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]