A meteorite approaching a planet of mass M (in the straight line passing through the centre of the planet) collides with an automatic space station orbiting the planet in a circular trajectory of radius R. The mass of the station is ten times as large as the mass of the meteorite. As a result of the collision, the meteorite sticks in the station which goes over to a new orbit with the minimum distance R/2 from the planet. Speed of the meteorite just before it collides with the planet is : .

Options

A. `sqrt((58GM)/(R))`
B. `sqrt((38GM)/(R))`
C. `sqrt((28GM)/(R))`
D. `sqrt((18GM)/(R))`

Explanation

As the space station is moving in circular orbit,
`(GM(10m))/(R^(2))=((10m)v_(0)^(2))/(R)`
`impliesv_(0)=sqrt((GM)/(R))` ..(i)
Let u be the velocity of meteorite.
Velocity of the space station after collision can be obtained from momentum conservation
`mu=(10m+m)v_(1)impliesv_(1)=(u)/(11)`
`10m,v_(0)=(10m+m)v_(2)impliesv_(2)=(10)/(11)v_(0)`
Let v be the velocity space stationa at closest distance
from angu7lar momentum conservation
`10mv_(0)xxR=11mv(R)/(2)impliesv=(20v_(0))/(11)`
from energy conservation
`(1)/(2)xx(11m)(v_(1)^(2)+v_(2)^(2))-(GM(11m))/(R)=(1)/(2)xx(11m)v^(2)-(GM.11m)/(R//2)`
`implies((u)/(11))^(2)+((10v_(0))/(11))^(2)-(2GM)/(R)=((20v_(0))/(11))^(2)-(4GM)/(R)`
`implies(u^(2))/(11^(2))=(400v_(0)^(2))/(11^(2))-(100v_(0)^(2))/(11^(2))-(2GM)/(R)`
`impliesu^(2)=(GM)/(R)(400-100-242)=58(GM)/(R)`
Ans `u=sqrt((58GM)/(R))`