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Imagine that a space probe could be fired as a projectile from the Earth's surface with an initial speed of 5.52 104 m/s relative to the Sun. What would its speed be when it is very far from the Earth (in m/s)? Ignore atmospheric friction, the effects of other planets, and the rotation of the Earth. (Consider the mass of the Sun in your calculations.)

Respuesta :

Answer:

[tex]v_f = 5.4 \times 10^4 m/s[/tex]

Explanation:

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

now we have

[tex]U_i = -\frac{GMm}{R} + \frac{1}{2}mv^2[/tex]

here we know that

M = mass of earth

m = mass of object

R = radius of earth

now when object reached far away from the earth then it will only have kinetic energy

[tex]U_f = \frac{1}{2}mv_f^2[/tex]

now by energy conservation

[tex]-\frac{GMm}{R} + \frac{1}{2}mv^2 = = \frac{1}{2}mv_f^2[/tex]

[tex]-\frac{GM}{R} + \frac{1}{2}v^2 = \frac{1}{2}v_f^2[/tex]

[tex]-\frac{(6.67 \times 10^{-11})(5.98\times 10^{24})}{6.37\times 10^6} + \frac{1}{2}(5.52 \times 10^4)^2 = \frac{1}{2}v_f^2[/tex]

[tex]-6.26 \times 10^7 + 1.52 \times 10^8 = \frac{1}{2}v^2[/tex]

[tex]v_f = 5.4 \times 10^4 m/s[/tex]

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