2} $$

Simplify the expression:

$$ g_p = G \frac{M_e}{10} \cdot \frac{4}{R_e^2} = \frac{4}{10} G \frac{M_e}{R_e^2} $$

Knowing that $ G \frac{M_e}{R_e^2} $ is the acceleration due to gravity on Earth $ g_e \approx 9.8 \, \mathrm{m/s}^2 $, we substitute this value into the equation:

$$ g_p = \frac{4}{10} \cdot 9.8 \, \mathrm{m/s}^2 = 3.92 \, \mathrm{m/s}^2 $$

Thus, the acceleration due to gravity on the planet is $3.92 \, \mathrm{m/s}^2$, which corresponds to:

Option D: $3.92 \mathrm{~m/s}^2$