A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. [2017]

Question

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 $v_{e} = 11.2 {km s}^{- 1}$ . The minimum 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) [2017]

Smart Achievers · Accepted Answer

(2)

Applying energy conservation

$\frac{1}{2} m V_{s}^{2} - \frac{G M_{e} m}{R_{e}} = \frac{G M_{e} m \times 3 \times 10^{5}}{2.5 \times 10^{4} R_{e}}$

$\frac{V_{s}^{2}}{2} = \frac{G M_{e}}{R_{e}} [1 + \frac{3 \times 10^{5}}{2.5 \times 10^{4}}]$

$V_{s} = \sqrt{13 (\frac{2 G M_{e}}{R_{e}})}$ $(∵ V_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}} = 11.2 km/s)$

$∴$ $V_{s} = \sqrt{13} \times 11.2 \approx 42 km/s$

Solution

1 $v_{s} = 22 {km s}^{- 1}$

2 $v_{s} = 42 {km s}^{- 1}$

3 $v_{s} = 62 {km s}^{- 1}$

4 $v_{s} = 72 {km s}^{- 1}$

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Solution

1 vs=22 km s-1

2 vs=42 km s-1

3 vs=62 km s-1

4 vs=72 km s-1

Ans. (2) Applying energy conservation 12mVs2-GMemRe=GMem×3×1052.5×104 Re Vs22=GMeRe[1+3×1052.5×104] Vs=13(2GMeRe) (∵Ve=2GMeRe=11.2 km/s) ∴ Vs=13×11.2≈42 km/s

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1 $v_{s} = 22 {km s}^{- 1}$

2 $v_{s} = 42 {km s}^{- 1}$

3 $v_{s} = 62 {km s}^{- 1}$

4 $v_{s} = 72 {km s}^{- 1}$