Ask Question
21 October, 17:47

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

+3
Answers (1)
  1. 21 October, 18:09
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers