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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2019 · 10 April · Evening Shift · Physics · Gravitation

A spaceship orbits around a planet at a height of 20km from its...

A spaceship orbits around a planet at a height of 20km20km 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 2424 hours around the planet?
[Given: Mass of planet =8×1022kg= 8 \times {10}^{22}kg ,
Radius of planet =2×106m,= 2 \times {10}^{6}m,
Gravitational constant G=6.67×10−11Nm2/kg2G = 6.67 \times {10}^{- 11}N{m}^{2} / k{g}^{2} ]

Options

  1. A1717
  2. B99
  3. C1313
  4. D1111

Correct answer

D. 1111

Step-by-step solution

T=2π(R+h)3GM=2π8.242×10186.67×10−11×8×1022T = 2\pi\sqrt{\dfrac{{( R + h )}^{3}}{GM}} = 2\pi\sqrt{\dfrac{8.242 \times {10}^{18}}{6.67 \times {10}^{- 11} \times 8 \times {10}^{22}}}
T≈7809secT \approx 7809sec
Number of revolutions in 24 hours:
n=864007809=11.06n = \dfrac{86400}{7809} = 11.06
≈\approx 11 revolutions
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