Chapter 9, Problem 5P

Olympus Mons is 26 kilometers above the surrounding surface of Mars. Calculate the surface gravity on Mars at the surface and on top of Olympus Mons (neglecting any effects on gravity from the mountain itself). What percentage of the surface value is the difference in the acceleration of gravity?

To determine

The surface gravity on Mars at the surface and on top of Olympus Mons and the percentage of the difference between the acceleration of gravity to the surface value.

Answer to Problem 5P

The surface gravity on Mars at the surface is $3.71{\text{ m/s}}^2$, the value on top of Olympus Mons is $3.66{\text{ m/s}}^2$ and the percentage of the difference between the acceleration of gravity to the surface value is $1.35\%$ .

Explanation of Solution

Write the equation for the acceleration of gravity at the surface.

  $g_s=\frac{GM}{R^2}$        (I)

Here, $g_s$ is the acceleration of gravity at the surface, $G$ is the universal gravitation constant, $M$ is the mass of the Mars and $R$ is the radius of Mars.

Write the equation for the acceleration of gravity on top of Olympus Mons.

  $g_t=\frac{GM}{{\left(R+h\right)}^2}$        (II)

Here, $g_t$ is the acceleration of gravity on top of Olympus Mons and $h$ is the height of the mountain from the surface of Mars.

Write the equation for the required percentage.

  $\text{percentage}=\frac{g_s-g_t}{g_s}\times 100\%$        (III)

Conclusion:

The value of $G$ is $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$, the mass of Mars is $6.42\times {10}^{23}\text{ kg}$ and the radius of Mars is $3396\times {10}^3\text{ m}$ .

Substitute $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$ for $G$, $6.42\times {10}^{23}\text{ kg}$ for $M$ and $3396\times {10}^3\text{ m}$ for $R$ in equation (I) to find $g_s$ .

  $\begin{array}{c}{g}_s=\frac{\left(6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2\right)\left(6.42\times {10}^{23}\text{ kg}\right)}{{\left(3396\times {10}^3\text{ m}\right)}^2}\\ =3.71{\text{ m/s}}^2\end{array}$

Substitute $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$ for $G$, $6.42\times {10}^{23}\text{ kg}$ for $M$, $3396\times {10}^3\text{ m}$ for $R$ and $26\text{ km}$ for $h$ in equation (II) to find $g_t$ .

  $\begin{array}{c}{g}_t=\frac{\left(6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2\right)\left(6.42\times {10}^{23}\text{ kg}\right)}{{\left(3396\times {10}^3\text{ m}+26\text{ km}\frac{1000\text{ m}}{1\text{ km}}\right)}^2}\\ =3.66{\text{ m/s}}^2\end{array}$

Substitute $3.71{\text{ m/s}}^2$ for $g_s$ and $3.66{\text{ m/s}}^2$ for $g_t$ in equation (III) to find the percentage.

  $\begin{array}{c}\text{percentage}=\frac{3.71{\text{ m/s}}^2-3.66{\text{ m/s}}^2}{3.71{\text{ m/s}}^2}\times 100\%\\ =1.35\%\end{array}$

Therefore, the surface gravity on Mars at the surface is $3.71{\text{ m/s}}^2$, the value on top of Olympus Mons is $3.66{\text{ m/s}}^2$ and the percentage of the difference between the acceleration of gravity to the surface value is $1.35\%$ .