Imagine that a new asteroid is discovered in
the solar system with a circular orbit and an
orbital period of 7 years.
What is the average distance of this object
from the Sun in Earth units?
Well, if this asteroid has a circular orbit with an orbital period of 7 years, it's clearly not in a rush to get anywhere! It's taking its sweet time, kind of like that person who stands in the middle of a busy sidewalk during rush hour.
Now, to calculate the average distance of this asteroid from the Sun in Earth units, we can use Kepler's Third Law. According to this law, the square of the orbital period of a planet (or in this case, an asteroid) is directly proportional to the cube of the average distance from the Sun.
So, let's do some math. Since the orbital period of this asteroid is 7 years, we square that to get 49. And since we're looking for the average distance, we can call it "r." Therefore, we have the equation:
49 = r^3
Now, we just need to solve for "r." Taking the cube root of both sides, we get:
r = 3.683
So, the average distance of this asteroid from the Sun in Earth units is approximately 3.683 times the average distance from the Earth to the Sun. Well, let's hope it doesn't get too close and start borrowing our sunscreen!
To calculate the average distance of the asteroid from the Sun in Earth units, we can use Kepler's Third Law, also known as the law of periods. According to this law, the square of the orbital period is proportional to the cube of the distance from the Sun.
The formula is: T^2 = (4π^2 / GM) r^3
Where:
T = orbital period
G = gravitational constant
M = mass of the Sun
r = distance from the Sun
Given that the orbital period (T) is 7 years, we need to determine the distance (r).
Let's assume the orbital period of Earth around the Sun is approximately 1 year, and the average distance of Earth from the Sun is approximately 93 million miles or 150 million kilometers.
Using this information, we can set up the following equation to compare the asteroid's orbit with that of Earth:
(7 years)^2 = (1 year)^2 × (r in Earth units)^3
Rearranging the equation, we get:
(r in Earth units)^3 = (7 years)^2 / (1 year)^2
Simplifying the equation, we get:
(r in Earth units)^3 = 49
Taking the cube root of both sides, we find:
r in Earth units ≈ ∛49
Therefore, the average distance of this asteroid from the Sun in Earth units is approximately 3.68.
To find the average distance of an object from the Sun in Earth units, we will use Kepler's Third Law of Planetary Motion. This law states that the square of the orbital period (P) is directly proportional to the cube of the average distance (r) between the object and the Sun.
Mathematically, it can be stated as:
P^2 = r^3
Given that the orbital period of the asteroid is 7 years, we can substitute this value into the equation:
(7 years)^2 = r^3
Simplifying further, we find:
49 years^2 = r^3
To calculate the average distance, we need to find the cube root of 49 years^2, which will give us the distance (r) in Earth units.
Taking the cube root of both sides, we have:
∛(49 years^2) = ∛r^3
∛(49 years^2) = r
Calculating the cube root of 49 years^2, we find that r ≈ 3.78.
Therefore, the average distance of this asteroid from the Sun in Earth units is approximately 3.78.
Doesn't Kepler's law apply?
T^2/r^3=constant=for Earth T^2/R^3
r^3=7^2*(92millionmiles)^3/1year^2
r= 92million miles * cuberoot(49)