Black Hole Identified #3

Classical Mechanics Level 3

The Escape Velocity From A Star Is Given By This Equation : V e = 2 GM R {\text{V}}_e=\sqrt{\frac{2\text{GM}}{R}} Where, V e = Escape Velocity \text{V}_e=\text{Escape Velocity} G = Gravitational Constant \text{G}=\text{Gravitational Constant} = 6.673 × 1 0 − 11 N m 2 k g − 2 =6.673 \times 10^{-11} \text{N}m^2kg^{-2} M = Mass of The Star \text{M}=\text{Mass of The Star} R = Radius of The Star \text{R}=\text{Radius of The Star} Now we have to determine the Escape Velocity (in km/s) from a star which has an enormous Mass of 6 × 1 0 30 k g 6 \times 10^{30} kg but a Radius of only 5.00475 k m 5.00475 km Clue: Plug the Mass and Radius in the Equation above. Don't forget to convert from kilometer to meter and vice versa!


The answer is 400000.

Muhammad Arifur Rahman by Muhammad Arifur Rahman , 23, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Given, Mass, M = 6 × 1 0 30 k g \text{M}=6\times 10^{30}kg Radius, R = 5.00475 k m = 5004.75 m \text{R}=5.00475km=5004.75m Then, v e = 2 × 6.673 × 1 0 − 11 × 6 × 1 0 30 5004.75 m s − 1 v_e=\sqrt{\frac{2\times6.673\times10^{-11}\times6\times10^{30}}{5004.75}}ms^{-1} = 400000000 m s − 1 =400000000ms^{-1} = 400000 k m s − 1 =400000kms^{-1}

And you did it!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...