Question

A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the asteroid's radius and then falls straight down toward the surface of the asteroid. The acceleration of the ball at the top of its path is:_________

a. at its maximum value for the ball's flight.

b. equal to the acceleration at the surface of the asteroid.

c. equal to one-half the acceleration at the surface of the asteroid.

d. equal to one-fourth the acceleration at the surface of the asteroid.

Answer 1

d. equal to one-fourth the acceleration at the surface of the asteroid.

Step-by-step explanation:

The explanation is attached as a picture with this answer

Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.

as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.

As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.

It is taken 2r for the top is the path. hence we obtain (1/4)g as our answer.

Answer 2

(d) equal to one-fourth the acceleration at the surface of the asteroid.

Step-by-step explanation:

From Newton's gravitational law, force between two objects (planet) is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

`F = (Gmm)/(r^2) = mg\\\\g = (Gm)/(r^2)`

Case 1: when the ball rises to a height equal to the asteroid's radius

height, h = r

`g = (Gm)/(r^2)`

Case 2: when the ball rises to a height equal to the asteroid's radius and falls straight down toward the surface of the asteroid.

At top of its path, total height is 2r.

h = 2r

`g' = (Gm)/((2r)^2) \\\\g' = (Gm)/(4r^2) \\\\g' = (1)/(4) ((Gm)/(r^2))\\\\g' = (1)/(4) (g)`

Therefore, the acceleration of the ball at the top of its path is equal to one-fourth the acceleration at the surface of the asteroid.

Option "d"