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A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5×109kg. It is approaching the Earth on a head-on course with a velocity of 660 m/s relative to the Earth and is now 4.0×106 km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

Respuesta :

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

[tex]-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2[/tex]

now we know that

[tex]v_0 = 660 m/s[/tex]

[tex]M = 5.98 \times 10^{24} kg[/tex]

[tex]m = 5 \times 10^9 kg[/tex]

[tex]r = 4 \times 10^9 m[/tex]

[tex]R = 6.37 \times 10^6 m[/tex]

now from above formula

[tex]GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2[/tex]

now we have

[tex]2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2[/tex]

now plug in all data

[tex]2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2[/tex]

[tex]v = 7934.2 m/s[/tex]