Question

A ball with weight of 1000 pounds is held 1000 above the earth. What is the ball's potential energy in kip-feet? one-kip=1000 pounds

Answer 1

I assume the ball weighs 10000 pounds force at that altitude. ?

Can't use PE = mgh as the value of g changes with altitude. At that heigh, it is significantly lower.

First we need the mass.
10000 lb force = 44452 newtons
1000 miles = 1.61e6 m
g value at that altitude is

g = Gm/r²
g is acceleration from mass attraction in m/s²
m is mass of earth or other body generating the a
r is radius of body in meters.
G = 6.67e-11 m³/kgs²
earth radius 6,371 km = 6.37e6 meters
earth mass M 5.974e24 kg
earth GM = 3.987e14
g = (3.987e14) / (6.37e6+ 1.61e6)² = 6.26 m/s²

mass = 44452/6.26 = 7098 kg


Gravitational potential energy (to surface)
from height h
E = [GmM/(R+h)] – [GmM/R]
E = GmM[(1/(R+h)) – (1/R)]

E = (3.987e14)(7098)[(1/(6.37e6+ 1.61e6)) – (1/6.37e6)]

you can do the rest.

edit:
E = (3.987e14)(7098)[(1/(6.37e6+ 1.61e6)) – (1/6.37e6)]
E = (3.987e14)(7098)[(3.17e-8] = 8.97e10 J

for curiosity, use E =mgh so see the difference.
E = (4336)(9.8)(1.61e6) = 7.16e10 J (false