Planets and Satellites: Kepler's Laws — Question 1

Section 13-6 · Physics pages 1992–1993

Question 1 Apsis distances locate the focus and determine the ellipse A negligible-mass satellite follows an ellipse about a fixed point source. Its minimum and maximum distances from the source are rp = 3R and ra = 9R. Use rp = a(1 −e), ra = a(1 + e) and b = a √ 1 −e2. M P A Orbital-plane schematic; the attracting source is at a focus, not the ellipse center. (a) Find semimajor axis a, eccentricity e and semiminor axis b. (b) Find the center-to-focus distance and explain why averaging distances over time is not how a is defined.
Show solutionHide solution
Question 1 - Solution Apsis distances locate the focus and determine the ellipse M P A (a) Derivation and calculation Add and subtract the apsis equations: rp + ra = 2a, ra −rp = 2ae. Therefore a = 3R + 9R 2 = 6R , e = 9R −3R 9R + 3R = 1 2 . b = a √ 1 −e2 = 6R √ 3/4 = 3 √ 3R . (b) Result and interpretation The focus lies distance c = ae = 3R = √ a2 −b2 from the ellipse center. Geometrically, rp = a −c and ra = a + c, so a = (rp + ra)/2 exactly. A time average instead has the definition ⟨r⟩t = 1 T ∫T 0 r(t) dt, which weights each location by residence time. Physical check The inferred dimensions satisfy b2 + c2 = a2 and reproduce both given apsis radii.

Question 1

Apsis distances locate the focus and determine the ellipse

A negligible-mass satellite follows an ellipse about a fixed point source. Its minimum and maximum distances from the source are rp=3Rr_p=3R and ra=9Rr_a=9R. Use rp=a(1−e)r_p=a(1-e), ra=a(1+e)r_a=a(1+e) and b=a1−e2b=a\sqrt{1-e^2}.

Diagram shown in the original SVG worksheet.

  1. Find semimajor axis aa, eccentricity ee and semiminor axis bb.

  2. Find the center-to-focus distance and explain why averaging distances over time is not how aa is defined.

Question 1 - Solution

Apsis distances locate the focus and determine the ellipse

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Add and subtract the apsis equations: rp+ra=2a,ra−rp=2ae.r_p+r_a=2a,\qquad r_a-r_p=2ae. Therefore a=3R+9R2=6R,e=9R−3R9R+3R=12.\boxed{a=\frac{3R+9R}{2}=6R},\qquad \boxed{e=\frac{9R-3R}{9R+3R}=\frac12}. b=a1−e2=6R3/4=33R.\boxed{b=a\sqrt{1-e^2}=6R\sqrt{3/4}=3\sqrt3R}.

(b) Result and interpretation

The focus lies distance c=ae=3R=a2−b2\boxed{c=ae=3R}=\sqrt{a^2-b^2} from the ellipse center. Geometrically, rp=a−cr_p=a-c and ra=a+cr_a=a+c, so a=(rp+ra)/2a=(r_p+r_a)/2 exactly. A time average instead has the definition ⟨r⟩t=1T∫0Tr(t)dt,\langle r\rangle_t=\frac1T\int_0^T r(t)\,dt, which weights each location by residence time.

Physical check

The inferred dimensions satisfy b2+c2=a2b^2+c^2=a^2 and reproduce both given apsis radii.