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A planet of mass \mathrm{3 \times 10^{29} gm} moves around a star with a constant speed of \mathrm{2 \times 10^6 \mathrm{~ms}^{-1} }in  a circle or radius \mathrm{1.5 \times 10^{14} \mathrm{~cm}}. The gravitational force exerted on the planet by the star is
 

Option: 1

\mathrm{6.67 \times 10^{22} dyne}


 


Option: 2

\mathrm{8 \times 10^{27} dyne}
 


Option: 3

\mathrm{8 \times 10^{26} \mathrm{~N}}
 


Option: 4

\mathrm{6.67 \times 10^{19} \mathrm{~N}}


Answers (1)

best_answer

Gravitational force supplies centripetal force

\mathrm{\therefore \quad F =\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{3 \times 10^{29} \times\left(2 \times 10^8\right)^2}{1.5 \times 10^{14}} \text { dyne } }

\mathrm{ =8 \times 10^{31} \text { dyne }=8 \times 10^{26} \mathrm{~N}}

Hence option 3 is correct.



 

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Gunjita

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