How many times further from the Sun than the Earth is an asteroid with an average orbital period of 7.43 years? (HINT: Let the distance to the Earth be R=1 AU)
To find out how many times further an asteroid is from the Sun compared to the Earth, we need to calculate the average distance of the asteroid from the Sun and divide it by the average distance of the Earth from the Sun.
Given that the distance to the Earth from the Sun is defined as 1 Astronomical Unit (1 AU), we will use this value as a reference.
We can use Kepler's third law of planetary motion to calculate the average distance of the asteroid from the Sun. The law states that the square of a planet's orbital period is proportional to the cube of its semi-major axis.
Let's denote the average distance of the asteroid from the Sun as "A". And the average orbital period of the asteroid as "T".
Using Kepler's third law, we can write the equation as:
(T_asteroid^2) / (T_earth^2) = (A_asteroid^3) / (A_earth^3)
Substituting the given values:
T_asteroid = 7.43 years
T_earth = 1 year
A_earth = 1 AU
We can solve for A_asteroid:
(7.43^2) / (1^2) = (A_asteroid^3) / (1^3)
54.9649 = (A_asteroid^3)
Taking the cube root of both sides to solve for A_asteroid:
A_asteroid = cube root of 54.9649
Calculating that, we find:
A_asteroid ≈ 3.748 AU
Therefore, the asteroid is approximately 3.748 times further from the Sun than the Earth.