Superman stands on top of a very tall mountain and throws a baseball horizontally with a speed such that the baseball goes into a circular orbit around the earth. while the baseball is in orbit, what is the magnitude of the acceleration of the ball?

The acceleration must be a little less than 9.8 m/s²

Explanation:

In this case, we know that the baseball is doing a circular orbit around the earth, therefore, the value of acceleration due of gravity would have to be 9.81 m/s²; however, this is not always true.

For a circular orbit, the change in velocity is a vector that is perpendicular to the direction of motion. That is, the acceleration is a vector pointing towards the center of the Earth, (so it is constantly changing even if the magnitude is not). For a circular orbit, the speed is not changing. That is, the magnitude of the velocity is constant, and only the direction is changing. Therefore, we use the value of gravity.

This value of gravity is only true when is taked from the surface of Earth, in other word, at sea level. However in this case, it's not. Superman throws the ball in the top of the very tall mountain. And alike the pressure, this value, decreases with the height. The gravity do the same thing. As we go up, the value of gravity is less than the usual.

Therefore, the value of the acceleration would have to be less of 9.81 m/s²

the components of a vector are the parts of the vectors that are to each other.

b perpendicular