A chord of length 64m is used to connect a 100\text{Kg} astronaut to spaceship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the chord?

Assume that the spaceship is orbiting near the earth surface. Assume that the spaceship and the astronaut fall on a straight line from the earth center. The radius of the earth is 6400\text{Km}

1 Answer
Jul 6, 2018

Assuming that the spaceship of mass M_s and the astronaut of mass m fall on a straight line from the center of earth having mass M_e in a circular orbit.
We know that Gravitational force provides the required centripetal force for the circular motion of angular frequency omega. For spaceship and astronaut as composite system, in the state of equilibrium we have

"Force of gravity"\ =\ "centripetal force"

M_sg=M_sRomega^2
where R is radius of the earth and g acceleration due to gravity near the earth surface =9.81\ ms^-1.

(We have ignored the mass of astronaut as it is much smaller than that of the spaceship)

=>omega=sqrt(g/R) .....(1)

Forces acting on the astronaut are

  1. Centrifugal force F_"Cf"=m(R+64)omega^2 ......(2)
  2. Gravitational force due to the earth
    F_"g"=(mM_eG)/(R+64)^2 .....(3)
    where G is Universal gravitational constant.

Net force will be tension in the chord.

:.T=F_"Cf"-F_"g"

Using (2) and (3)

T=m(R+64)omega^2-(mM_eG)/(R+64)^2

Using (1) and substituting gR^2 in place of M_eG in the second term of RHS we get

T=m(R+64)g/R-(mgR^2)/(R+64)^2
=>T=mg+mg64/R-mg((R+64)/R)^-2

Expanding using binomial theorem and ignoring higher order terns as 64\ "<<"\ R we get

T=mg+mg64/R-mg(1-2(64)/R)
=>T=3mg64/R

Inserting given values we get

=>T=3xx100xx9.81xx64/(6.4xx10^6)
=>T=2.94xx10^-2\ N