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)

Difficult

Solution

$\begin{array}{l} \frac{m V^{2}}{2} - \frac{G M M_{E}}{R_{E}} - \frac{G M M_{S}}{(x + R_{E})} = 0 \\ \frac{V_{0}^{2}}{2} - \frac{G M_{E}}{R_{E}} - \frac{G M_{S}}{x + R_{E}} = 0 \\ V_{0} = \sqrt{\frac{2 G M_{E}}{R_{E}} + \frac{2 G M_{S}}{x + R_{E}}} \\ = \sqrt{V_{e}^{2} + \frac{2 G M_{E} \times 3 \times 10^{5}}{R_{E} (2.5 \times 10^{4})}} \\ = \sqrt{V_{e}^{2} + V_{0}^{2} \frac{3 \times 10^{2} \times 2}{5}} \\ = V_{e} \sqrt{13 \times 10} \\ = 11.2 \times 3.6 \\ V_{0} = 40.38 {kms}^{- 1} \end{array}$

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