JEE Main 202310 Apr, Evening shiftPhysicsHardGravitation159× since 2002The time period of a satellite, revolving above earth's surface at a height equal to R\mathrm{R}R will be (Given g=π2 m/s2,R=g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=g=π2 m/s2,R= radius of earth)AOption A. 32R\sqrt{32 R}32RBOption B. 4R\sqrt{4 \mathrm{R}}4RCOption C. 8R\sqrt{8 R}8RDOption D. 2R\sqrt{2 R}2RCheckSkip · show solution
