Chapter 5, Problem 80GP

To determine

To find: The value of acceleration due to gravity at equator of Jupiter.

Answer to Problem 80GP

  $2.34g_{\text{E}}$

Explanation of Solution

Given Data:

  $m=1.9\times {10}^{27}$ kg

  $R_E=7.1\times {10}^4$ km

and $T=9$ hr $55$ min

Formula Used:

An object moving in a circular path at constant speed the net force must point to the center of the

circle, thus he acceleration due to gravity $\text{'}g\text{'}$ is reduced to due to centripetal acceleration.

Apply the Newton laws by choosing the inward direction as positive, thus the net force is

  $\Sigma F_{\text{R}}=m{g}_{\text{j}}-F_{\text{N}}=m{a}_{\text{R}}$

Rewrite the above equation.

  $\begin{array}{l}{F}_{\text{N}}=m{g}_j-m{a}_{\text{R}}\\ =m\left(g_j-\frac{v^2}{R_j}\right)\end{array}$

From equation (1), acceleration due to gravity on the surface of the Jupiter is

  $g_j=\frac{G{M}_{\text{j}}}{R_{\text{j}}^2}$

Thus, the normal force is

  $F_{\text{N}}=m\left(\frac{G{M}_{\text{j}}}{R_{\text{J}}^2}-\frac{v^2}{R_{\text{j}}}\right)$

The linear speed of the person on the planet is

  $v=\frac{2\pi R_{\text{j}}}{T}$

  $F_{\text{N}}=m\left(\frac{G{M}_{\text{j}}}{R_{\text{J}}^2}-\frac{{\left(2\pi R_{\text{j}}\right)}^2}{T^2{R}_{\text{J}}}\right)$

The perceived acceleration at the equator is given as,

  $\begin{array}{l}m{g}^{\prime }=m\left(\frac{G{M}_1}{R_j^2}-\frac{{\left(2\pi R_j\right)}^2}{T^2{R}_j}\right)\\ g^{\prime }=\left(\frac{G{M}_{\text{J}}}{R_{\text{J}}^2}-\frac{{\left(2\pi R_{\text{j}}\right)}^2}{T^2{R}_{\text{J}}}\right)\end{array}$

Calculation:

Substituting the values

  $\begin{array}{l}{g}^{\prime }=\left(\frac{G{M}_{\text{J}}}{R_{\text{J}}^2}-\frac{{\left(2\pi R_{\text{j}}\right)}^2}{T^2{R}_{\text{J}}}\right)\\ g^{\prime }=\frac{\left(6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.9\times {10}^{27}\text{kg}\right)}{{\left(7.1\times {10}^7\text{m}\right)}^2}-{\left(\frac{2\pi }{35700\text{s}}\right)}^2\left(7.1\times {10}^7\text{m}\right)\\ =22.95{\text{m/s}}^2\end{array}$

In term of $g$ we need to divide with $9.8\text{m}/{\text{s}}^2$ .

  $\begin{array}{l}{g}^{\prime }=\left(22.95\text{m}/{\text{s}}^2\right)\left(\frac{1g}{9.8\text{m}/{\text{s}}^2}\right)\\ =(2.34)g\end{array}$ .

Conclusion:

Therefore, the acceleration experienced by a person at the equator of the Jupiter planet is $2.34g_{\text{E}}$