Gravitation

JEE Main level, single correct, medium

190 of 223

Question 190 of 223

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} ]

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Answer: (D) 1111

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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