An elliptical satellite orbit has nearest and farthest distances 2R and 4R from Earth’s centre. Radius of curvature at the nearest point is aR/b (in lowest terms). Find a − b.
Physics · Mixed Chapter
JEE Mains

Question

An elliptical satellite orbit has nearest and farthest distances 2R and 4R from Earth’s centre. Radius of curvature at the nearest point is aR/b (in lowest terms). Find a − b.

G1 Physics · Accepted Answer

Step 1 — Semimajor axis and eccentricity: rp=2R, ra=4R ⇒ semimajor axis A=(2R+4R)/2=3R. Eccentricity e=(ra−rp)/(ra+rp)=(4−2)/(4+2)=1/3. Step 2 — Orbital speed at closest approach (vis-viva): vp²=GM(2/rp−1/A)=GM(1/R−1/3R)=2GM/(3R). Step 3 — Acceleration perpendicular to velocity at periapsis: an=GM/rp²=GM/(4R²). Step 4 — Radius of curvature: ρ=vp²/an = [2GM/(3R)]/[GM/(4R²)] = 8R/3. Step 5 — Read a and b: a=8, b=3 ⇒ a−b=5. Answer: 5.