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Transcript

Proportionality between the velocity V and radius r In circular motion with a constant centripetal force. Applying Newton’s 2nd Law • 𝐹 = 𝑚𝑎 • In this case the only force on the body is the force of gravity and the acceleration of the body is centripetal. • 𝐹𝐺 = 𝐹𝐶 A centripetal force is required for circular motion. This is given by the equation: v2 Fc m1 ( ) r The centripetal force is supplied by the force of gravity between the two bodies which is given by the equation: Gm1m2 FG r2 Set the two forces equal to one another. FC FG 2 Gm1m2 v m1 2 r r Gm1m2 v2 m1 2 r r m1 = mass of the satellite m2 = mass of object being orbited V = velocity of the satellite G = universal gravitational constant r = radius of the circular motion Mass of the satellite cancels out. The speed of the satellite depends on radius r and mass m of the object being orbited. Orbital Velocity Solving for v Gm v r Where: v = orbital velocity (m/s) G = universal gravitational constant = 6.67 x 10-9 m = mass of the object being orbited (kg) r = radius of circular motion (m) Period of Satellite • We know that x x v or t t v For a circular motion, the distance x travelled by the satellite in one revolution is the circumference of the circle. That would be x 2r Solving for t x t v or 2r t v But we also know that This gives us t Gm v r 2r Gm r Squaring both sides 2 2 4 r 2 t Gm r We end up with orbital period t of a satellite. This is the time it takes for a satellite to complete one orbit. t 2 3 r Gm Again, m here is the mass of the object being orbited NOT the mass of the satellite. Quick Check # 1: • What is orbital velocity of the earth around the sun? The sun has a mass of 1.99 x 1030kg, the mean distance from the earth to the sun is 1.50 x 1011 m. Quick Check #1 Solution v 2 N.m 30 6.67 x10 11 (1.99 x 10 ) kg 2 Gm kg r 1.50 x 1011 m 2 m 8 4 m v 8.85 x10 2 2.97 x 10 s s Quick Check #2 • A satellite is in a low earth orbit, some 250 km above the earth's surface. rearth is 6.37 x 106 m and mearth = 5.98 x 1024 kg. Find the period of the satellite in minutes. Quick Check #2 Solution r is the distance from the center of the earth to the satellite; thus r = rearth + 250 km = 6.37 x 106 m + 250 x 103 m T 2 3 r Gm 2 6.67 x 10 1 min T 5357 s 60 s 6.62 x 106 m 3 m m2 24 5.98 x 10 kg 2 2 s kg 11 kg 89.3 min