Question

A planet has mass equal to mass of the earth but
radius one fourth of radius of the earth. Then
escape velocity at the surface of this planet will be
(1) 11.2 km/s
(2) 22.4 km/s
(3) 5.6 km/s
(4) 44.8 km/s

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A planet has mass equal to mass of the earth but radius one fourth of radius of the earth. Then escape velocity at the surface of this planet will be?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Masses of Planets are constant.
• Radius of Planet is ¼ of the radius of Earth.

$\huge{\bold{\underline{Explanation:-}}}$

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Escape velocity:-

Escape velocity is the velocity required by a body escapes from Earth Gravitational field.

we Know, the Expression for it.

$\large{\boxed{\tt v = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}}}$

As the question says that Mass is Constant, we need to take the First equation as Both Universal Gravitational constant (G) & Mass of earth (M) are constant.

$\large{\tt \hookrightarrow v = \sqrt{\dfrac{2GM}{R}}}$

From this we can clearly see that Escape velocity is Inversely proportional to the square root of Radius of Planet.

$\large{\hookrightarrow{\underline{\underline{\tt v \propto \sqrt{\dfrac{1}{R}}}}}}$

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Assumptions:-

• Velocity of Earth = ${\sf v_e}$
• Velocity of Planet = ${\sf v_p}$
• Radius of Earth = ${\sf R_e}$
• Radius of Planet = ${\sf R_p}$

According to the Statement.

Radius of Planet is ¼ of the radius of Earth.

Therefore,

$\large{\tt \hookrightarrow R_p = \dfrac{1}{4} . R_e}$

$\large{\tt \hookrightarrow 4 \times R_p = R_e}$

$\large{\tt \hookrightarrow R_e = 4R_p}$

Now, From Above obtained relation,

$\large{\boxed{\tt v \propto \sqrt{\dfrac{1}{R}}}}$

Now,

$\large{\tt \hookrightarrow \dfrac{v_e}{v_p} = \sqrt{\dfrac{R_p}{R_e}}}$

Substituting the values,

$\large{\tt \hookrightarrow \dfrac{v_e}{v_p} = \sqrt{\dfrac{R_p}{4R_p}}}$

$\large{\tt \hookrightarrow \dfrac{v_e}{v_p} = \sqrt{\cancel{\dfrac{R_p}{4R_p}}}}$

$\large{\tt \hookrightarrow \dfrac{v_e}{v_p} = \sqrt{\dfrac{1}{4}}}$

$\large{\tt \hookrightarrow \dfrac{v_e}{v_p} = \dfrac{1}{2}}$

As we Know Escape velocity at Earth is 11.2 Km/sec.

Substituting it,

$\large{\tt \hookrightarrow \dfrac{v_e}{v_p} = \dfrac{1}{2}}$

$\large{\tt \hookrightarrow \dfrac{11.2}{v_p} = \dfrac{1}{2}}$

Cross - Multiplying,

$\large{\tt \hookrightarrow 1 \times v_p = 11.2 \times 2}$

$\large{\tt \hookrightarrow v_p = 11.2 \times 2}$

$\huge{\boxed{\boxed{\tt v_p = 22.4 \: Km/sec}}}$

So, the Escape velocity at the surface of Planet is 22.4 Km/sec (Option- 2).

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