{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dynamics_Part26

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: at these two points on the orbit is a n = v 2 ρ where v is the satellite’s speed and ρ is the radius of curvature of the orbit at the apogee and the perigee. We know from Newton’s Universal Law of Gravitation that the normal force at these two points, F n , is F n = GMm r 2 = ma n = mv 2 ρ = ⇒ 1 ρ = 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 ρ Combining with the result above, there follows 1 ρ = 1 2 w 1 r A + 1 r P W...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online