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A star `2.5` times the mass of the sun is reduced to a size of `12km` and rotates with a speed of `1.5rps.` Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun `=2xx10^(30) kg`) |
Answer» Acceleration due to gravity, `g=(GM)/(R^(2))` `=(6.67xx10^(-11)xx2.5xx2xx10^(30))/((12000)^(2))=2.3xx10^(12) ms^(-2)` Centrifugal acceleration `=r omega^(2)=r(2pif)^(2)=12000(2pixx1.5)^(2)=1.1xx10^(6) ms^(-2)` Since, `g gt r omega^(2)`, the body will remain stuck with the surface of the star. |
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