If an asteroid is 4 au from the sun, what is the period of revolution around the sun for the asteroid?

8 years

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”:

(1)

In other words: this law states a relation between the orbital period of a body (moon, planet, satellite, comet, asteroid) orbiting a greater body in space (the Sun, for example) with the size of its orbit.  

However, if is measured in years (Earth years), and is measured in astronomical units (equivalent to the distance between the Sun and the Earth: ), equation (1) becomes:

(2)

This means that now both sides of the equation are equal.

Knowing and isolating from (2):

(3)

(4)

Finally:

This is the period of the asteroid

newton's first law of motion

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft