A cord of length `64 m` is used to connected a `100 kg` astronaut to spaceship whose mass is much larger than that of the astronuat. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surface. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. the radius of the earth is `6400 km`.

Answer

`T=3xx10^(-2) N`

Explanation

Centripetal force for orbiting of surface ship is provided by gravitational force.
`(GMM_(e))/(R^(2))=Mromega^(2)`
`omega=sqrt((GM_(e))/(R^(3)))...(1)`
For orbiting of astronaut
`F_(s)+T=m(R+L)omega^(2)`
`(GM_(e)m)/((R+L)^(2))+T=m(R+L)((GM_(e))/R^(3)))`
`T=m(R+L)(GM_(e))/(R^(3))-(GM_(e)m)/((R+L)^(2))`
`T=mg+(mgL)/R-mg(1+1/R)^(-2)`
`T=mg+(mgL)/R-mg(1-(2L)/R)`
`T=mg+(mgL)/R-mg+(2mgL)/R`
`T=(3mgL)/RimpliesT=(3xx100x10xx64)/(64xx10^(5))`
`T=3xx10^(-2)N`