Hi, I’m having a really difficult time in physics, and I need some help on this homework assignment. Here are the questions and the work I’ve done. I really don’t know how to approach them, and at this point, I’m just kinda throwing equations at them haphazardly… 5. A 175 kg weather surveillance satellite of is launched into orbit with an altitude of 250 km. The mass of the earth is 5.97 x 1024 kg. The radius of the earth is 6.38 x 103 km. a. Calculate the tangential velocity required for this orbit. mg = mv^2 / r g = v^2 / r v = sqrt(gr) = sqrt( 9.8m/s^2 * (6.38*10^6 m + 250000m) ) v = 8060.6m/s b. Calculate the period of this orbit in minutes. T^2 = (4PI^2 / GM[E]) r^3 T^2 = (4PI^2 / (6.67 * 10^-11) * 5.97 x 1024 kg) * (6.38 x 106 m + 250000m)^3 T = 3.2*10^28s/rev 3.2*10^28s/rev * (1min / 60s) = 5.33*10^26 min/rev ?? c. What is the weight of the satellite? mg = G( Mm / r^2 ) I can’t use this formula, there are two m’s… v^2 = G( m/r ) m = rv^2 / G m = ( (6.38*10^6 m + 250000m) * (8060.6m/s)^2 ) / (6.67*10^-11) This isn’t right… 9. In an Atwood machine, one block has a mass of 500g and the other a mass of 460g. The pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.0cm. When released from rest, the heavier block is observed to fall 75cm in 5.0s with no slippage of the cord on the pulley. d = v*t + (1/2)*a*t^2 d-v*t = (1/2)*a*t^2 a = 2*( d -v * t)/ t^2 ) a = 2*( ( .75m -0 * 5s) / (5s)^2 ) = .06m/s^2 a. Calculate the acceleration of the two blocks. a[M] = -.06m/s^2 a[m] = .06m/s^2 b. Calculate the tension in the cord supporting the heavier block. T = mg+ma = (.46kg) (9.8m/s^2) + (.46kg) (.06m/s^2) = 4.54N c. Calculate the tension in the cord supporting the lighter block. Note the answer to b and c is not the same. T – T = Mg – mg – Ma – ma T – 4.54N = (.5kg) (9.8m/s^2) – (.46kg) * (9.8m/s^2) – (.5kg) * (.06m/s^2) – (.46kg) * (.06m/s^2) T – 4.54N = .33N T = 4.87N d. Calculate the angular acceleration of the pulley. α = a / r = .06m/s^2 / .05m = 1.2rad/s^2 e. Calculate the moment of inertia of the pulley. T – T = I (a / r^2) I = (T – T) / (a / r^2) I = (.33N) / (.06m/s^2 / (.05m)^2) = .014 kg*m^2 Any help would be greatly appreciated. Thanks, dusty…….