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 - Elena, 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)
and from the second result (7.83•10^10 J for g= 5.05 m/s²)
by ~ 47% (with opposite sign).