Chapter 4, Problem 54Q

To determine

The data given is in agreement with Newton’s form of Kepler’s third law.

Answer to Problem 54Q

It is found that the data for all the three satellites are in agreement with Newton’s form of Kepler’s third law

Explanation of Solution

Given:

$P^2=\frac{4{\pi }^2{a}^3}{G(m_1+m_2)}$

The universal constant of gravitation $,G=6.67\times {10}^{11}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}$

Formula used:

The Kepler’s 3^rd Law, written in Newton’s form gives

$P^2=\frac{4{\pi }^2{a}^3}{G(m_1+m_2)}$

Calculation:

The Kepler’s 3^rd Law, written in Newton’s form gives

$\begin{array}{l}{P}^2=\frac{4{\pi }^2{a}^3}{G(m_1+m_2)}\\ \therefore m_1+m_2=\frac{4{\pi }^2{a}^3}{G{P}^2}\\ P=(3.551\text{days)(}\frac{\text{86400}\text{s}}{\text{1}\text{day}})=3.068\times {10}^5\text{s}\end{array}$

The mass of the Jupiter is $1.90\times {10}^{27}\text{kg}$ and the value of $m_1+m_2$ is equal to Jupiter’s mass

Therefore, the given data is correct.

Europa:

To calculate the total mass

$m_1+m_2=\frac{4{\pi }^2{a}^3}{G{P}^2}$

Converting the sidereal P from days to seconds, we get

$P=(3.551\text{days)(}\frac{\text{86400}\text{s}}{\text{1day}})=3.068\times {10}^5\text{s}$

We know that,

$a=6.709\times {10}^8\text{m},G=6.67\times {10}^{11}N{m}^2/k{g}^2\&P=3.068\times {10}^5\text{s}$

Therefore,

$\begin{array}{l}{m}_1+m_2=\frac{4{\pi }^2{(6.709\times {10}^8\text{m})}^3}{(6.67\times {10}^{-11}N{m}^2/k{g}^2){(3.068\times {10}^5\text{s})}^2}\\ m_1+m_2=(5.916\times {10}^{11})\frac{{(1.070\times {10}^9)}^3}{{(6.18\times {10}^5)}^2}\text{kg}\\ m_1+m_2=1.90\times {10}^{27}\text{kg}\end{array}$

Ganymede:

Total Mass,

$m_1+m_2=\frac{4{\pi }^2{a}^3}{G{P}^2}$

$P=(7.15\text{5}\text{days)}\left(\frac{\text{86400s}}{\text{1day}}\right)=6.18\times {10}^5\text{s}$

Substituting the values,

$a=1.070\times {10}^{8}m,G=6.67\times {10}^{11}\&P=6.18\times {10}^{5}s$

$\begin{array}{l}{m}_1+m_2=\frac{{\text{4π}}^{\text{2}}{\text{(1}{\text{.070×10}}^{\text{8}}\text{m)}}^{\text{3}}}{\text{(6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{)(6}{\text{.18×10}}^{\text{5}}{\text{s)}}^{\text{2}}}\\ m_1+m_2=(5.916\times {10}^{11})\frac{{\text{(1}{\text{.070×10}}^{\text{9}}\text{)}}^{\text{3}}}{{\text{(6}{\text{.18×10}}^{\text{5}}\text{)}}^{\text{2}}}\text{kg}\\ m_1+m_2=1.90\times {10}^{\text{27}}\text{kg}\end{array}$

Calisto:

Total mass,

$m_1+m_2=\frac{4{\pi }^2{a}^3}{G{P}^2}$

$P=(16.689\text{days)}\left(\frac{\text{86400s}}{\text{1day}}\right)=1.44\times {10}^6\text{s}$

Substituting,

$a=1.883\times {10}^9\text{m},G=6.67\times {10}^{11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\&P=1.44\times {10}^{\text{6}}\text{s}$

$\begin{array}{l}{m}_1+m_2=\frac{{\text{4π}}^{\text{2}}{\text{(1}{\text{.883×10}}^{\text{9}}\text{m)}}^{\text{3}}}{\text{(6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{)(1}{\text{.44×10}}^{\text{6}}{\text{s)}}^{\text{2}}}\\ m_1+m_2=(5.916\times {10}^{11})\frac{{\text{(1}{\text{.070×10}}^{\text{9}}\text{)}}^{\text{3}}}{{\text{(1}{\text{.44×10}}^{\text{6}}\text{)}}^{\text{2}}}\text{kg}\\ m_1+m_2=1.90\times {10}^{27}\text{kg}\end{array}$

Conclusion:

Thus, it is proved that the data is in agreement with Newton’s form of Kepler’s third law.