Dynamics_Part26

Dynamics_Part26 - at these two points on the orbit is a n =...

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Problem Set 3: Problem 5. Problem: A satellite is in an elliptic orbit about the Earth. The distances of the satellite from Earth’s center are r A and r P at the apogee and perigee of the orbit, respectively. Verify that the radius of curvature for the orbit, ρ , at the apogee and perigee is given by 1 ρ = 1 2 w 1 r A + 1 r P W Solution: In general, we know that the shape of the orbit is given by 1 r = GM h 2 (1 + 6 cos θ ) At the apogee, θ = 180 o , and at the perigee, θ =0 o . Therefore, 1 r A = GM h 2 (1 6 )a n d 1 r P = GM h 2 (1 + 6 ) Summing these two equations yields 1 r A + 1 r P =2 GM h 2 Now, at the apogee and the perigee, the radial direction is normal to the path. Consequently, the acceleration at these two points on the orbit is
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Unformatted text preview: at these two points on the orbit is a n = v 2 where v is the satellites speed and is the radius of curvature of the orbit at the apogee and the perigee. We know from Newtons Universal Law of Gravitation that the normal force at these two points, F n , is F n = GMm r 2 = ma n = mv 2 = 1 = GM r 2 v 2 But, specific angular momentum is h = rv , so that GM r 2 v 2 = GM h 2 = GM h 2 = 1 Combining with the result above, there follows 1 = 1 2 w 1 r A + 1 r P W...
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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