# Europa, a moon of Jupiter has an orbital diameter of 1.34 \times 10^9 m and a period of 3.55...

## Question:

Europa, a moon of Jupiter has an orbital diameter of {eq}1.34 \times 10^9 {/eq} m and a period of 3.55 days. What is the mass of Jupiter?

## Planets:

When a planets move round the Jupiter due to the gravitational attraction of the Jupiter, then the relation between the time period of the orbit and the radius of the orbit that is given by: {eq}\displaystyle T^2 = \frac{4 \pi^2 a^3}{G M} {/eq} where:

• {eq}T {/eq} is time taken to compute one orbit that is called time period.
• {eq}a {/eq} is the distance between the planet and Jupiter that is called the radius of the orbit.
• {eq}M {/eq} is the mass of the Jupiter.
• {eq}G = 6.67 \times 10^{-11} {/eq} is the universal gravitational constant.

Given:

• The time period of the orbit is: {eq}T = 3.55 \ \rm days = 3.55 \times 24 \times 60 \times 60 \ \rm sec = 306,720 \ \rm sec {/eq} and the diameter of the Jupiter is: {eq}d = 1.34 \times 10^{9} \ \rm m {/eq}.

As we know the radius of the orbit equals the distance between the satellite and the center of the Jupiter, therefore we can write:

\begin{align*} \displaystyle a &= R_J + r &\text{(Where } R_J \text{ is the radius of the Jupiter and } r \text{ is the distance between the satellite and earth)}\\ a &= \frac{d}{2} + 0.71492 \times 10^8 &\text{(As we know the distance between the moon and Jupiter)}\\ a &= \frac{1.34 \times 10^{9}}{2} + 0.71492 \times 10^8 \\ a &= 6.7 \times 10^8 + 0.71492 \times 10^8 \\ a &= 7.41492 \times 10^8 \ \rm m \end{align*}'

Now we will compute the mass of the Jupiter.

As we know the relation between the time period and radius of the orbit is:

\begin{align*} \displaystyle T^2 &= \frac{4 \pi^2 a^3}{G M} &\text{(Where } T = 306,720 \ \rm sec, G = 6.67 \times 10^{-11}, \text{ and } a = 7.41492 \times 10^8 \text{)}\\ 306,720^2 &= \frac{4 \times 3.14^2 \times (7.41492 \times 10^8)^3}{6.67 \times 10^{-11} \times M} &\text{(Plugging in all given values)}\\ M &= \frac{4 \times 3.14^2 \times (7.41492 \times 10^8)^3}{6.67 \times 10^{-11} \times 306,720^2 } \\ &= \frac{16,078.247 \times 10^{24}}{6.275} \\ &= 2,562.27 \times 10^{24} \\ M \ &\boxed{ = 2.562 \times 10^{27} } \ \rm kg \end{align*} 