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QJEEM19Phe0c7: JEE Main 2019 Physics - Class 11 JEEM19Phe0c7 Solution: A spaceship orbits around a planet at a height of 20 km from…

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]

Multiple Choice Options

  • A 13
  • B 9
  • C 17
  • D 11

Step-by-Step Verified Solution

mV2r=GMmr2{{m{V^2}} \over r} = {{GMm} \over {{r^2}}}

V=GMrV = \sqrt {{{GM} \over r}}

n=VT2πr=GMrT2πrn = {{VT} \over {2\pi r}} = \sqrt {{{GM} \over r}} {T \over {2\pi r}}

=(GMr3)×T2π=6.67×10−11×8×1022(202×104)3×T2π= \left( {\sqrt {{{GM} \over {{r^3}}}} } \right) \times {T \over {2\pi }} = \sqrt {{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}} \over {{{\left( {202 \times {{10}^4}} \right)}^3}}}} \times {T \over {2\pi }}

=24×36002×3.146.67×8×1011(202)3×1012= {{24 \times 3600} \over {2 \times 3.14}}\sqrt {{{6.67 \times 8 \times {{10}^{11}}} \over {{{\left( {202} \right)}^3} \times {{10}^{12}}}}}

=24×36002×3.14×1242.8=24×360074.51=11= {{24 \times 3600} \over {2 \times 3.14 \times 1242.8}} = {{24 \times 3600} \over {74.51}} = 11

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