A rocket is launched normal to the surface of earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3 × 105 times heavier than the Earth and is at a distance 2.5 × 104 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 ( 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

A rocket is launched normal to the surface of earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3 × 105 times heavier than the Earth and is at a distance 2.5 × 104 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 ( 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) (A) vs = 22 km s-1 (B) vs = 42 km s-1 (C) vs = 62 km s-1 (D) vs = 72 km s-1

Tardigrade Admin · Accepted Answer

(1/2) mv es 2-( GMm / R )-( G (3 × 105 M ) × m /2.5 × 104 R )=0 ⇒ V es =√13 v e=40.3 km / s

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

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

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

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

$\frac{1}{2} m v_{es}^{2} - \frac{GM m}{R} - \frac{G ( 3 \times 1 0 ^{5} M ) \times m}{2.5 \times 1 0 ^{4} R} = 0$
$\Rightarrow V_{es} = 13 v_{e} = 40.3 km / s$

vs​=22kms−1

vs​=42kms−1

vs​=62kms−1

vs​=72kms−1

21​mves2​−RGMm​−2.5×104RG(3×105M)×m​=0 ⇒Ves​=13​ve​=40.3km/s

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

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

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

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

$\frac{1}{2} m v_{es}^{2} - \frac{GM m}{R} - \frac{G ( 3 \times 1 0 ^{5} M ) \times m}{2.5 \times 1 0 ^{4} R} = 0$
$\Rightarrow V_{es} = 13 v_{e} = 40.3 km / s$