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Posted by on Wednesday, May 9, 2012 at 8:19pm.

What is the change in gravitational potential energy of a 6200 kg satellite that lifts off from Earth's surface into a circular orbit of altitude 2500 km? What percent error is introduced by assuming a constant value of g and calculating the change in gravitational potential energy from m⋅g⋅Δh?

  • physics - , Thursday, May 10, 2012 at 10:48am

    ΔPE = PE( at the height h) –PE 9at the Earth’s surface) =
    = m•g•h – 0 = m•g•h = 6200•9.8•2500000 = 10.52•10^11 J.
    Really, the acceleration due to gravity at the height h = 2500000 m is
    g1 = G•M/(R+h)²,
    where the gravitational constant G =6.67•10^-11,
    Earth’s mass is M = 5.97•10^24 kg,
    Earth’s radius is R = 6.378•10^6 m.
    g1 = 5.05 m/s².
    For this value of acceleration due to gravity
    m•g1 •h = 6200•5.05•2500000 =7.83•10^10 J.
    Therefore, if we don’t use calculus, we have to use
    the average value of g =(9.8+5.05) /2 ≈ 7.43 m/s²
    The potential energy will be
    m•g•h =6200•7.43•2500000= 1.15•10^11 J.
    This result differs from our first result (using 9.8 m/s²)
    by ~11%.
    and from the second result (7.83•10^10 J for g= 5.05 m/s²)
    by ~ 47% (with opposite sign).

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