The minimum and maximum distances of a satellite from the centre of the earth are 2R and 4R respectively where R is the radius of earth and M is the mass of the earth find radius of curvature at the point of minimum distance.

Answer

`(8R)/(3)`

Explanation

COAM: `mv_(1)(2R)=mv_(2)(2R)impliesv_(1)=2v_(2)` .(i)
COME: `(-GMm)/(2R)+(1)/(2)mv_(1)^(2)=-(GMm)/(4R)+(1)/(2)mv_(2)^(2)` ..(ii)
`impliesv_(1)=sqrt((2GM)/(3R))`
`therefore` radius of curvature at perigee `=(v_(1)^(2))/(g_(1))`
`impliesR_(P)=(2GM)/(3R)xx(4R^(2))/(GM)=(8R)/(3)`