Question

explian variation in acceleration due to gravity (1) due to depth (2) due to shape of earth.

and also explain condition of weightlessness on earth surface?
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Answer 1

$\huge\bf{\red{\mid{\overline{\underline{Your \: Answer}}}\mid}}$

1. $\small{\underline{\underline{\mathbf{Due\: to\: depth-}}}}$

✏ Consider the above digram.

Assuming that the density of earth remains same throughout the volume.

At earth surface :

$g = \frac{4}{3} \times \pi \: g r_{e}p........(i)$

At a depth d inside the earth :

✡ For point p only mass of the inner sphere is effective :

$g_{d} = \frac{gm}{ {r}^{2} }$

then,

$= > g_{d} = \frac{g}{ {r}^{2} } \times \frac{ m_{e} {r}^{3} }{ r_{e} }$

$= > \frac{g m_{e} }{ { r_{e}}^{2} } \times \frac{r}{ r_{e} }$

$= > g_{d} = g(1 - \frac{d}{ r_{e} } )$

★$\bold{valid \: for \: any\: depth}$

2. $\small{\underline{\underline{\mathbf{Due\:to\:shape\:of\:earth}}}}$

✏ consider the above digram

✡

$r_{p} < r_{e}( r_{e} = r_{p} + 21km)$

$= > g_{p} = \frac{g m_{e} }{ { r_{p} }^{2} }$

$= > g_{e} = \frac{ gm_{e} }{( { r_{p} + 2100}^{2} )}$

★ by putting values,

$g_{p} - g_{e}$

$\boxed{0.02\: m/s}$

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$\small{\underline{\underline{\mathbf{Condition\: of\: weightlessness\: on\: earth\: surface}}}}$

✏ If apparent weight of body is zero then angular speed of earth can be calculated as,

mg' =

$mg - m r_{e} { \alpha }^{2} { \cos }^{2} \gamma$

$\gamma = \frac{1}{ \cos \alpha } \times \sqrt{ \frac{g}{ r_{e} } }$

★ but, at equator it is 0°

the value is, $\boxed{0.00125 \: rad/s}$

Note: - if earth will to rotate with 17 times of it present angular speed then bodies lying on equator would fly off into the space. time period of earth rotation in this case would be 1.4 h.

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Answer 2

Step-by-step explanation:

Variation of Acceleration due to Gravity with depth

Let’s say, a body of mass m is resting at point A , where A is at a depth of h from the earth’s surface.

Distance of point A from the centre of the earth = R – h,

where R is the radius of the earth.

Mass of inner sphere = (4/3). Pi. (R-h)^3. p

Here p is the density.

Now at point A, the gravitational force on the object of mass m is

F = G M m/ (R-h)^2 

= G. [(4/3). Pi. (R-h)^3. p] m/(R-h)^2 

= G. (4/3). Pi. (R-h). p. m

Again at point A, the acceleration due to gravity (say g2) = F/m = G. (4/3). Pi. (R-h). p  _________________ (7)

Now we know at earth’s surface, g = (4/3) Pi R p G

Taking the ratio, again,

g2/g 

= [G. (4/3). Pi. (R-h). p ] / [(4/3) Pi R p G] 

= (R-h) / R = 1 – h/R.

=> g2 = g (1 – h/R)  ______ (8)

So as depth h increases, the value of acceleration due to gravity falls.

In the next paragraph we will compare these 2 equations to get a clearer picture.

Equations – formula- Comparison

g1 = g (1 – 2h/R) 

at a height h from the earth’s surface

g2 = g (1 – h/R)  

at a depth h from the earth’s surface

1) Now from eqn 6 and 8 we see that both g1 and g2 are less than g on the earth’s surface.

2) We also noticed that, g1 < g2

And that means:

1) value of g falls as we go higher or go deeper.

2) But it falls more when we go higher.