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 $3 \times 10^{5}$ times heavier than the Earth and is at a distance $2.5 \times 10^{4}$ times larger than the radius of the Earth. The escape velocity from Earth’s gravitational field is $v_{e} = 11.2 k m s^{- 1}$ . The maximum initial velocity $(v_{s})$ required for the rocket to be able to leave the sun-Earth system is closest to
(ignore the rotation and revolution of the Earth and the presence of any other planet)

Question

Infinity Learn · Accepted Answer

To escape, total energy of rocket $$=0mV022$$−GmMERE−$$GmMSx+RE=0$$⇒$$V022$$−GMERE−$$GMSx+RE=0$$⇒$$V0=2GMERE+2GMSx+RE=Ve2+2GME$$×$$3$$×$$105RE2.5$$×$$104=Ve2+Ve23$$×$$10$$×$$25=Ve13=11.2$$×$$3.6$$⇒$$V0=40.38kms$$−$$1$$

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