Question

) The escape velocity for a rocket on earth is 11. 2
km/sec. Its value on planet where acceleration due to
gravity is double that on the earth and diameter of
the planet is twice that of earth will be in km/sec.
a) 11.2
b) 5.6
C) 22.4
d) 53.6​

Answer 1

Formulae to be used:-

• $\sf g = \dfrac{GM}{R^2}$ where g is the acceleration due to gravity, G is the universal gravitational constant, M is the mass and R is the radius of a planet.
• $\sf V_{esc} = \sqrt{\dfrac{2GM}{R}}$ where $\sf V_{esc}$ is the escape velocity on a planet, G is the universal gravitational constant, M is the mass and R is the radius of a planet.

Answer:-

(Note: In this solution subscript E means for Earth and X means for the unknown planet X)

On Earth:-

$\sf V_{esc(E)} = \sqrt{\dfrac{2GM_E}{R_E}}\quad \quad\dots (1)$

As it is given that, escape velocity on Earth is 11.2 km/sec,

$\longrightarrow \sf V_{esc(E)} = 11.2\:km/sec\quad\quad\dots(2)$

From (1) and (2),

$\sf \longrightarrow 11.2 = \sqrt{\dfrac{2GM_E}{R_E}}$

Multiplying numerator and denominator with $\sf R_E$ inside the square root in R.H.S.,

$\longrightarrow \sf 11.2 = \sqrt{\dfrac{2GM_E R_E}{R_E {\:}^2}}$

Using $\sf g_E = \dfrac{GM_E}{R_E {\:}^2},$

$\longrightarrow \sf \sqrt{2g_E R_E} = 11.2\:km/sec \quad\quad \dots (3)$

On planet X:-

As it is given that, acceleration due to gravity in planet X is twice the acceleration due to gravity on Earth,

$\sf g_X = 2g_E\quad\quad\dots (4)$

And, as it is given that, diameter of planet X is twice the diameter of Earth,

$\sf d_X = 2d_E$

$\longrightarrow \sf R_X = 2R_E\quad\quad\dots (5)$

So,

$\sf V_{esc(X)} = \sqrt{\dfrac{2GM_X}{R_X}}$

Multiplying numerator and denominator inside the square root with $\sf R_X$ in R.H.S.,

$\longrightarrow \sf V_{esc(X)} = \sqrt{\dfrac{2GM_X R_X}{R_X {\:}^2}}$

Using $\sf g_X = \dfrac{GM_X}{R_X {\:} ^2} ,$

$\longrightarrow \sf V_{esc(X)} = \sqrt{2g_X R_X}$

From (4) and (5),

$\longrightarrow \sf V_{esc(X)} = \sqrt{2 \times 2g_E \times 2R_E}$

$\longrightarrow \sf V_{esc(X)} = 2\sqrt{2g_E R_E}$

From (3),

$\longrightarrow \sf V_{esc(X)} = 2 \times 11.2\:/sec$

$\longrightarrow \sf \underline{\underline{\sf{\green{ V_{esc(X)} = 22.4\: km/sec}}}}$