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Step 1: Recall the formula for gravitational acceleration
The acceleration due to gravity on the surface of a planet of mass $M$ and radius $R$ is given by:
$g = \frac{G M}{R^2}\,,
$
where $G$ is the universal gravitational constant.
Step 2: Express the masses of the planets in terms of density and radius
The mass $M$ of a spherical planet can be written using its density $\rho$ and volume:
$M = \rho \times \text{Volume} = \rho \times \frac{4}{3}\pi R^3\,.
For planet A (radius $R$ and density $\rho$):
$M_A = \rho \cdot \frac{4}{3}\,\pi\,R^3\,.
For planet B (radius $1.5R$ and density $\rho/2$):
$M_B = \left(\frac{\rho}{2}\right) \cdot \frac{4}{3}\,\pi \,(1.5R)^3\,.
Step 3: Compare the masses $M_B$ and $M_A$
Expand $(1.5R)^3 = 1.5^3 \times R^3 = 3.375\,R^3$. Hence,
$M_B = \frac{\rho}{2} \times \frac{4}{3}\,\pi \times 3.375 \,R^3
= \frac{4}{3} \,\pi \,\rho \,R^3 \times \frac{3.375}{2}
= \left(\frac{3.375}{2}\right) \,M_A
= 1.6875 \,M_A\,.
Step 4: Write expressions for $g_A$ and $g_B$
$g_A = \frac{G\,M_A}{R^2}
\quad\text{and}\quad
g_B = \frac{G\,M_B}{(1.5R)^2}\,.
Substitute $M_B = 1.6875\,M_A$ into the expression for $g_B$:
$g_B = \frac{G \left(1.6875\,M_A\right)}{(1.5R)^2} = \frac{1.6875\,G\,M_A}{2.25\,R^2}\,,
$
since $(1.5R)^2 = 2.25\,R^2.$
Step 5: Compute the ratio $g_B/g_A$
$\displaystyle \frac{g_B}{g_A}
= \frac{\frac{1.6875\,G\,M_A}{2.25\,R^2}}{\frac{G\,M_A}{R^2}}
= \frac{1.6875}{2.25}
= 0.75
= \frac{3}{4}\,.
$
Step 6: State the final ratio
Therefore, the ratio of the acceleration due to gravity at the surface of planet B to that of planet A is
$\frac{g_B}{g_A} = \frac{3}{4}\,,$
which can be written as 3 : 4.
