Planets and Satellites: Kepler's Laws — Question 8

Section 13-6 · Physics pages 2006–2007

Question 8 An algebraically valid ellipse can intersect the central body A proposed exterior satellite orbit around a spherical planet of radius R has semimajor axis a = 2R and eccentricity e = .60. Ignore atmospheric drag but require the trajectory not enter the solid planet. R rp = .8R To scale in the orbital plane: a = 2R, e = .60. The periapsis lies inside the planet. (a) Test whether the proposed orbit is physically admissible. (b) Find the largest eccentricity allowed at fixed a = 2R, and explain the grazing equality case.
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Question 8 - Solution An algebraically valid ellipse can intersect the central body R rp = .8R (a) Derivation and calculation The closest source distance is rp = a(1 −e) = 2R(1 −.60) = .8R . Since rp −R = −.2R < 0, the proposed trajectory penetrates the solid planet and is not an admissible exterior satellite orbit. (b) Result and interpretation At fixed a = 2R, impose clearance: a(1 −e) ≥R =⇒ e ≤1 −R a . e ≤.50 . Equality gives rp = R, grazing the ideal surface. Strictly positive clearance requires e < .50; period alone does not constrain this geometry. Physical check The forbidden orbit is a valid point-source ellipse mathematically, but its exterior-field and collision assumptions fail inside the planet.

Question 8

An algebraically valid ellipse can intersect the central body

A proposed exterior satellite orbit around a spherical planet of radius RR has semimajor axis a=2Ra=2R and eccentricity e=.60e=.60. Ignore atmospheric drag but require the trajectory not enter the solid planet.

Diagram shown in the original SVG worksheet.

  1. Test whether the proposed orbit is physically admissible.

  2. Find the largest eccentricity allowed at fixed a=2Ra=2R, and explain the grazing equality case.

Question 8 - Solution

An algebraically valid ellipse can intersect the central body

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The closest source distance is rp=a(1−e)=2R(1−.60)=.8R.r_p=a(1-e)=2R(1-.60)=\boxed{.8R}. Since rp−R=−.2R<0,r_p-R=-.2R<0, the proposed trajectory penetrates the solid planet and is not an admissible exterior satellite orbit.

(b) Result and interpretation

At fixed a=2Ra=2R, impose clearance: a(1−e)≥R⇒e≤1−Ra.a(1-e)\ge R\quad\Longrightarrow\quad e\le1-\frac Ra. e≤.50.\boxed{e\le.50}. Equality gives rp=Rr_p=R, grazing the ideal surface. Strictly positive clearance requires e<.50e<.50; period alone does not constrain this geometry.

Physical check

The forbidden orbit is a valid point-source ellipse mathematically, but its exterior-field and collision assumptions fail inside the planet.