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Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of 100 kg and they are 100 m apart. How long will it be before the relative to one another. How long will it be before the gravitational attraction brings them 1 cm closer together ? |
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Answer» `2.52` DAYS `=(Gm_(1)m_(2))/(r^(2))xx(1)/(m_(1))=(Gm_(2))/(r^(2))=(6.67xx10^(-11)xx100)/((100)^(2))` `=6.67xx10^(-13)m s^(-1)` `a^(2)`=acceleration of second astronaut `=(Gm_(1))/(r^(2))=(6.67xx10^(-11)xx100)/((100)^(2))=6.67xx10^(13)" m "s^(-1)` Net acceleration of approach `a=a_(1)+a_(2)=2xx6.67xx10^(-13)" m s^(-1)` Now, `s=(1)/(2)at^(2)1xx10^(-2)=(1)/(2)xx6.67xx10^(-13)xx2xxt^(2)` Solving, we GET `t=1.41` days. |
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