1.

An astronaut orbiting in a spaceship round the earth , has centripetal acceleration2.45m//s^2 Find the height of the spaceship . (Take R=6400 km)

Answer»

Solution :As centripetalacceleration equals to acceleration due to gravity at that gravity at that height , then
`a=g_h=(GM)/r^2=(GM)/(R+h)^2 =(gR^2)/(R+h)^2`
`rArr a=(gR^2)/(R+h)^2 =(R/(R+h))^2 = a/g =2.45/9.8=1/4`
`rArr R/(R+h)=1/2 rArr R+h=2R ` R=h=6400 KM
`THEREFORE` SPACESHIP is at a height 6400 km.


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