Respuesta :
Answer:
v = 1.078 10⁴ m / s
Explanation:
To solve this exercise we can use conservation of mechanical energy
Starting point. Just clearing
Em₀ = K + U = ½ m v² - G m M / [tex]R_{e}[/tex]
Final point. At r = 14R_{e}
Em_{f} = U = -G m M / 14R_{e}
how energy is conserved
Em₀ = [tex]Em_{f}[/tex]
½ m v² - G m M /R_{e} = -G m M / 14R_{e}
½ v² = G M / R_{e} (-1/14 + 1)
v² = 2 G M / R_{e} 13/14
we calculate
v² = 2 6.67 10⁻¹¹ 5.98 10²⁴ / 6.37 10⁶ 13/14
v = √ (1,16 10⁸)
v = 1.078 10⁴ m / s
The launch speed of a projectile that rises above the earth will be
v = 1.078 10⁴ m / s
What is speed?
Speed is defned as the movement of any object with respect to the time the formula of speed is given as the ratio of the distance with time.
To solve this exercise we can use conservation of mechanical energy
Starting point. Just clearing
[tex]E_{mo} = K + U = \dfrac{1}{2}m v^2 - \dfrac{G m M} { r}[/tex]
Final point. At[tex]r = 14R_{e}[/tex]
[tex]Em_{f} = U = \dfrac{-G m M }{14R_{e}}[/tex]
how energy is conserved
[tex]E_{mo }= \dfrac{1}{2} m v^2 -\dfrac{-G m M }{R_{e}} = \dfrac{-G m M }{ 14R_{e}}[/tex]
[tex]\dfrac{1}{2} v^2 = \dfrac{G M }{ R_{e} (\dfrac{-1}{14} + 1)}[/tex]
[tex]v^2 = \dfrac{2 G M} { R_{e} \dfrac{13}{14}}[/tex]
we calculate
[tex]v^2 = \dfrac{2 6.67 \times10^{-11}\times 5.98 \times 10^{24}} { 6.37 \times 10^{6} \times \dfrac{13}{14}}[/tex]
[tex]v = \sqrt{ (1,16\times 10^8)}[/tex]
v = 1.078 10⁴ m / s
Hence the launch speed of a projectile that rises above the earth will be
v = 1.078 10⁴ m / s
To know more about speed follow
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