Satellites: Orbits and Energy — Question 2

Section 13-7 · Physics pages 2014–2015

Question 2 Raising a circular orbit adds total energy while reducing kinetic energy A satellite moves between circular orbits of radii r and 3r around the same fixed source. Compare the final and initial states; details of the transfer are not specified. Use E = −GMm/(2r) and K = GMm/(2r). rorbit energy K E U r 3r Circular-orbit endpoint energies: K > 0, E = −K, and U = −2K; schematic vertical scale. (a) Find the changes in E, K and U. (b) Explain how positive supplied energy is consistent with a lower final speed.
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Question 2 - Solution Raising a circular orbit adds total energy while reducing kinetic energy rorbit energy K E U r 3r (a) Derivation and calculation Let C = GMm/r. The endpoint energies are Ei = −C/2, Ki = C/2, Ui = −C, Ef = −C/6, Kf = C/6, Uf = −C/3. Subtracting, ∆E = C/3, ∆K = −C/3, ∆U = 2C/3 . (b) Result and interpretation The speed ratio and energy balance are vf/vi = √ r/(3r) = 1/ √ 3 , ∆K + ∆U = −C/3 + 2C/3 = C/3 = ∆E. The potential-energy increase exceeds the supplied total energy because kinetic energy decreases. Endpoint energies alone do not specify a burn sequence. Physical check The higher orbit is less tightly bound: its negative total energy has smaller magnitude.

Question 2

Raising a circular orbit adds total energy while reducing kinetic energy

A satellite moves between circular orbits of radii rr and 3r3r around the same fixed source. Compare the final and initial states; details of the transfer are not specified. Use E=−GMm/(2r)E=-GMm/(2r) and K=GMm/(2r)K=GMm/(2r).

Diagram shown in the original SVG worksheet.

  1. Find the changes in EE, KK and UU.

  2. Explain how positive supplied energy is consistent with a lower final speed.

Question 2 - Solution

Raising a circular orbit adds total energy while reducing kinetic energy

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Let C=GMm/rC=GMm/r. The endpoint energies are Ei=−C/2,Ki=C/2,Ui=−C,E_i=-C/2,\quad K_i=C/2,\quad U_i=-C, Ef=−C/6,Kf=C/6,Uf=−C/3.E_f=-C/6,\quad K_f=C/6,\quad U_f=-C/3. Subtracting, ΔE=C/3,ΔK=−C/3,ΔU=2C/3.\boxed{\Delta E=C/3,\qquad\Delta K=-C/3,\qquad\Delta U=2C/3}.

(b) Result and interpretation

The speed ratio and energy balance are vf/vi=r/(3r)=1/3,\boxed{v_f/v_i=\sqrt{r/(3r)}=1/\sqrt3}, ΔK+ΔU=−C/3+2C/3=C/3=ΔE.\Delta K+\Delta U=-C/3+2C/3=C/3=\Delta E. The potential-energy increase exceeds the supplied total energy because kinetic energy decreases. Endpoint energies alone do not specify a burn sequence.

Physical check

The higher orbit is less tightly bound: its negative total energy has smaller magnitude.