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(i) Define orbital velocity and establish an expression for it. (ii) Calculate the value of orbital velocity for an artifical satellite of earth orbiting at a height of 1000km (Mass of the earth =6xx10^(24)kg, radius of the earth =6400km). |
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Answer» Solution :`(i)` `(i)` Definition : It is the HORIZONTAL velocity that has to be imported to a satellite at the determined height so that it makes a circular orbit around the planet is called orbital velocity. `(ii)` Let us assume that a satellite of mass `'m'` moves around the earth in a circular orbit of radius `'r'` with uniform speed `'v'`. Let the satellite be at a height `'b'` from the surface of the earth. Here `r-R+h`, where `'R'` is the radius of the earth. `(III)` The centripetal force required to KEEP the satellite in circular orbit is `F=(mv^(2))/(r )` `=(mv_(0)^(2))/(R+h)` `(IV)` The gravitational force between the earth and the satellite is `F=(GMm)/(r^(2))=(GMm)/((R+h)^(2))` for stable motion. `(mv_(0)^(2))/(R+h)=(GMm)/((R+h)^(2))` `v_(0)=sqrt((GM)/(R+h))` `:'g=(GM)/(R )impliesv_(0)=sqrt((GR^(2))/(R+h))` If satellite is at a height of few hundred kilometers (i.e. say `200km`) `R+h` will be `R`. `:.v_(0)=sqrt(gR)` `(ii)` Given : height `h=1000km` Mass of Earth `M=6xx10^(2)kg` Radius of Earth `R=6400km` Orbital velocity `v_(0)=?` Formula : `v_(0)=sqrt((GM)/(R+h))` `v_(0)=sqrt((6.67xx10^(-11)xx6xx10^(24))/(6400+1000))` `=sqrt((40.02xx10^(13))/(7400))` `=sqrt((40.02xx10^(13))/(7.4xx10^(3)))=sqrt((40.02xx10^(10))/(7.4))` `v_(0)=sqrt((40.02)/(7.4))xx10^(5)=(6.32613xx10^(5))/(2.7202)` `=232561.209` `=2.325xx10^(5)km//s` `=2.325xx10^(5)xx10^(3)m//s` `=2.325xx10^(8)m//s`. |
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