Question #4929a

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Question #4929a

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Answer 1 · 2017-03-14T10:25:49

If I understood your answer correctly I would say the following:

As per Law of Universal Gravitation the force of attraction $F_{G}$ $F_G$ between two bodies of masses $M_{1} and M_{2}$ $M_1 and M_2$ located at a distance $r$ $r$ from each other is given by the expression

$F_{G} = G \frac{M_{1} . M_{2}}{r^{2}}$ $F_G =G (M_1.M_2)/r^2$ .......(1)

Where $G$ $G$ is the proportionality constant.
It has the value $6.67408 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ $6.67408 xx 10^-11 m^3 kg^-1 s^-2$

We see that if $M_{e}$ $M_e$ is the mass of earth and $m$ $m$ be the mass of an object of interest then
$F_{G} = m (G \frac{M_{e}}{r^{2}})$ $F_G =m(G M_e/r^2)$

Comparing it with equation of Newton's Second law of motion

$F = m a$ $F=ma$

we observe that
Acceleration due to gravity $g = G \frac{M_{P}}{r^{2}}$ $g=G M_P/r^2$

When the object initially located at some height above the surface of earth is brought down, the distance $r$ $r$ between the two decreases. The value of $g$ $g$ increases; $r$ $r$ being in the denominator of the relationship. When the object touches the surface of earth $r = R_{e}$ $r=R_e$ , radius of earth. Here we have
$g = G \frac{M_{P}}{R_{e}^{2}} = 9.81 m s^{- 2}$ $g=G M_P/R_e^2=9.81ms^-2$ , a constant.

As the object is taken below the surface of earth, we see that the distance between the two decreases.

However, it can be proved that attractive force between part earth mass of spherical shell above the object and the object is zero.

As such only the mass of sphere below the location of the object and the object needs to considered in the force equation.

This leads to decrease in the value of acceleration due to gravity.

Hope it helps. Feel free to raise any questions.

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Question #4929a

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