JEE Main 2019PhysicsGravitationMediumMCQ

JEE Main 2019Gravitation Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
[Given: Mass of planet  =8×1022 kg ,
Radius of planet =2×106 m,
Gravitational constant  G=6.67×10-11 Nm2/kg2 ]

Choose an option

Show full solutionCorrect option: D
Correct answer
D11

Step-by-step explanation

T=2π(R+h)3GM=2π 8.242×10186.67×10-11×8×1022 
T7809 sec
Number of revolutions in 24 hours:
n=864007809=11.06
11 revolutions

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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.