12. An Asteroid passes by Earth every 17 years. The asteroid orbits the Sun, not the Earth. Remember these: M(Earth) 5.97 x 1024kg, R(Earth) 6400 km, M(Sun)- 1.99 x 1030kg, R(Sun) 700, 000 km, and the distance between the Earth and the Sun is 1.5 x 1011m. Answer the following: What is the radius of its orbit? What is the closest it gets to Earth? What is its angular velocity? Solution Time period of the Asteroid, T = 17 years = 17*365*24*60*60 = 5.361*10^8 s let r is the radius of the orbit of the asteroid. we know, T = 2*pi*r^1.5/sqrt(G*Msun) T^2 = 4*pi^2*r^3/(G*M_sun) r^3 = G*M_sun*T^2/4*pi^2 r = (G*M_sun*T^2/4*pi^2)^(1/3) = (6.67*10^-11*1.99*10^30*(5.361*10^8)^2/(4*pi^2))^(1/3) = 9.886*10^11 m <<<<<<<---------Answer distance between the earth and the asteroid = 9.886*10^11 - 1.5*10^11 8.386*10^11 m <<<<<<<---------Answer Angular velocity, w = 2*pi/T = 2*pi/(5.361*10^8) = 1.172*10^-8 rad/s <<<<<<<---------Answer .