The period of the earth around the sun is
1 year and its distance is 150 million km from
the sun. An asteroid in a circular orbit around
the sun is at a distance 475 million km from
the sun.
What is the period of the asteroid’s orbit?
Answer in units of year
Time for asteroid = (RadiusEarthOrbit/RadiusAsteroidOrbit)^3/2 * Time for Earth
Ta= (475/150)^3/2 *1 = 5.54 yr
To calculate the period of the asteroid's orbit, we can use Kepler's third law, which states that the square of the period of an object's orbit is equal to the cube of its average distance from the Sun.
Let's assume the period of the asteroid's orbit is T (in years) and its average distance from the Sun is R (in kilometers).
According to the information given:
Period of Earth's orbit (T₁) = 1 year
Distance of Earth from the Sun (R₁) = 150 million km
Period of asteroid's orbit (T) = ?
Distance of asteroid from the Sun (R) = 475 million km
Using Kepler's third law, we can set up the following equation:
(T/T₁)² = (R/R₁)³
Substituting the values:
(T/1 year)² = (475 million km / 150 million km)³
Simplifying the equation:
(T)² = (475/150)³
(T)² = (475/150)³
(T)² = (19/6)³
Now, we can solve for T:
T = √((19/6)³)
Calculating the value:
T ≈ 1.802 years
Therefore, the period of the asteroid's orbit is approximately 1.802 years.
To find the period of the asteroid's orbit, we can use Kepler's Third Law, which states that the square of the period of an object's orbit is proportional to the cube of its average distance from the sun.
Let's denote the period of the asteroid's orbit as T_a and its average distance from the sun as a_a. Similarly, for the Earth, the period is T_e and the average distance is a_e.
According to Kepler's Third Law:
(T_a^2) / (T_e^2) = (a_a^3) / (a_e^3)
We are given:
T_e = 1 year
a_e = 150 million km
a_a = 475 million km
Plugging in the values, we get:
(T_a^2) / (1^2) = (475^3) / (150^3)
Simplifying the equation, we have:
T_a^2 = (475^3) / (150^3)
To solve for T_a, we take the square root of both sides:
T_a = √[(475^3) / (150^3)]
Calculating this, we find that the period of the asteroid's orbit, T_a, is approximately 14.158 years.