JEE Advanced · PhysicsHard

$A satellite is in an elliptical orbit around a planet of mass M . The distance of closest approach is r_{1} and the farthest distance is r_{2} . The speed of the satellite at the closest approach is v_{1} . The speed at the farthest distance is v_{2} . The ratio v_{1} / v_{2} is:$

Show correct answer & step-by-step solution

Correct answer: B — $r_{2} / r_{1}$

Solution

$Angular momentum L = m v r sin (θ) is conserved.$
$At periapsis and apoapsis, the velocity is purely tangential, so θ = 9 0^{\circ} .$
$Thus, m v_{1} r_{1} = m v_{2} r_{2} .$
$Rearranging gives v_{1} / v_{2} = r_{2} / r_{1} .$
$Hence the answer is (B).$