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A body is taken to a height of $R_{e}$ above the surface of the earth, where $R_{e}$ is the radius of the earth. Which of the following is the acceleration due to gravity at $R_{e}$ ? (Given gravitation constant $G$ is $6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ and the radius of the sun is $6.96 \times 10^{8}$ , mass of the sun is $1.99 \times 10^{30}$ $kg$ )

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2.45

m s^{- 2}

245

m s^{- 2}

0.245

m s^{- 2}

0.0245

m s^{- 2}

answer is A.

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Detailed Solution

Acceleration due to gravity at a height of

R_{e}

from the surface of the earth which is equal to the radius of earth is

2.45

m s^{- 2}

R_{e}

is the height at which the body is placed which is equal to the radius of earth.
As per Newton's universal law of gravitation we know that force of attraction between 2 bodies is given by

F = \frac{G m_{1} m_{2}}{r^{2}} - - - (i)

Where

G

is the gravitational constant

m_{1}

is the mass of first body

m_{2}

is the mass of the second body

r

is the distance between two bodies.
At the same time we know that,

F = mg - - - (ii)

From (i) and (ii) we can write

\frac{G {m_{2} m}_{1}}{r^{2}} = m_{2} g

\frac{G m_{2} m_{1}}{r^{2}} = m_{1} g

Mass of the earth

M = 6 \times 10^{24} kg

Radius of the earth

R_{e} = 6.4 \times 10^{6} m

m

is the mass of the body

g_{e}

is the acceleration due to gravity on the surface of the earth

g_{R_{e}}

is the acceleration due to gravity at a height of

R_{e}

from the surface of the earth.
Here, we need to find the acceleration due to gravity at a height of

R_{e}

from the surface of the earth.
If the object is on the surface of the earth
We can write from equation (i)

\frac{G m_{1} m_{2}}{r^{2}} = \frac{GMm}{{{(R}_{e})}^{2}} - - - (a)

Also from (ii)

F = m g_{e} - - - (b)

Equating (a) and (b)

\frac{GMm}{{{(R}_{e})}^{2}} = m g_{e}

= > \frac{GM}{{{(R}_{e})}^{2}} = g_{e} - - - (c)

If the body is at a distance of

R_{e}

from the surface of earth
New distance is

R_{e} + R_{e} = 2 R_{e}

New acceleration due to gravity is

g_{R_{e}}

From equation (c) we can write

\frac{GM}{{(2 R_{e})}^{2}} = g_{R_{e}} - - - (d)

Dividing equation (d) with (c) we get

\frac{g_{R_{e}}}{g_{e}} = \frac{\frac{GM}{{(2 R_{e})}^{2}}}{\frac{GM}{{{(R}_{e})}^{2}}}

= > \frac{g_{R_{e}}}{g_{e}} = \frac{GM}{{4 {(R}_{e})}^{2}} \times \frac{{{(R}_{e})}^{2}}{GM}

= > \frac{g_{R_{e}}}{g_{e}} = \frac{1}{4} \times \frac{1}{1}

= > g_{R_{e}} = \frac{g_{e}}{4}

We know that the acceleration due to gravity on the surface of the earth is

9.8 m s^{- 2}

Hence

g_{R_{e}} = \frac{9.8}{4}

{= > g}_{R_{e}} = 2.45 m s^{- 2} .

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