# Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one...

## Question:

Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one revolution every 18h. The other satellite has orbital radius, 3.9r. How long does the second satellite take to make one revolution around Jupiter?

## Kepler's Third Law

The Kepler's third law of planetary motion states the square of the time period of revolution around a star by a planet or around a planet by a satellite, is directly proportional to the cube of the mean distance from the star or planet. Mathematically

{eq}\begin{align} T^2 \propto a^3 \end{align} {/eq}

Where T is the time period and a is the radius of the path.

Data Given

• Radius of the orbit of satellite one {eq}a_1 = r {/eq}
• Time period of satellite one {eq}T_1 = 18 \ \rm h {/eq}
• Radius of the orbit of satellite two {eq}a_2 = 3.9 \ r {/eq}

Now using Kepler's third law for satellite one

{eq}\begin{align} T_1^2 \propto a_1^3 \end{align} {/eq}

For satellite two

{eq}\begin{align} T_2^2 \propto a_2^3 \end{align} {/eq}

Dividing the second equation with first we get

{eq}\begin{align} \frac{T_2^2}{T_1^2} = \frac{a_2^3}{a_1^2} \end{align} {/eq}

Because the proportionality constant get cancelled from both the equations

{eq}\begin{align} T_2 =\left ( \frac{a_2}{a_1} \right ) ^{\frac{3}{2}} T_1 \end{align} {/eq}

{eq}\begin{align} T_2 =\left ( \frac{3.9 \ r}{r} \right ) ^{\frac{3}{2}} \times 18 \ \rm h \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ T_2 = 138.6 \ \rm h }} \end{align} {/eq}