If Neptune’s mass were reduced, what could be done to maintain the same force of gravitational attraction between Neptune and the Sun?

i dont no

Explanation:

===>  Distance fallen from rest in free fall =

                                         (1/2) (acceleration) (time²)

                               (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²):   (122.5 m / 4.9 m/s²)  =  time²

                                                           (122.5/4.9) s²  =  time²

Take the square root of each side:    5.0 seconds

===> (Accelerating at 9.8 m/s², he will be dropping at
                               (9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'.  We'll need this number for the last part.)

===> With no air resistance, the horizontal component of velocity
doesn't change.

Horizontal distance = (10 m/s) x (5.0 s)  =  50 meters .

===>  Impact velocity =  (10 m/s horizontally) + (49 m/s vertically)

                                 = √(10² + 49²)  =  50.01 m/s  arctan(10/49)

                                 =    50.01 m/s   at  11.5° from straight down,
                                                            away from the base of the cliff.