Respuesta :
The speed of a space probe when it is very far from the Earth is 5.4×10⁴m/s.
As per energy conservation we can say, initial total energy of the object at the surface of earth = final total energy of the object at highest height.
[tex]Ui=\frac{-GMm}{R} +\frac{1}{2}mv^2[/tex]
Where,
M = mass of earth
m = mass of object
R = radius of earth
When object reached far away from the earth then it will only have kinetic energy,
[tex]Uf=\frac{1}{2}mvf^2[/tex]
Now by energy conservation,
[tex]\frac{-GMm}{R} +\frac{1}{2}mv^2=\frac{1}{2}mvf^2[/tex]
[tex]\frac{-GM}{R} +\frac{1}{2}v^2=\frac{1}{2}vf^2[/tex]
[tex]\frac{-(6.67*10^-11)(5.98*10^24}{6.37*10^6} +\frac{1}{2}(5.52*10^4)^2=\frac{1}{2}vf^2[/tex]
[tex]-6.26*10^7+1.52*10^8=\frac{1}{2}vf^2[/tex]
[tex]vf=5.4*10^4m/s[/tex]
To learn more about space probe projectile: https://brainly.com/question/13027259
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