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# Engineering

A spaceship orbits around a planet at a hight 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\times 10^{22} Kg,\, \, Radius\, of \, \, planet\, \, =2\times 10^{6}m \\ Gravitational\, \, Constant\, \, G=6.67\times 10^{-11}Nm^{2}Kg^{2})

  • Option 1)

    9

  • Option 2)

    17

  • Option 3)

    13

  • Option 4)

    11

Answers (1)
P Plabita

As we

\left ( T=time, r= radius ,G= gravitational \, \, force\, \, ,m=mass\, \, of \, \, planet\right )

T= 2\pi \sqrt \frac{r^3}{GM}=2\pi \sqrt{\frac{\left ( 2.02\times 10^{6} \right )^{3}}{\frac{20}{3}*10^{-11}\times 8\times 10^{22}}}

 

= 2\pi \sqrt{\frac{\left ( 8\times 10^{18} \right )}{\frac{20}{3}*10^{-11}\times 8\times 10^{22}}}

=7800sec.

Now , No of revolution =\frac{24\times 3600}{7800}=11.07=11sec.


Option 1)

9

Option 2)

17

Option 3)

13

Option 4)

11

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