JEE Main 2019 — Gravitation Question with Solution
From: JEE Main 2019 (Online) 10th January Morning Slot
Question
A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -
Choose an option
Show full solutionCorrect option: A
Correct answer
Amv2
Step-by-step explanation
At height r from center of earth. orbital velocity
=
By energy conservation
KE of 'm' + = 0 + 0
(At infinity, PE = KE = 0)
KE of 'm' = = m = mv2
=
By energy conservation
KE of 'm' + = 0 + 0
(At infinity, PE = KE = 0)
KE of 'm' = = m = mv2
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Gravitation chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, covering the Gravitation chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.