Given that Earth orbits the Sun with a semimajor axis of 1.000 AU and an approximate orbital period of 365.24 days, determine the mass of the Sun. (1 AU = 1.496 ✕ 1011 m)
Just plug your numbers into the formula from Kepler's 3rd law:
T^2 = 4π^2 a^3/(GM)
where a is the semimajor axis of the orbit.
I have laid this all out for you.
Now Steve has done the same.
Please try yourself now.
To determine the mass of the Sun using the orbital information of Earth, we can apply Kepler's third law of planetary motion. Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit.
The equation for Kepler's third law is:
T^2 = (4π^2 / G * (M1 + M2)) * r^3
Where:
T is the orbital period of Earth (in seconds)
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
M1 is the mass of the Sun (what we are trying to find)
M2 is the mass of Earth (approximately 5.972 × 10^24 kg)
r is the semimajor axis of Earth's orbit (in meters)
First, we need to convert the semimajor axis of Earth's orbit from Astronomical Units (AU) to meters:
1 AU = 1.496 × 10^11 meters
r = 1.000 AU * 1.496 × 10^11 meters/AU
r ≈ 1.496 × 10^11 meters
Next, we need to convert the orbital period of Earth from days to seconds:
365.24 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute
T ≈ 31,556,926 seconds
Now we can rearrange the equation to solve for M1 (the mass of the Sun):
M1 = (T^2 * G * (M2 + M1)) / (4π^2 * r^3)
Substituting the known values:
M1 = (31,556,926 s^2 * 6.67430 × 10^-11 m^3 kg^-1 s^-2 * (5.972 × 10^24 kg + M1)) / (4π^2 * (1.496 × 10^11 m)^3)
Simplifying the equation and expanding the expression, we get:
M1 = 2.968583 × 10^29 kg + (M1 * 9.871578 × 10^-12) / (2.805281 × 10^33)
Next, move (M1 * 9.871578 × 10^-12) to the left side:
M1 - (M1 * 9.871578 × 10^-12) / (2.805281 × 10^33) = 2.968583 × 10^29 kg
Combining like terms gives:
M1 - (M1 * 9.871578 × 10^-12) / (2.805281 × 10^33) = 2.968583 × 10^29 kg
Multiplying both sides by the denominator on the left side:
M1 * (1 - 9.871578 × 10^-12 / (2.805281 × 10^33)) = 2.968583 × 10^29 kg
Simplifying further:
M1 * (2.805281 × 10^33 - 9.871578 × 10^-12) / (2.805281 × 10^33) = 2.968583 × 10^29 kg
To find M1, divide both sides by (2.805281 × 10^33 - 9.871578 × 10^-12):
M1 = (2.968583 × 10^29 kg) / (2.805281 × 10^33 - 9.871578 × 10^-12)
Calculating this on a calculator or computer will give the mass of the Sun.