Question

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Answer 1

To calculate the orbital period for Jupiter's moon Io, use Kepler's Third Law and plug in the values for the radius and mass of Jupiter. The orbital period for Io is approximately 1.763 days.

Step-by-step explanation:

To calculate the orbital period for Jupiter's moon Io, we can use Kepler's Third Law. This law states that the period of an orbit is related to the radius of the orbit by the formula T2 = 4π2 / GM, where T is the period, G is the gravitational constant, and M is the mass of the planet.

In this case, the orbital radius of Io is given as 4.22×105 km. Converting this to meters, we get 4.22×108 m. The mass of Jupiter is given as 1.9×1027 kg. Plugging these values into the formula, we can calculate the orbital period for Io.

Using the formula T2 = 4π2 / GM, we have T2 = (4 * 3.14162 ) / (6.67 × 10-11) * (1.9 x 1027) / (4.22 x 108)

Simplifying the equation gives us T ≈ 1.763 days

Answer 2

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period
`T` of a body (moon, planet, satellite) orbiting a greater body in space with the size
`a` of its orbit.

This Law is originally expressed as follows:

`T^(2) =(4\pi^(2))/(GM)a^(3)` (1)

Where;

`G` is the Gravitational Constant and its value is
`6.674(10^(-11))(m^(3))/(kgs^(2))`

`M=1.9(10^(27))kg` is the mass of Jupiter

`a=4.22(10^(5))km=4.22(10^(8))m` is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

`T=\sqrt{(4\pi^(2))/(GM)a^(3)}` (2)

`T=\sqrt{(4\pi^(2))/(6.674(10^(-11))(m^(3))/(kgs^(2))1.9(10^(27))kg)(4.22(10^(8))m)^(3)}`

`T=\sqrt{(2.966(10^(27))m^(3))/(1.268(10^(17))m^(3)/s^(2))}`

`T=\sqrt{2.339(10^(10))s^(2)}`

Then:

`T=152938.0934s` (3)

Which is the same as:

`T=42.482h`

Therefore, the answer is:

The orbital period of Io is 42.482 h