Calculate the orbital period for a planet that is 4 AU from the Sun.
To calculate the orbital period for a planet, we can use Kepler's third law of planetary motion. According to this law, the square of the orbital period (T) of a planet is proportional to the cube of its average distance from the Sun (r).
The equation can be represented as: T^2 = k * r^3
To find the value of k, we need to consider the known values for the Earth's orbital period (T_Earth) and distance from the Sun (r_Earth).
T_Earth^2 = k * r_Earth^3
Now, let's find the value of k.
The average distance of the Earth from the Sun is about 1 AU, which is approximately 149.6 million kilometers (km). The Earth's orbital period is about 365.25 days, which is roughly 31,557,600 seconds (s).
Substituting these values into the equation, we get:
(31,557,600 s)^2 = k * (149.6 million km)^3
Simplifying this equation gives us:
(996,677,721,600 s^2) = k * (3.5460736 × 10^25 km^3)
Now, we can solve for the value of k:
k = (996,677,721,600 s^2) / (3.5460736 × 10^25 km^3)
k ≈ 2.96 × 10^-19 (s^2/km^3)
Once we have the value of k, we can calculate the orbital period for a planet at a given distance (r) from the Sun.
Let's calculate the orbital period for a planet that is 4 AU from the Sun.
First, convert the distance from AU to kilometers:
1 AU ≈ 149.6 million km
So, 4 AU ≈ 4 * 149.6 million km
4 AU ≈ 598.4 million km
Now, substitute this value into the equation:
T^2 = (2.96 × 10^-19 s^2/km^3) * (598.4 million km)^3
T^2 ≈ (2.96 × 10^-19 s^2/km^3) * (2.1382290944 × 10^23 km^3)
Simplifying this equation gives us:
T^2 ≈ 6.36 × 10^4 s^2
Finally, we can solve for the orbital period (T) by taking the square root:
T ≈ √(6.36 × 10^4 s^2)
T ≈ 252.3 seconds
Therefore, the orbital period for a planet that is 4 AU from the Sun is approximately 252.3 seconds.