JEE Main 2019PhysicsGravitationMediumMCQ

JEE Main 2019 — Gravitation Question with Solution

JEE Main 2019 (09 Jan Shift 2)

Question

The energy required to take a satellite to a height h above the Earth surface (radius of Earth =6.4×103 km ) is E1, and the 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

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Show full solutionCorrect option: C
Correct answer
C3.2×103km

Step-by-step explanation

Usurface +E1=Uh

KE of satellite is zero at earth surface and at height h

-GMemRe+E1=-GMemRe+h

E1=GMem1Re-1Re+h

E1=GMemRe+h×hRe

Gravitational attraction FG=maC=mv2Re+h

⇒mv2Re+h=GMemRe+h2

mv2=GMemRe+h

E2=mv22=GMem2Re+h

E1=E2

hRe=12⇒h=Re2=3200 km

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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.