Satellites: Orbits and Energy — Question 1

Section 13-7 · Physics pages 2012–2013

Question 1 A bound circular orbit has positive kinetic energy and negative total energy A satellite of mass m moves in a circular orbit of radius r about a fixed spherical source. Write µ = GM and choose potential zero at infinity. R r v Spherical central body; radii are measured from its center. (a) Derive its speed, kinetic energy, potential energy and total energy. (b) Find the additional orbital energy required to escape with zero speed at infinity, and compare it with the kinetic energy already present.
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Question 1 - Solution A bound circular orbit has positive kinetic energy and negative total energy R r v (a) Derivation and calculation Circular force balance gives mv2 c r = µm r2 , vc = √ µ/r . With zero potential at infinity, K = 1 2mv2 c = µm 2r , U = −µm r , E = K + U = −µm 2r , U = −2K. (b) Result and interpretation Marginal escape requires Ef = 0: ∆E = Ef −Ei = µm 2r = Ki . At fixed radius, this raises the speed to v2 esc = 2µ/r = 2v2 c. Initial kinetic energy is positive, but the larger negative potential keeps the original circular orbit bound. Physical check The final escape energy is zero with zero speed at infinity, consistent with the chosen potential reference.

Question 1

A bound circular orbit has positive kinetic energy and negative total energy

A satellite of mass mm moves in a circular orbit of radius rr about a fixed spherical source. Write μ=GM\mu=GM and choose potential zero at infinity.

Diagram shown in the original SVG worksheet.

  1. Derive its speed, kinetic energy, potential energy and total energy.

  2. Find the additional orbital energy required to escape with zero speed at infinity, and compare it with the kinetic energy already present.

Question 1 - Solution

A bound circular orbit has positive kinetic energy and negative total energy

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Circular force balance gives mvc2r=μmr2,vc=μ/r.\frac{mv_c^2}{r}=\frac{\mu m}{r^2},\qquad \boxed{v_c=\sqrt{\mu/r}}. With zero potential at infinity, K=12mvc2=μm2r,U=−μmr,K=\frac12mv_c^2=\frac{\mu m}{2r},\quad U=-\frac{\mu m}r, E=K+U=−μm2r,U=−2K.\boxed{E=K+U=-\frac{\mu m}{2r}},\qquad U=-2K.

(b) Result and interpretation

Marginal escape requires Ef=0E_f=0: ΔE=Ef−Ei=μm2r=Ki.\boxed{\Delta E=E_f-E_i=\frac{\mu m}{2r}=K_i}. At fixed radius, this raises the speed to vesc2=2μ/r=2vc2.v_{\mathrm{esc}}^2=2\mu/r=2v_c^2. Initial kinetic energy is positive, but the larger negative potential keeps the original circular orbit bound.

Physical check

The final escape energy is zero with zero speed at infinity, consistent with the chosen potential reference.