will the period be 31536000s. if it is, is this right:Sirus said:I believe you need the mass of the sun, unless you are allowed to give your answer as a function of that variable. Use this formula:
[tex]T=\frac{2\pi r^{\frac{3}{2}}}{\sqrt{Gm_{sun}}}[/tex]
where G is the universal gravitational constant and T is the orbital period.
oppsSirus said:Check your period calculation again. Your method looks correct.
An asteroid's orbital radius is the distance between the center of the asteroid and the center of the object it is orbiting, such as a planet or the sun. It is typically measured in astronomical units (AU), which is the average distance between the Earth and the sun.
An asteroid's orbital radius is calculated using Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet or asteroid is directly proportional to the cube of its average distance from the sun.
The average orbital radius of an asteroid in our solar system is approximately 2.8 AU, or 2.8 times the distance between the Earth and the sun.
An asteroid's orbital radius and speed are inversely proportional, meaning that as the orbital radius increases, the speed decreases and vice versa. This is due to the conservation of angular momentum, as the asteroid must travel a longer distance in the same amount of time, resulting in a slower speed.
Yes, an asteroid's orbital radius can change over time due to various factors such as gravitational interactions with other objects in the solar system or collisions with other asteroids. However, these changes are usually small and occur over long periods of time.