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Friday, 9 October

JEE Main 2025 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 2 April 2025, Shift 2, Physics Q47

Question 47 of 75 in this shift, Physics question 22 of 25, Section B.

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is __________ ×1010\times 10^{10} J. (Mass of earth =6×1024= 6\times10^{24} kg, Radius of earth =6.4×106= 6.4\times10^{6} m, Gravitational constant =6.67×10−11 Nm2kg−2= 6.67\times10^{-11}\ \mathrm{Nm^2kg^{-2}})

Official answer

3

NTA final key.

Chapter
Gravitation
Topic
Orbital Mechanics
Idea tested
Energy of Orbiting Bodies

Same idea in other shifts

Asked 2× in all
  1. 9 Apr 2024, Shift 1 · Q36An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5318.5…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.