Respuesta :
Answer:
vₐ = v_c [tex]( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )[/tex]
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 = [tex]\sqrt{\frac{2GM}{R} }[/tex]
now indicates that the mass and radius of the planet changes slightly
M ’= M + ΔM = M ( [tex]1+ \frac{\Delta M}{M}[/tex] )
R ’= R + ΔR = R ( [tex]1 + \frac{\Delta R}{R}[/tex] )
we substitute
vₐ = [tex]\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }[/tex]
let's use a serial expansion
√(1 ±x) = 1 ± ½ x +…
we substitute
vₐ = v_ c ( [tex](1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )[/tex])
we make the product and keep the terms linear
vₐ = v_c [tex]( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )[/tex]
The minimum escape speed, vc , for an object launched from the surface of the planet will be [tex]v_a=v_c(1+\dfrac{1}{2}(\dfrac{\Delta M}{M}-\dfrac{\Delta R}{R})[/tex]
What is escape velocity of the planet?
The escape velocity is defined as the velocity required to send the object out of the gravitational influence of the earth.
To calculate the escape velocity let's use the conservation of energy
starting point. On the surface of the planet
[tex]E_{mo} = K + U = \dfrac{1}{2} m v_c^2 - \dfrac{G Mm} { R}[/tex]
final point. At a very distant point
[tex]E_{mf} = U = \dfrac{- G Mm }{ R_2}[/tex]
energy is conserved
[tex]E{mo} = E{mf}[/tex]
[tex]\dfrac{1}{2}m v_c^2 - \dfrac{G Mm} {R} = \dfrac{- G Mm }{ R_2}[/tex]
[tex]v_c^2 = 2 G M (\dfrac{1} {R} - \dfrac{ 1 }{R_2})[/tex]
if we consider the speed so that it reaches an infinite position R₂ = ∞
[tex]v_c = \sqrt{\dfrac{2GM}{R}[/tex]
now indicates that the mass and radius of the planet changes slightly
[tex]M ’= M + \Delta M = M(1+\dfrac{\Delta M}{M})[/tex]
[tex]R ’= R + \Delta R = R (1+\dfrac{\Delta R}{R} )[/tex]
we substitute
[tex]vₐ = \sqrt{\dfrac{2GM}{R} }\dfrac{\sqrt{1+\dfrac{\Delta M}{M}}} {\sqrt{1+\dfrac{\Delta R}{R}}}[/tex]
let's use a serial expansion
√(1 ±x) = 1 ± ½ x +…
we substitute
[tex]v_a=v_c(1+\dfrac{1}{2}\dfrac{\Delta M}{M})(1-\dfrac{1}{2}\dfrac\Delta R}{R})[/tex]
Hence the minimum escape speed, vc , for an object launched from the surface of the planet will be [tex]v_a=v_c(1+\dfrac{1}{2}(\dfrac{\Delta M}{M}-\dfrac{\Delta R}{R})[/tex]
To know more about escape velocity follow
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