I need help on this physics problem! Please and thanks!?

The gravitational field strength at an altitude of 200 km above the surface of the moon is ___ N/kg.  A satellite in orbit at that distance would have a speed of ___  km/s and the work needed to raise a 2500 kg satellite from the surface of the moon to that height is ___ x 10^ ___  J.

1 Answer

Relevance
  • NCS
    Lv 7
    3 weeks ago

    Our moon?

    M = 7.35E22 kg

    R = 1.74E6 m

    G = 6.674E-11 N·m²/kg²

    at 200 km altitude,

    g = GM / (R+h)²

    g = 6.674E-11N·m²/kg² * 7.35E22kg / (1.74E6m + 200E3m)²

    g = 1.30 m/s² ◄

    For orbit, centripetal acceleration = gravitational acceleration

    v²/(R+h) = g

    v² = 1.30m/s² * (1.74E6m + 200E3m) = 2.53E6 m²/s²

    v = 1590 m/s = 1.59 km/s◄

    For the work required, a good approximation can be had by "averaging" the gravity over the distance. At the surface,

    g' = 1.62 m/s²

    work = mgh = 2500kg * ½*(1.30+1.62)m/s² * 200E3m

    work = 7.3E8 J

    But "satellite" implies orbit, and so in addition to lifting the satellite to that height, you've got to give it the velocity found earlier. If they meant "the work required to put a satellite in orbit at that altitude, then the most accurate method is this:

    At the surface,

    GPE = -GmM / d

    GPE = -6.674E-11N·m²/kg² * 2500kg * 7.35E22kg / 1.74E6m

    GPE = -7.05E9 J

    At 200 km,

    GPE' = 6.674e−11N·m²/kg² * 2500kg * 7.35E22kg / 1.94E6m

    GPE' = -6.32E9 J

    The difference is the work required to lift the object to that height:

    work = 7.27E8 J ◄

    which is quite close to our earlier approximation.

    If in fact they want the work to put it in orbit, then add the KE:

    KE = ½ * 2500kg * (1590m/s)² = 3.16E9 J

    for a total of 3.89E9 J ◄

    Hope this helps!

Still have questions? Get answers by asking now.