When a 3.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.64 cm. If th
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Answer:
vₐ = v_c
Explanation:
To calculate the escape velocity let's use the conservation of energy
starting point. On the surface of the planet
Em₀ = K + U = ½ m v_c² - G Mm / R
final point. At a very distant point
Em_f = U = - G Mm / R₂
energy is conserved
Em₀ = Em_f
½ m v_c² - G Mm / R = - G Mm / R₂
v_c² = 2 G M (1 /R - 1 /R₂)
if we consider the speed so that it reaches an infinite position R₂ = ∞
v_c =
now indicates that the mass and radius of the planet changes slightly
M ’= M + ΔM = M ( )
R ’= R + ΔR = R ( )
we substitute
vₐ =
let's use a serial expansion
√(1 ±x) = 1 ± ½ x +…
we substitute
vₐ = v_ c ( )
we make the product and keep the terms linear
vₐ = v_c