Question

solve this problem..​

Answer 1

Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of Earth R, i.e. h << R. Orbital velocity of a satellite,

Answer 2

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

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

• Radius of Earth = R.
• Height is very small compared to Radius of Earth.

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

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Orbital velocity:-

Orbital velocity is a Velocity with which a Body should be projected, so that it moves in a circular orbit (or) path.

As we know, From the Expression of Orbital velocity.

$\large{\boxed{\tt v = \sqrt{\dfrac{GM}{R + h}}}}$

But the question states that the height is small when compared to Radius of Earth.

[h << R]

Therefore, "h" can be neglected.

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

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Total Energy of a body on the earth's surface,

$\large{\boxed{\tt T.E = K.E + P.E}}$

If body is at Rest the K.E = 0.

$\large{\tt \hookrightarrow T.E = P.E}$

$\large{\tt \hookrightarrow T.E = \dfrac{- GMm}{R}}$

Negative sign indicates that the Body is bounded to Earth.

But the Body to Escape the earth's atmosphere this Energy should be positive.

And, this Energy is supplied in the Form of Kinetic energy.

Now,

$\large{\boxed{\tt K.E = |T.E|}}$

$\large{\tt \hookrightarrow \dfrac{1}{2} m v'^2 = \left| \dfrac{- GMm}{R}\right|}$

$\large{\tt \hookrightarrow \dfrac{1}{2} m v'^2 = \dfrac{GMm}{R}}$

$\large{\tt \hookrightarrow \dfrac{1}{2} \cancel{m} v'^2 = \dfrac{GM\cancel{m}}{R}}$

$\large{\tt \hookrightarrow \dfrac{1}{2} v'^2 = \dfrac{GM}{R}}$

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

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

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Change in velocity,

$\large{\boxed{\tt \Delta v = v' - v}}$

Substituting the values,

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

$\large{\tt \hookrightarrow \Delta v = \sqrt{\dfrac{GM}{R}} \bigg[ \sqrt{2} - 1\bigg]}$

From Universal Gravitational constant (G) & Acceleration due to gravity (g) relation.

⇒ GM = gR²

⇒ GM/R = gR

Substituting,

$\large{\tt \hookrightarrow \Delta v = \sqrt{gR} \times \bigg[ \sqrt{2} - 1\bigg]}$

$\huge{\boxed{\boxed{\tt \Delta v = \sqrt{gR} [ \sqrt{2} - 1] }}}$

So, Option-1 is correct.

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