The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consider an asteroid in a circular orbit of radius 5.7 1011 m. Find the period of the orbit.
To find the period of the orbit, we can use Kepler's third law of planetary motion, which relates the period of an orbit to the radius of the orbit. The formula is given by:
T^2 = (4π^2 / G) * r^3,
where T is the period of the orbit, G is the gravitational constant (approximately 6.67 x 10^-11 m^3 / (kg * s^2)), and r is the radius of the orbit.
In this case, the radius of the circular orbit is given as 5.7 x 10^11 m. Plugging these values into the formula, we get:
T^2 = (4π^2 / G) * (5.7 x 10^11)^3.
Simplifying this equation further, we can calculate the square of the period:
T^2 = (4π^2 / G) * 1.47 x 10^34.
Now, we can solve for T by taking the square root of both sides of the equation:
T = √[(4π^2 / G) * 1.47 x 10^34].
Calculating this expression will give us the period of the orbit.