If a small planet were located 7 times as far as the Earth's distance from the Sun, it period is ___ earth years?
Use Kepler's third law.
(Period,y)^2/(Distance,a.u.)^3 = 1
(for objects orbiting the sun)
The distance is 7 a.u in this case.
(Period,y)^2 = 7^3 = 343
Period = 18.5 years
Well, if a small planet is located 7 times as far from the Sun as Earth, it's in for quite a "long-distance relationship"! In this scenario, we can use Kepler's Third Law of Planetary Motion, which tells us that the square of the period of a planet is proportional to the cube of its average distance from the Sun.
Since the distance has increased 7 times, the period squared should be multiplied by 7 cubed, which equals 343. Therefore, the small planet's period is approximately 343 Earth years. In human terms, that's one long orbit around the Sun! So, get ready to celebrate your birthday once every few centuries on this planet! 🎉
To find the period of a planet's orbit, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution is proportional to the cube of the average distance from the Sun.
Let's assume Earth's distance from the Sun is approximately 93 million miles, or 150 million kilometers.
If the small planet is located 7 times as far as Earth's distance from the Sun, its distance from the Sun would be 7 * 150 million kilometers = 1.05 billion kilometers.
Now, let's use the formula to find the period:
(T_small planet)^2 / (T_earth)^2 = (d_small planet)^3 / (d_earth)^3
where T_small planet and T_earth are the periods of revolution of the small planet and Earth, respectively, and d_small planet and d_earth are the corresponding distances from the Sun.
Substituting the values:
(T_small planet)^2 / (1 year)^2 = (1.05 billion km)^3 / (150 million km)^3
(T_small planet)^2 / 1 = (1.05^3) / (0.15^3)
(T_small planet)^2 = (1.157625) / (0.003375)
(T_small planet)^2 = 342.76296296
Taking the square root, we get:
T_small planet ≈ 18.5267 years
Therefore, the period of the small planet would be approximately 18.53 Earth years.
To determine the period of a planet, we can use Kepler's third law, also known as the law of periods. According to this law, the square of a planet's period (in Earth years) is proportional to the cube of its average distance from the Sun.
In this case, since the small planet is located 7 times Earth's distance from the Sun, we can make the following proportion:
(Period of the small planet)² : (Period of Earth)² = (Distance of small planet)³ : (Distance of Earth)³
Let's denote the period of the small planet as "P" and the period of Earth as "1" (since it's in Earth years). The distance of the small planet will be "7" (7 times Earth's distance). Plugging these values into the proportion, we get:
P² : 1² = 7³ : 1³
P² : 1 = 343 : 1
P² = 343
P = √343
Therefore, the period of the small planet is approximately 18.5 Earth years.