Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is$4.9\times {10}^{22}\mathrm{kg}$and making the assumption that its mass per unit volume is the same as Earth’s.

The mass of the Europa is $m_E=4.9\times {10}^{22}\mathrm{kg}$.

In order to estimate the acceleration due to gravity at the surface of Europa, the formula of acceleration due to gravity along with the object’s mass, its radius and gravitational constant is applied.

The relation of density is given by,

$\begin{array}{c}\rho ={\rho }_E\\ \frac{m}{V}=\frac{m_E}{V_E}\\ \frac{m}{\frac{4}{3}\pi r^3}=\frac{m_E}{\frac{4}{3}\pi {r_E}^3}\\ r_E=r{\left[\frac{m_E}{m}\right]}^{1/3}\end{array}$

Here, $\rho$and ${\rho }_E$are the density of Earth and Europa, $V$and $V_E$are the volume of Earth and Europa, m is the mass of Earth, r is the radius of the Earth and $r_E$is the radius of the Europa.

The relation of gravitational acceleration is given by,

$g_E=\frac{G{m}_E}{{r_E}^2}$

Here, G is the gravitational constant.

On plugging the values in the above relation.

$\begin{array}{c}{g}_E=\frac{G{m}_E}{{\left[r{\left(\frac{m_E}{m}\right)}^{1/3}\right]}^2}\\ g_E=\left(\frac{Gm}{r^2}\right)\times \left(\frac{{m_E}^{1/3}}{m^{1/3}}\right)\\ g_E=g\times \left(\frac{{m_E}^{1/3}}{m^{1/3}}\right)\end{array}$

Here, g is the Earth’s gravitational acceleration.

Substitute the known values in the above relation.

$\begin{array}{l}{g}_E=\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\times \left(\frac{{\left(4.9\times {10}^{22}\mathrm{kg}\right)}^{1/3}}{{\left(5.98\times {10}^{24}\mathrm{kg}\right)}^{1/3}}\right)\\ g_E=1.98\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $g_E=1.98\mathrm{m}/{\mathrm{s}}^2$is the required acceleration due to gravity.