A rocket is launched normal to the surface of the earth, away from the sun, along the line joining the sun and the earth. The sun is `3 xx 10^(5)` times heavier than the earth and is at a distance `2.5 xx 10^(4)` times larger than the radius of the earth. the escape velocity from earth's gravitational field is `u_(e) = 11.2 kms^(-1)`. The minmum initial velocity `(u_(e)) = 11.2 kms^(-1)`. the minimum initial velocity `(u_(s))` required for the rocket to be able to leave the sun-earth system is closest to (Ignore the rotation of the earth and the presence of any other planet

Options

A. `v_(s) = 72 Km s^(-1)`
B. `v_(s) = 22 Km s^(-1)`
C. `v_(s) = 42 Km s^(-1)`
D. `v_(s) = 62 Km s^(-1)`

Answer

C

Explanation

Let mass of satellite be m and mass of sun be

Given, `v_(s) = 11.2 km//s, M_(s) = 3 xx 10^(5) M_(e)`
Total energy of the system remains constant
`(1)/(2) mv_(e)^(2) - (GmM_(e))/(R ) - (GM_(s)m)/(2.5 xx 10^(4)R) = 0`
`v_(e)^(2) = (2GM_(e))/(R ) + (2GM_(s))/(2.5 xx 10^(4)R)`
`= (2GM_(e))/(R ) + (6 xx 10^(5) GM_(e))/(2.5 xx 10^(4)R) = (GM_(e))/(R )(2+24)`
`= sqrt(13)(sqrt((2GM_(e))/(R ))) = (11.2 xx sqrt(13)) km//s`
= 40.4 km/s
The minimum velocity required from the option is 42 m/s.