Question

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

Answer 1

The radius of its orbit is
`R \approx 1.899* 10^(9)\,m`.

Step-by-step 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 = (2\pi)/(T)` (2)

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

`a_(net) = (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:

`(4\pi^(2)\cdot R)/(T^(2)) = (G\cdot m)/(R^(2))`

`R^(3) = (G\cdot m \cdot T^(2))/(4\pi^(2))`

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

If we know that
`G = 6.674* 10^(-11)\,(m^(3))/(kg\cdot s^(2))`,
`m = 1.90* 10^(27)\,kg` and
`T = 1460160\,s`, then the radius of the orbit of Callisto is:

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

`R \approx 1.899* 10^(9)\,m`