A satellite in elliptical orbit has maximum speed v_max at perigee (closest) and minimum v_min at apogee. The ratio v_max/v_min equals (r_apogee/r_perigee):
A√(r_apogee/r_perigee)
Br_perigee/r_apogee
C1
Dr_apogee/r_perigee
Answer & Solution
Correct answer: D. r_apogee/r_perigee
Angular momentum conservation: m v_max r_perigee = m v_min r_apogee. So v_max/v_min = r_apogee/r_perigee. The satellite moves faster when closer to Earth.
Related questions
Tycho Brahe recorded his observations using:Kepler's three laws describe the motion of:The low escape speed of the moon explains why it has no:The moon's escape speed compared with the earth's is smaller by about:The escape speed for the moon works out to about:The law of periods uses which measurement of the ellipse?By the law of periods, the square of the period is proportional to:The law of areas explains why a planet moves slower when it is: