(a) One of the moons of Jupiter, named Io, has an orbital radius of 4.22 108 m and a period of 1.77 days. Assuming the orbit is circular, calculate the mass of Jupiter.
(b) The largest moon of Jupiter, named Ganymede, has an orbital radius of 1.07 109 m and a period of 7.16 days. Calculate the mass of Jupiter from this data.
(c) Are your results to parts (a) and (b) consistent?
Explain.
To solve these problems, we can use Kepler's third law of planetary motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
(a) To calculate the mass of Jupiter using information about its moon Io, we can use the formula:
T^2 = (4π^2 / GM) * r^3
Where:
T is the period of the moon's orbit (in seconds),
G is the gravitational constant (6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)),
M is the mass of Jupiter (in kilograms),
r is the orbital radius of the moon (in meters).
Given:
T = 1.77 days = 1.77 * 24 * 60 * 60 = 152,928 seconds (approximated)
r = 4.22 * 10^8 m
Rearranging the equation, we can solve for M:
M = (4π^2 / G) * r^3 / T^2
So, plugging in the values:
M = (4 * π^2 / G) * (4.22 * 10^8)^3 / (152928)^2
Calculating this expression yields the mass of Jupiter.
(b) To calculate the mass of Jupiter using information about its moon Ganymede, we use the same formula and rearrangement as before:
T = 7.16 days = 7.16 * 24 * 60 * 60 = 618,624 seconds (approximated)
r = 1.07 * 10^9 m
M = (4 * π^2 / G) * (1.07 * 10^9)^3 / (618624)^2
Calculating this expression yields the mass of Jupiter.
(c) To determine if the results from parts (a) and (b) are consistent, we compare the masses obtained. If the values are approximately the same, the results are consistent.
Explanation: If the calculations are done correctly, the masses obtained in parts (a) and (b) should be consistent. This is because the calculations are based on Kepler's third law, which holds true regardless of the moon used to derive the mass of Jupiter.
To calculate the mass of Jupiter using the given data, we can make use of Kepler's third law of planetary motion. According to this law, the square of the orbital period of a planet (in seconds) is directly proportional to the cube of the semi-major axis of its orbit (in meters). Mathematically, it can be expressed as:
T^2 ∝ R^3
Where T is the orbital period of the planet in seconds and R is the semi-major axis of its orbit in meters. The constant of proportionality can be written as:
T^2 = k * R^3
where k is a constant.
(a) For the moon Io with an orbital radius of 4.22 * 10^8 m and a period of 1.77 days, we need to convert the period to seconds. One day has 24 hours, and one hour has 60 minutes, and finally, one minute has 60 seconds. So, the period in seconds is:
1.77 days * 24 hours * 60 minutes * 60 seconds = 152,928 seconds
Now we can plug in the values into Kepler's third law equation:
152,928 seconds^2 = k * (4.22 * 10^8 m)^3
Now we can use this equation to find the value of k.
(b) For the moon Ganymede with an orbital radius of 1.07 * 10^9 m and a period of 7.16 days, we need to convert the period to seconds:
7.16 days * 24 hours * 60 minutes * 60 seconds = 618,336 seconds
Now we can plug in the values into Kepler's third law equation:
618,336 seconds^2 = k * (1.07 * 10^9 m)^3
Again, we need to find the value of k.
Once we have the values of k for both scenarios, we can calculate the mass of Jupiter using a common value for the constant of proportionality. Since the constant k represents the same value for both cases, the mass of Jupiter calculated using the data from part (a) and part (b) should also be the same.
(c) To check if the results from parts (a) and (b) are consistent, we can compare the masses obtained from both scenarios. If they are close to each other, it indicates consistency.
Please note that without numerical values for k from either scenario, we cannot provide an exact mass for Jupiter.
A) 1.90 x 10^27
B) 1.90 x 10^27
C)yes