JEE Main 2019 — Gravitation Question with Solution
From: JEE Main 2019 (Online) 9th January Evening Slot
Question
The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is
Choose an option
Show full solutionCorrect option: B
Correct answer
B3.2 103 km
Step-by-step explanation
Energy required to move a satellite from earth surface to height h is,
E1 = Uh Usurface
=
= GMm
=
We know, for sattelite at height h.
Centrifigual force = Gravitational force
2 =
=
Kinetic energy (E2) =
Given that,
E1 = E2
h =
h =
= 3.2 103 km
E1 = Uh Usurface
=
= GMm
=
We know, for sattelite at height h.
Centrifigual force = Gravitational force
2 =
=
Kinetic energy (E2) =
Given that,
E1 = E2
h =
h =
= 3.2 103 km
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.