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 ms^(-1)`. The minmum initial velocity `(u_(e)) = 11.2 ms^(-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. `upsilon_(s) = 22 km s^(-1)`
B. `upsilon_(s) = 72 km s^(-1)`
C. `upsilon_(s) = 42 km s^(-1)`
D. `upsilon_(s) = 62 km s^(-1)`

Answer

C

Explanation

Given, `upsilon_(e) = 11.2 km s^(-1) = sqrt((2Gm_(e))/(R_(e)))`
Here, `M_(s) = 3 xx 10^(5)M_(e)`, Let `r` be the distance of rocket from the. Then `r = 2.5 xx 10^(4) R_(e)`. Let `m` be the mass of the rocket.
According to law of conservation of mechanical energy when rocket is to leave the sun earth system
`(1)/(2) m upsilon_(s)^(2) + ((-GM_(e)m)/(R_(e))) + ((-GM_(s)m)/(r)) = 0`
or `upsilon_(s)^(2) = (2GM_(e))/(R_(e)) + (2GM_(s))/(r )`
`= (2GM_(e))/(R_(e)) + (2Gxx 3 xx 10^(5)M_(e))/(2.5 xx 10^(4)R_(e))`
`upsilon_(s) = sqrt((2GM_(e))/(R_(e))(1 + (3 xx 10^(5))/(2.5 xx 10^(4)))) = sqrt((2GM_(e))/(R_(e)) xx 13)`
`= (11.2) sqrt(3) ~~ 42 km s^(-1)`