A) 0.189 N
The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

where
G is the gravitational constant
8.7×10^13 kg is the mass of the asteroid
m = 130 kg is the mass of the man
R = 2.0 km = 2000 m is the radius of the asteroid
Substituting into the equation, we find

B) 2.41 m/s
In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

where
v is the speed of the astronaut
Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

Your answer is going to be Appellate jurisdiction.
Answer:
It is an SI unit
Explanation:
The metre is defined as the length of the path travelled by light in a vacuum in 1299 792 458 of a second. The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole
I think is between A&B
I think I would answer B