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 earths gravitational field is $V_{c} = 11 .2 k m s^{- 1}$ . The minimum initial $(V_{c})$ 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)

(V_{c}) = 22 k m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(V_{c}) = 42 k m s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(V_{c}) = 62 k m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(V_{c}) = 72 k m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $(V_{c}) = 42 k m s^{- 1}$
$\frac{1}{2} {m V}_{e}^{2} - \frac{{G M}_{e} m}{R_{e}} - \frac{{G M}_{e} m \times 3 \times 10^{5}}{2.5 \times 10^{4} R_{e}} = 0$
$\frac{{V e}^{2}}{2} = \frac{{G M}_{m}}{R_{e}} [\frac{1 + 3 \times 10^{5}}{2.5 \times 10^{4}}]$
$V_{e} = \sqrt{\frac{13 ({2 G M}_{e})}{R_{e}}} = \sqrt{13} \times 11.2 \approx 42$
The minimum initial $V_{s} = 42 k m / s$ .

Suggest Corrections

Similar questions

Q. 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 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)

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 $v_{e} = 11.2 k m / s$ . 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 $)$

Q. A rocket is fired from the earth towards the sun. At what distance from the earths centre is the gravitational force on the rocket zero? Mass of the sun =

2 \times 10^{30} k g

, mass of the earth =

6 \times 10^{24} k g

. Neglect the effect of other planets etc. (orbital radius =

1.5 \times 10^{11} m

Q. A rocket is launched straight up from the surface of the earth. When its altitude is one fourth of the radius of the earth, its fuel runs out and therefore it coasts. The minimum velocity which the rocket must have when it starts to coast if it is to escape from the gravitational pull of the earth is [escape velocity on surface of earth is

11.2 k m / s

]

Q. A rocket is fired from the Earth towards the Sun. At what distance from the Earth's center, the gravitational force on the rocket is zero? Mass of the Sun

= 2 \times 10^{30} k g

and mass of the Earth

= 6 \times 10^{24} k g

Join BYJU'S Learning Program

The correct option is B (Vc)=42kms−112mV2e−GMemRe−GMem×3×1052.5×104Re=0Ve22=GMmRe[1+3×1052.5×104]Ve=√13(2GMe)Re=√13×11.2≈42The minimum initial Vs=42km/s.