{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dynamics_Part26

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

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

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
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