Respuesta :
To find the period of Ganymede revolving around Jupiter, we can use Kepler's third law of planetary motion, which states:
(T1^2 / R1^3) = (T2^2 / R2^3)
Where:
- T1 and T2 are the periods of the two objects (Moon and Ganymede in this case).
- R1 and R2 are the average distances from the respective objects to their primary bodies (Earth and Jupiter in this case).
Given:
- Period of the Moon around the Earth (T1) = 27.3 days
- Radius of Ganymede's orbit around Jupiter (R2) = 3 times the radius of the Moon's orbit around the Earth
- Mass of Jupiter (MJ) = 318 times the mass of the Earth (ME)
First, let's find the radius of Ganymede's orbit around Jupiter relative to the Earth's orbit around the Moon:
R2 = 3 * R1
Next, let's find the ratio of the periods:
(T2^2 / R2^3) = (T1^2 / R1^3)
Substituting the known values:
(T2^2 / (3 * R1)^3) = (27.3^2 / R1^3)
Solving for T2:
T2^2 = (27.3^2 / R1^3) * (3 * R1)^3
T2^2 = (27.3^2 / R1^3) * 27
T2^2 = (27.3^2 * 27) / R1^3
T2^2 = (27.3^2 * 27) / R1^3
T2^2 = (27.3^2 * 27) / R1^3
T2^2 ≈ 219.542
Taking the square root of both sides:
T2 ≈ √219.542 ≈ 14.8 days
However, this gives us the period of Ganymede's orbit around Jupiter in terms of the Moon's orbital period around the Earth. Since the Moon's period is 27.3 days, we need to divide this value by the square root of Jupiter's mass relative to Earth's mass to get the period of Ganymede's orbit around Jupiter:
T2 ≈ 14.8 days / √(318) ≈ 8 days
So, the period of Ganymede revolving around Jupiter is approximately 8 days.
(T1^2 / R1^3) = (T2^2 / R2^3)
Where:
- T1 and T2 are the periods of the two objects (Moon and Ganymede in this case).
- R1 and R2 are the average distances from the respective objects to their primary bodies (Earth and Jupiter in this case).
Given:
- Period of the Moon around the Earth (T1) = 27.3 days
- Radius of Ganymede's orbit around Jupiter (R2) = 3 times the radius of the Moon's orbit around the Earth
- Mass of Jupiter (MJ) = 318 times the mass of the Earth (ME)
First, let's find the radius of Ganymede's orbit around Jupiter relative to the Earth's orbit around the Moon:
R2 = 3 * R1
Next, let's find the ratio of the periods:
(T2^2 / R2^3) = (T1^2 / R1^3)
Substituting the known values:
(T2^2 / (3 * R1)^3) = (27.3^2 / R1^3)
Solving for T2:
T2^2 = (27.3^2 / R1^3) * (3 * R1)^3
T2^2 = (27.3^2 / R1^3) * 27
T2^2 = (27.3^2 * 27) / R1^3
T2^2 = (27.3^2 * 27) / R1^3
T2^2 = (27.3^2 * 27) / R1^3
T2^2 ≈ 219.542
Taking the square root of both sides:
T2 ≈ √219.542 ≈ 14.8 days
However, this gives us the period of Ganymede's orbit around Jupiter in terms of the Moon's orbital period around the Earth. Since the Moon's period is 27.3 days, we need to divide this value by the square root of Jupiter's mass relative to Earth's mass to get the period of Ganymede's orbit around Jupiter:
T2 ≈ 14.8 days / √(318) ≈ 8 days
So, the period of Ganymede revolving around Jupiter is approximately 8 days.