91Ó°ÊÓ

The mean orbital radii ‘a’ and periods ‘T’ for four moons of Jupiter.

Plot the straight line graph of log a vs log T from the given data. Then, find its slope. Using Kepler’s third law and the intercept of the line, find the mass of Jupiter. According to Kepler’s third law, the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits.

The formula is as follows:

According to Kepler’s third law,${\mathrm{T}}^2=\frac{4{\mathrm{π}}^2}{\mathrm{GM}}{\mathrm{a}}^3$

where T is time, G is gravitational constant, M is mass and a is the radius.

The graph of log a against log T is given below,

The slope of the graph of log a against log T is,

$\frac{\text{AB}}{\text{BC}}=\frac{1.8}{2.7}$

$\frac{\text{AB}}{\text{BC}}=\frac{2}{3}$

Kepler’s third law gives,

${\mathrm{T}}^2=\frac{4{\mathrm{Ï€}}^2}{\mathrm{GM}}{\mathrm{a}}^3$

${\mathrm{a}}^3=\frac{\mathrm{GM}}{4{\mathrm{Ï€}}^2}{\mathrm{T}}^2$

Taking the log of both sides,

$3\mathrm{loga}=\log \left(\frac{\mathrm{GM}}{4{\mathrm{Ï€}}^2}\right)+2\mathrm{logT}$

$\mathrm{loga}=\frac{1}{3}\log \left(\frac{\mathrm{GM}}{4{\mathrm{Ï€}}^2}\right)+\frac{2}{3}\mathrm{logT}$

This equation suggests that the graph for log a against log T is a straight line and 2/3 is the slope of the graph.

Therefore, the value of the slope found in the graph is exactly the same as the value found in Kepler’s third law.

Hence, the slope of the graph of log a against log T is$\frac{2}{3}.$

The intercept of the graph is,

$\frac{1}{3}\log \left(\frac{\mathrm{GM}}{4{\mathrm{Ï€}}^2}\right)=5.17$

So,

$\log \left(\frac{\mathrm{GM}}{4{\mathrm{Ï€}}^2}\right)=3(5.17)$

$\left(\frac{\mathrm{GM}}{4{\mathrm{Ï€}}^2}\right)=\text{Antilog}\left(15.51\right)$

$\begin{array}{c}\mathrm{M}=\frac{3.23×{10}^{15}\left(4\right){\left(3.142\right)}^2}{6.67×{10}^{−11}\text{ }}\mathrm{kg}\\ =1.9×{10}^{27}\text{ }\mathrm{kg}\end{array}$

Hence, the mass of Jupiter from the intercept of the graph with the y-axis is $1.9×{10}^{27}\text{kg}$.

Therefore, using Kepler’s third law, the mass of the planet can be found.