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=62 km s^(-1)`
B. `v_s=42 km s^(-1)`
C. `v_s=72 km s^(-1)`
D. `v_s=22 km s^(-1)`

Answer

B

Explanation

For earth, `V_e=sqrt((2GM)/R)`
For Sun +Earth,
Potential energy = `(-"GM"/R-(3xx10^5)/(2.5xx10^4)"GM"/R)m`
`=(GMm)/R (1+12) = (13GMm)/R therefore 1/2mv^2 =(13GMm)/R`
`v=sqrt((2GM)/R) sqrt13=sqrt13v_e therefore v_e`=40.4 km/s , 42 km/s