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A RESEARCH SATELLITE OF MASS 200 KG CIRCLES IN AN ORBIT of averge radius of 3R/2 WHERE R is the radius of the earth. Aasuuming that the gravitational pull on the mass of 1 kg on the earth's surface is 10N, the pull on the satelite is --------------

8 years ago

accleration due to gravity at surface(gs)= GM/R2

accleration due to gravity at a height (gh) = GM/(R+H)2

force on a particle of 1kg at surface  = mGM/R2 =GM/R2 = 10N     ......1           (given)

force on satellite of mass 200kg at a height H  = mGM/(R+H)2

F = 200GM/(R+H)2  .............2

dividing 2 by 1

F/10 = 200/(1+H/R)2

=2000(4/25)                      (H=3R/2)

=320N

8 years ago

let us take mass of earth as Me then 10=GMe.1/R^2   -----1
now mass of satellite =200 kg let us take force on it as F then,
F=GMe.200/(3R/2)^2-------2
dividing 2by 1
F/10=200/9/4
F/10=800/9 hence f=8000/9=888.8=889N
4 years ago
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