Chapter 5, Problem 34P

To determine

The acceleration due to gravity at the surface of Europa.

Answer to Problem 34P

Solution:

The acceleration due to gravity at the surface of Europa is $1.49{\text{m/s}}^{\text{2}}$ .

Explanation of Solution

Given:

Mass of one of the moons of Jupiter is $m=4.9\times {10}^{22}\text{kg}$

Density is the same as Earth’s.

Formula used:

Acceleration due to gravity on the surface is given by:

  $g=\frac{Gm}{r^2}$

Where, m is the mass of the body and r is its radius.

Volume of a sphere is:

  $V=\frac{4}{3}\pi r^3$

  $\begin{array}{l}\rho =\frac{m}{V}\\ V=\frac{m}{\rho }\end{array}$

Where, $\rho$ is the density.

Calculations: Earth’s mean density is $5.515{\text{g/cm}}^{\text{3}}=5515{\text{kg/m}}^{\text{3}}$

Therefore, Europa has a volume of

  $\begin{array}{l}V=\frac{m}{\rho }\\ V=\frac{4.9\times {10}^{22}\text{kg}}{5515{\text{kg/m}}^{\text{3}}}\\ V=8.88\times {10}^{18}{\text{m}}^{\text{3}}\\ \frac{4}{3}\pi r^3=8.88\times {10}^{18}{\text{m}}^{\text{3}}\\ r^3=3.77\times {10}^{18}{\text{m}}^{\text{3}}\\ r=1.48\times {10}^6\text{m}\end{array}$

So, the radius of Europa is $1.48\times {10}^6\text{m}$

Acceleration,

  $\begin{array}{l}g=Gm/r^2\\ g=\frac{(6.67\times {10}^{-11})(4.9\times {10}^{22})}{{\left(1.48\times {10}^6\right)}^2}\\ g=1.49{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Hence, the acceleration due to gravity at the surface of Europa is $1.49{\text{m/s}}^{\text{2}}$ .