Satellites: Orbits and Energy — Question 5

Section 13-7 · Physics pages 2020–2021

Question 5 A pure orbital-plane change needs an impulse even when speed is unchanged At a node, a satellite's velocity is redirected through angle i while keeping speed vc and radius fixed. Both the old and new orbits are circular about the same spherical source.⃗ v1⃗ v2 ∆⃗v i Velocity-space diagram at the common node; both velocity vectors have equal length. (a) Derive the magnitude of the velocity increment. (b) For i = 60◦, compare ∆v with vc and determine the change in orbital energy.
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Question 5 - Solution A pure orbital-plane change needs an impulse even when speed is unchanged⃗ v1⃗ v2 ∆⃗v i (a) Derivation and calculation Let the two velocity vectors have equal magnitude vc and angle i. Then |∆⃗v|2 = |⃗vf −⃗vi|2 = 2v2 c(1 −cos i). Using 1 −cos i = 2 sin2(i/2), ∆v = 2vc sin(i/2) . (b) Result and interpretation At i = 60◦, ∆v = 2vc sin 30◦= vc . The orbital energy change is nevertheless ∆E = 1 2m(v2 f −v2 i ) + [U(r) −U(r)] = 0 . An impulse changes momentum direction even at constant speed; zero net orbital-energy change does not imply zero propulsion cost. Physical check The result tends to zero for i →0 and to 2vc for a complete velocity reversal.

Question 5

A pure orbital-plane change needs an impulse even when speed is unchanged

At a node, a satellite's velocity is redirected through angle ii while keeping speed vcv_c and radius fixed. Both the old and new orbits are circular about the same spherical source.

Diagram shown in the original SVG worksheet.

  1. Derive the magnitude of the velocity increment.

  2. For i=60∘i=60^\circ, compare Δv\Delta v with vcv_c and determine the change in orbital energy.

Question 5 - Solution

A pure orbital-plane change needs an impulse even when speed is unchanged

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Let the two velocity vectors have equal magnitude vcv_c and angle ii. Then |Δv→|2=|v→f−v→i|2=2vc2(1−cos⁡i).|\Delta\vec v|^2=|\vec v_f-\vec v_i|^2 =2v_c^2(1-\cos i). Using 1−cos⁡i=2sin⁡2(i/2)1-\cos i=2\sin^2(i/2), Δv=2vcsin⁡(i/2).\boxed{\Delta v=2v_c\sin(i/2)}.

(b) Result and interpretation

At i=60∘i=60^\circ, Δv=2vcsin⁡30∘=vc.\boxed{\Delta v=2v_c\sin30^\circ=v_c}. The orbital energy change is nevertheless ΔE=12m(vf2−vi2)+[U(r)−U(r)]=0.\Delta E=\tfrac12m(v_f^2-v_i^2)+[U(r)-U(r)]=\boxed{0}. An impulse changes momentum direction even at constant speed; zero net orbital-energy change does not imply zero propulsion cost.

Physical check

The result tends to zero for i→0i\to0 and to 2vc2v_c for a complete velocity reversal.