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)
G = 6.674*10^-11
F =G m M/r^2 = m v^2/r
G M= v^2 r
M = v^2 r/G
but v = 2 pi r/(365.24*24*3600)
To determine the mass of the Sun using the given information, we can use Kepler's third law of planetary motion. The law states that the square of the orbital period of a planet is directly proportional to the cube of its average distance from the Sun.
The equation for Kepler's third law is expressed as:
T^2 = (4π^2 / G * M) * 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)
- M is the mass of the Sun (what we want to solve for)
- r is the average distance between the Sun and Earth in meters
To calculate the mass of the Sun, we need to rearrange the equation to solve for M:
M = (T^2 * G) / (4π^2) * r^3
Let's substitute the given values into the equation:
T = 365.24 days * 24 hours * 60 minutes * 60 seconds
T = 31,557,600 seconds
r = 1.000 AU * 1.496 × 10^11 m / AU
r = 1.496 × 10^11 m
Now, let's calculate M:
M = (31,557,600^2 * 6.67430 × 10^-11) / (4π^2) * (1.496 × 10^11)^3
By simplifying the equation and performing the calculations, we will find the mass of the Sun.