Comet-Problem

A comet (Sphere of Ice)(Using weight of water 62.4 Lbs per cubic foot) approaching the earth at a 60 degree angle... traveling at 30 miles per second...and losing weight at the rate of 1000 lbs per second per each 1 lb per square inch atmospheric pressure...Assume it starts losing weight when the atmospheric pressure is 1 lb per square inch at 56412 feet altitude.What size sphere will lose all it's weight by the time it hits the earth? Find the diameter of the sphere. Volume of a sphere = (4/3)*Pi*R*R*R where R = the Radius & Pi = 3.14159 At Sea Level Atm Pressure = 14.7 psi At 18,000 ft = 7.35 psi At 27,480 ft = 4.90 psi At 56,412 ft = 1.00 psi Enter-01: Enter Angle A = 60 Degrees Enter-02: Enter Speed of Comet V = 30 Miles/sec Enter-03: Enter Diameter of Comet D = 5 Feet Enter-04: Enter Lbs Lost/sec = 1000 lbs Result-01: Speed = 158,400 ft/sec (b1) Result-02: Distance traveled in atmosphere at 60 deg angle = 65139 feet (b5) Result-03: => Lbs Lost in First 8566 feet (b9 = 54.07) Result-04: => Lbs Lost in Next 24,841 feet (b11 = 156.82) Result-05: => Lbs Lost in Next 10,946 feet (b14 = 169.31) Result-06: => Lbs Lost in Next 18,000 feet (b17 = 964.43) Result-07: => Total Lbs Lost = (b18 = 1344.65) Result-08: => Total Lbs/62.4 = Cubic Feet (b19 = 21.549) Result-09: => Diameter of Comet = Feet (b22 = 3.452) Result-10: => Diameter that hits the earth = (b27 = 4.376) This Calculates the Angular Velocity of your Telescope Needed to Capture the view of a Rocket going Straight up At any given altitude. Enter-01: Enter Y = 4000 feet = Altitude of Rocket Enter-02: Enter X = 3000 feet = Distance you are from the launch site. Enter-03: Enter Speed of Rocket V = 880 ft/sec Result-01: Radians/sec = 66/625 = .1056 Rad/Sec Result-02: Degrees/Sec = (Rad/sec)*(180/Pi) = 6.0504 Deg/Sec Since Tan-A = Y/X => dA/dT = (1/(X*(Sec-A)*(Sec-A)))*(dY/dT) And Sec-A = Z/X = 5000/3000 when Y = 4000 => (Sec-A)*(Sec-A) = 25/9 = K => 1/K = 9/25 => (1/X)*(1/K) = (1/3000)*(9/25)*880 = .11 radians/sec