Newtons Laws EX 35

Io has an orbit of radius $r=4.22\times 10^{8}m$ and the period $T=1.77days=1.53\times 10^{5}s$. From the period and the radius, we can find the speed of Io:

$v={2\pi r \over T}={2\pi (4.22\times 10^{8}) \over 1.53\times 10^{5}}=1.73\times 10^{4}m/s$

Mass of Jupiter

Then we can use the Newton's second law to find the mass of Jupiter:

${GM_{J}m_{I} \over r^{2}}=m_{I}{v^{2} \over r}$

Cancel out $r$ and $m_{I}$ and substitute known values into the equation.

${(6.67\times 10^{-11})M_{J} \over 4.22\times 10^{8}}=(1.73\times 10^{4})^{2}$

From this equation we compute $M_{J}=1.9\times 10^{27}kg$.

Radius of Europa's orbit

The period of Europa is $T=3.55days=3.07\times 10^{5}s$. We substitute the formula for $v={2\pi r \over T}$ into the Newton's law

${GM_{J}m_{E} \over r^{2}}=m_{E}{({2\pi r \over T})^{2} \over r}$

and then cancel out the mass of Europa ($m_{E}$) and substitute the known values of other constants to obtain:

${(6.67\times 10^{-11})(1.9\times 10^{27}) \over r^{3}}={4\pi ^{2} \over (3.07\times 10^{5})^{2}}$

We solve to find $r=6.71\times 10^{8}m$.