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JEE Advance - Physics (2017 - Paper 2 Offline - No. 6)

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 $${V_c} = 11.2km\,{s^{ - 1}}.$$. The minimum initial velocity $$\left( {{v_s}} \right)$$ required for the rocket to be able to leave the sun-earth system is closest to (Ignore the the rotation and revoluation of the earth and the presence of any other planet)
$${v_s} = 22\,km\,{s^{ - 1}}$$
$${v_s} = 42\,km\,{s^{ - 1}}$$
$${v_s} = 62km\,{s^{ - 1}}$$
$${v_s} = 72km{s^{ - 1}}$$

Explanation

JEE Advanced 2017 Paper 2 Offline Physics - Gravitation Question 11 English Explanation

Given : Mass of the sun, Ms = 3 $$\times$$ 105 $$\times$$ mass of the earth = 3 $$\times$$ 105 Me

Distance between the sun and the earth,

d = 2.5 $$\times$$ 104 $$\times$$ radius of the earth = 2.5 $$\times$$ 104 Re,

Escape speed, ve = 11.2 km s$$-$$1

$$\therefore$$ $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} = 11.2$$ km s$$-$$1

Minimum velocity required for the rocket to escape the given earth $$-$$ sun system = vs

Using energy conservation for the given situation,

$${1 \over 2}mv_s^2 = {{G{M_e}m} \over {{R_e}}} + {{G{M_s}m} \over {d + {R_e}}} = {{G{M_e}m} \over {{R_e}}} + {{G \times 3 \times {{10}^5}{M_e}m} \over {(2.5 \times {{10}^4}{R_e} + {R_e})}}$$

$$\therefore$$ $$v_s^2 = {{2G{M_e}} \over {{R_e}}} + {{24G{M_e}} \over {{R_e}}}$$ ($$\because$$ 2.5 $$\times$$ 104 >> 1)

$$ \Rightarrow v_s^2 = v_e^2 + 12v_e^2 = 13v_e^2$$

or $${v_s} = \sqrt {13} $$ ve = 40.38 km s$$-$$1 $$\approx$$ 42 km s$$-$$1

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