A small satellite revolves around a heavy planet in a circular orbit. At certain point in its orbit a sharp impulse acts on it and instantaneously increases its kinetic energy to 'k' `(lt2)` times without change in its direction of motion show that in its subsequent motion the ratio of its maximum and minimum distance from the planet is `(k)/(2-k)` assuming the mass of the satellite is negligibly small as compared to that of the planet.

Answer

n/a

Explanation

Orbital velocity `v_(0)=sqrt((GM)/(r_(1)))`
After impulse `v_(1)=sqrt(k)v_(0)`
COAM: `mv_(1)r_(1)=mv_(2)r_(2)` ..(i)
COME: `(-GMm)/(r_(1))+k((1)/(2)mv_(0)^(2))`
`=(-GMm)/(r_(2))+((1)/(2)mv_(2)^(2))` …(ii)
Solving equation (i) and (ii) `(r_(2))/(r_(1))=(k)/(2-k)`