What is the value of acceleration due to gravity at a height equal to half the radius of Earth?

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Answer

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Hint: Here we have to apply the formula and concept of acceleration due to gravity.Acceleration due to gravity is the acceleration of the mass due to gravitational force. The SI unit is given by $m{s^{ - 2}}$ . It is a vector quantity since it has both magnitude and direction. Acceleration due to gravity is given by $g$ .

Complete step by step answer:

Gravity is the power by which the planet draws the body to its core. Let us consider two bodies of masses ${m_a}$ and ${m_b}$ . With regard to the application of equivalent forces on two bodies, the mass in terms of body $b$ is given as:${m_b} = {m_a}\left( {\dfrac{{{a_A}}}{{{a_B}}}} \right)$The above mass is called the inertial mass of a body.Under the influence of gravity the mass is given by:${m_B} = \left( {\dfrac{{{F_B}}}{{{F_A}}}} \right) \times {m_A}$The above mass is called the gravitational mass.We know that according to universal law of gravitation:$F = \dfrac{{GMm}}{{{r^2}}}$Also, $F = ma$Equating the two equations we get:$a = \dfrac{{GM}}{{{r^2}}}$If $g$ is the acceleration due to gravity then:$g = \dfrac{{GM}}{{{r^2}}}$The acceleration due to gravity depends on the mass and radius.Now, if the mass is at a height $h$ from the surface of the earth, the force $F$ is given by:$F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$So, the acceleration due to gravity is given by:${g_h} = \dfrac{{GM}}{{\left[ {{R^2}{{\left( {1 + \dfrac{h}{R}} \right)}^2}} \right]}}$ …… (i)The acceleration due to gravity on surface of earth is given as:$g = \dfrac{{GM}}{{{R^2}}}$ …… (ii)On dividing equation (ii) by (i), we get:$\begin{gathered} {g_h} = g{\left( {\dfrac{{1 + h}}{R}} \right)^{ - 2}} \\ {g_h} = g\left( {1 - \dfrac{{2h}}{R}} \right) \\ \end{gathered} $Here according to question:$h = \dfrac{R}{2}$Thus,$ {g_h} = g\left( {1 - \frac{{2 \times \dfrac{R}{2}}}{R}} \right) \\ = g\left( {1 - 1} \right) \\ = 0 \\ $

Hence, the value of acceleration is zero.

Note:

Here, we have to take the height as $\dfrac{R}{2}$ , otherwise we shall get a wrong answer.Also, we have to correctly remember the formula for acceleration due to gravity.It is a vector quantity since it has both magnitude and direction.

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CONCEPT:

Acceleration Due to Gravity:

The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
We know that when a force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.
The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g.

The acceleration due to gravity on the surface of the earth is given by,

$⇒g = \frac{{GM}}{{{R^2}}}$

The acceleration due to gravity above the surface of the earth is given as,

$⇒ g_{h}=\frac{g}{\left ( 1+\frac{h}{R} \right )^2}$

If h <<< R

$⇒ g_{h}=g\left ( 1-\frac{2h}{R} \right )$

The acceleration due to gravity below the surface of the earth is given as,

$⇒ g_{d}=g\left ( 1-\frac{d}{R} \right )$

CALCULATION:

The gravitational acceleration of the earth at a height equal to half of the radius of the earth is given as (h = R/2),

$⇒ g_{1}=\frac{g}{\left ( 1+\frac{h}{R} \right )^2}$

$⇒ g_{1}=\frac{g}{\left ( 1+\frac{R}{2R} \right )^2}$

$⇒ g_{1}=\frac{4g}{9}$ -----(1)

The gravitational acceleration of the earth at a depth equal to half of the radius of the earth is given as (d = R/2),

$⇒ g_{2}=g\left ( 1-\frac{d}{R} \right )$

$⇒ g_{2}=g\left ( 1-\frac{R}{2R} \right )$

$⇒ g_{2}=\frac{g}{2}$ -----(2)

By equation 1 and equation 2,

$⇒ \frac{g_{1}}{g_2}=\frac{4g}{9}\times\frac{2}{g}$

$⇒ \frac{g_{1}}{g_2}=\frac{8}{9}$

Hence, option 2 is correct.

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