The biggest asteroid (Asteroid 2005 YU55 400 m wide) to cruise by the Earth in 3

Question

The biggest asteroid (Asteroid 2005 YU55 400 m wide) to cruise by the Earth in 35 years made its closest approach on Tuesday November 11/08/2011 at 6:28 p.m. (PST). According to NASA, at the point of closest approach, it was no closer than 324,600 km from the center of the earth. Let us assume that Asteroid 2005 YU55 has a mass of 10 billion kg. Calculate the gravitational force of attraction between the asteroid and the earth the point of closest approach. Assume that the asteroid goes into circular orbit at a distance of 324,600 km from the center of the earth. Find the acceleration and speed of the asteroid in its circular orbit. (please work out and explain your answer)

Explanation / Answer

a) F = G M m/r^2 =6.67E-11*10E9*5.97E24/(324600E3)^2=3.78E7 b) F = ma a = F/m = 3.78E7/10E9=.00378 m/s^2 a = v^2/r v = sqrt(ar) = sqrt(.00378*324600E3)=1108 m/s

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