Question

Please help i forgot how to solve this, thank you.

Answer 1

Answer:

The radius of its orbit is $R \approx 1.899\times 10^{9}\,m$.

Explanation:

Let suppose that Callisto rotates around Jupiter in a circular path and at constant speed, then we understand that net acceleration of this satellite is equal to the centripetal acceleration due to gravity of Jupiter. That is:

$\omega^{2}\cdot R = a_{net}$ (1)

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of the orbit, measured in meters.

$a_{net}$ - Net acceleration, measured in meters per square second.

In addition, angular speed can be described in terms of period ($T$), measured in seconds:

$\omega = \frac{2\pi}{T}$ (2)

And the net acceleration by the Newton's Law of Gravitation:

$a_{net} = \frac{G\cdot m}{R^{2}}$ (3)

Where:

$G$ - Gravitation constant, measured in cubic meters per kilogram-square second.

$m$ - Mass of Jupiter, measured in kilograms.

Now we apply (2) and (3) in (1) to derive an expression for the radius of the orbit:

$\frac{4\pi^{2}\cdot R}{T^{2}} = \frac{G\cdot m}{R^{2}}$

$R^{3} = \frac{G\cdot m \cdot T^{2}}{4\pi^{2}}$

$R = \sqrt[3]{\frac{G\cdot m\cdot T}{4\pi^{2}} }$ (4)

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$, $m = 1.90\times 10^{27}\,kg$ and $T = 1460160\,s$, then the radius of the orbit of Callisto is:

$R = \sqrt[3]{\frac{(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} )\cdot (1.90\times 10^{27}\,kg)\cdot (1460160\,s)^{2}}{4\pi^{2}} }$

$R \approx 1.899\times 10^{9}\,m$