Final answer:
Gravitational acceleration at the top of the tower can be approximately assumed to be the same as on the asteroid's surface, 0.03 m/s². The speed to attain circular orbit at the top of the tower would be calculated by equating centripetal force to gravitational force, considering the asteroid's mass and radius.
Step-by-step explanation:
The question asks about gravitational acceleration at the top of a 100 m tower on an asteroid and the speed required to make a ball orbit the asteroid returning to the prince at the top of the tower.
The gravitational acceleration can be deduced from the provided information that when the spacecraft lands on the asteroid, it has a velocity of 5 m/s, and upon reaching 100 m closer to the surface, the velocity increases to 8 m/s. Using the equation v^2 = u^2 + 2as (where v is final velocity, u is initial velocity, a is acceleration, and s is distance), one can calculate that the gravitational acceleration is approximately 0.03 m/s² on the surface of the asteroid.
To achieve a circular orbit, the ball needs to be thrown at a speed where centripetal force equals gravitational force. The centripetal force is given by F = (mv²)/r (where m is mass, v is speed, and r is radius), and gravitational force is F = GMm/r² (where G is a gravitational constant, M is the mass of the asteroid, and m is the mass of the ball). Balancing these forces, we get the orbital speed v = sqrt(GM/r). With an assumed similar composition, we expect the mass M and consequently the gravitational constant G to be significantly less compared to Earth, specifically adjusted for the asteroid's size and mass to maintain a low orbit just above the surface.
Thus, the answer is given as an ordered pair with gravitational acceleration first, followed by orbital speed.