An asteroid revolves around the Sun with a mean orbital radius twice that of Earth's. Predict the period of the asteroid in Earth years. (6.38*10^6 radius) I don't understand how to solve this problem.
wouldn't Keplers law work:
Period= radius^(3/2)
To solve this problem, we need to use Kepler's third law of planetary motion, which states that the square of a planet's orbital period is directly proportional to the cube of its mean distance from the Sun. In equation form, it can be written as:
T^2 = (r^3) / G * (M + m)
Where:
T = Orbital period
r = Mean distance from the Sun
G = Gravitational constant
M = Mass of the Sun
m = Mass of the asteroid (which we can assume negligible compared to the Sun)
In this case, we are given that the mean orbital radius of the asteroid is twice that of Earth's radius, which is approximately 6.38 * 10^6 km. So, we can substitute this value into the equation:
T^2 = ((2 * 6.38 * 10^6)^3) / G * M
Now, to find the period of the asteroid in Earth years, we need to convert the units appropriately. Here's the step-by-step process:
1. Convert the radius of the asteroid from km to meters:
(2 * 6.38 * 10^6 km) * (1000 meters / 1 km) = (2 * 6.38 * 10^9 meters)
2. Calculate the square of the orbital period:
T^2 = ((2 * 6.38 * 10^9)^3) / G * M
3. Calculate the value of G * M:
G * M = 6.67 * 10^-11 m^3 kg^-1 s^-2 * (1.98 * 10^30 kg) → approximately 1.32 * 10^20 m^3 s^-2
4. Substitute the values into the equation and solve for T^2:
T^2 = ((2 * 6.38 * 10^9)^3) / (1.32 * 10^20)
5. Take the square root of T^2 to find the orbital period T in seconds:
T = √((2 * 6.38 * 10^9)^3) / (1.32 * 10^20)
6. Convert the period from seconds to Earth years:
T (in years) = T (in seconds) / (365.25 * 24 * 60 * 60)
By following these steps and performing the calculations, you should be able to determine the period of the asteroid in Earth years.
To solve this problem, we can make use of Kepler's third law of planetary motion, which states that the square of the period of an object's orbit is proportional to the cube of its mean orbital radius.
Let's denote the period of the asteroid as T and its mean orbital radius as r. We know that the mean orbital radius of the asteroid is twice that of Earth's, so we can express it as 2 times the mean orbital radius of Earth, which is approximately 1.496 x 10^8 kilometers.
Now, we can set up a proportion using the formula mentioned above:
(T_asteroid)^2 / (r_asteroid)^3 = (T_earth)^2 / (r_earth)^3
Plugging in the given values, we have:
(T_asteroid)^2 / (2 * 1.496 x 10^8)^3 = (1 year)^2 / (1.496 x 10^8)^3
Simplifying, we have:
(T_asteroid)^2 = (1 year)^2 * (2 * 1.496 x 10^8)^3
Now, we can take the square root of both sides to solve for T_asteroid:
T_asteroid = sqrt[(1 year)^2 * (2 * 1.496 x 10^8)^3]
Using a calculator, we can evaluate the expression to find the value of T_asteroid.