How fat is Jupiter

This is a fairly simple problem, but has some interesting results:

Note that we are told that this is a circular orbit. From this case, we can deduce, for some angular velocity $\omega$ , radius of orbit $R$ , we have centripetal acceleration: $a_c = \omega^2R$ We also have some gravitational acceleration (note that we assume Callisto's mass is negligible in comparison, hence it has no effect on the motion of Jupiter, per se): $a_G=\frac{GM}{R^2}$ Setting these two equal (for circular motion due to gravity): $\frac{GM}{R^2}=\omega^2R$ This gives us, then: $M=\frac{\omega^2R^3}{G}$ Note that, for some time $T$ : $\omega = \frac{2\pi}{T}$ Of course, then: $M\approx\frac{\left(\frac{2\pi}{16.7\cdot86400\text{ s}}\right)^2(1.9\cdot10^9\text{ m})^3}{6.673\cdot10^{-11}\text{ m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2}}\approx 1.94883\cdot10^{27}\text{ kg}$ Which is our answer.

We know that, the mean orbital velocity of the natural satellite i.e $v$ can be expressed as $v=\frac{2 \times \Pi \times r}{P}$ according to Kepler's Third Law where, $P$ is the time period of revolution of Callisto and $r$ is the semi major axis, here it is the radius since the trajectory is circular. We can equate the centripetal force exerted by Jupiter on Callisto by Jupiter with force of attraction between Callisto and Jupiter. $\frac{G \times M_J \times M_C}{r^{2}}=\frac{M_C \times v^{2}}{r}$ where, $M_J$ is the mass of Jupiter. $M_C$ is the mass of Callisto. $G$ is the Newton's Gravitational Constant. $v$ is the mean orbital velocity of Callisto. $\Rightarrow M_J=\frac{v^{2} \times r}{G}$ On substituting the values of $G,v,r$ we get, $\Rightarrow M_J=\boxed{1.95 \times 10^{27}kg}$

First step: Convert radius into meters. $r=1.9 \times {10}^9$ m.

Second step: Convert time into seconds. $T=16.7 \times 86400=1442880$ seconds.

$m$ is the mass of moon and $M$ is the mass of Jupiter.

The centripetal force on the moon must be equal to the force on the moon due to mass of Jupiter.

$\frac{GMm}{r^2}=\frac{mv^2}{r}$ .

Cancelling out $m$ and $r$

$M=\frac{v^2\times r}{G}$

velocity $v$ of the moon is equal to $\frac{2\pi r}{T}$

Plugging in the values we get $M=\boxed{1.95 \times {10}^{27}}$ .

You can use the period $T$ and radius $R$ to find the orbital speed $v=\frac{2\pi R}{T}$

Next, the gravitational force between Jupiter and Callisto will be equal to the mass of Callisto times its centripetal acceleration: $\frac{G M_{J} M_{C}}{R^2} = \frac{M_{C} v^2} {R}$

Putting all of that together, $M_{J} = \frac{R v^2}{G} = \frac{R^3 4\pi^2}{G T^2}=1.96\times10^{27}$ kg

As given by the problem, the moon is in circular orbit, so the net centripetal force on the moon is $\text{F}_\text{net} = \frac{\text{M}_\text{Callisto} \times \text{V}_\text{Callisto}^2}{\text{R}_\text{Callisto-Jupiter}}$ However, this net centripetal force is caused by the gravitational force between the two bodies. Thus the two forces are equal, and $\text{F}_\text{net} = G\frac{\text{M}_\text{Callisto} \times \text{M}_\text{Jupiter}}{\text{R}_\text{Callisto-Jupiter}} = \frac{\text{M}_\text{Callisto} \times \text{V}_\text{Callisto}^2}{\text{R}_\text{Callisto-Jupiter}}$ where G is the Newtonian gravitational constant. Simplifying the equation, we have that $G\frac{\text{M}_\text{Callisto} \times \text{M}_\text{Jupiter}}{\text{R}_\text{Callisto-Jupiter}} = \frac{\text{M}_\text{Callisto} \times \text{V}_\text{Callisto}^2}{\text{R}_\text{Callisto-Jupiter}}$ $\text{M}_\text{Jupiter} = \frac{\text{V}_\text{Callisto}^2}{G}$ $\text{M}_\text{Jupiter} = \frac{(\frac{2 \pi \cdot \text{R}_\text{Callisto-Jupiter}}{\text{T}_\text{Callisto}})^2}{G}$

Where $\text{T}_\text{Callisto}$ is the period of orbit of Callisto. All that's left is to substitute the numbers! Converting all given numbers to SI units, we have

$\text{M}_\text{Jupiter} = \frac{(\frac{2 \pi \cdot (1.6 \times 10^9 \text{m})}{1.443 \times 10^6 \text{s}})^2}{G}$

I'll save the calculation for you and just let you know that the final answer is $\text{M}_\text{Jupiter} = \boxed{1.95 \times 10^{27} \text{kg}}$ .

By the Kepler laws we have that: $$ $\frac{P^2}{r^3_{med}} = \frac{4\pi^2}{G m}$ being $$ $P \rightarrow$ Period ( $1442880$ seconds) $$ $r \rightarrow$ Medium radius ( $1,9 . 10^9$ meters) $$ $G \rightarrow$ Gravitational constant ( $6,67384 . 10^{-11} \frac{m^3}{kg \times s}$ ) $$ $m \rightarrow$ Mass of Jupter $$ Rearranging, we have: $$ $m = \frac{P^2 G}{r^3_{med} 4\pi^2 }$ , then substituing the values: $$ $m = \frac{4 \times \pi^2 \times (1,9 . 10^9)^3}{1442880^2 \times 6,67384.10^{-11}}$ $$ $m = 1,94887.10^{27} kg$

We must think of how we can use the given info to come up with 2 equations that will eventually give the mass of Jupiter. We have to cancel out the mass of Callisto and have to make good use of the period.

The idea is to use 2 equations: Newton's law of universal gravitation and centriprital accelaration Therefore, $F = G\frac{m_{j}m_{c}}{R^{2}} = m_{c}\frac{v^{2}}{R}$ , where $m_{j}$ = mass of Jupiter, $m_{c}$ = mass of Callisto, $v$ =velocity of Callisto around Jupiter

So, rearranging the equation, we get $m_{c} = \frac{v^{2}R}{G}$ , $G$ = gravitational constant = $6.67\times10^{-11}$ (why didnt they give the value?)

By plugging in the values, we find $V = \frac{2 \pi R}{16.7\times24\times60\times60}$ , $R = 1.9\times10^{6}\times10^{3}$ and finally $m_{c} = 1.95\times10^{27}$