Question 195: Gravitation

Question

The height ‘h’ at which the weight of a body will
be the same as that at the same depth ‘h’ from
the surface of the earth is (Radius of the earth
is R and effect of the rotation of the earth is
neglected)

Accepted Answer

M = mass of earth
M
1
= mass of shaded portion
R
e
= Radius of earth
M
1
= $${M \over {{4 \over 3}\pi {R^3}}}.{4 \over 3}\pi {\left( {R - h} ight)^3}$$
= $${{M{{\left( {R - h} ight)}^3}} \over R^3}$$
Given, Weight of body is same at P and Q
$$ 	herefore $$ mg
P
= mg
Q
$$ \Rightarrow $$ g
P
= g
Q
$$ \Rightarrow $$ $${{G{M_1}} \over {{{\left( {R - h} ight)}^2}}} = {{GM} \over {{{\left( {R + h} ight)}^2}}}$$
$$ \Rightarrow $$ $${{GM{{\left( {R - h} ight)}^3}} \over {{R^3}{{\left( {R - h} ight)}^2}}} = {{GM} \over {{{\left( {R + h} ight)}^2}}}$$
$$ \Rightarrow $$ (R Ã¢ÂÂ h) (R + h)
2
= R
3
$$ \Rightarrow $$ R
3
Ã¢ÂÂ hR
2
+ h
2
R Ã¢ÂÂ h
3
+ 2R
2
h Ã¢ÂÂ 2Rh
2
= R
3
$$ \Rightarrow $$ R
2
h
Ã¢ÂÂ Rh
2
Ã¢ÂÂ h
3
= 0
$$ \Rightarrow $$ R
2
Ã¢ÂÂ Rh Ã¢ÂÂ h
2
= 0
$$ \Rightarrow $$ h
2
+ Rh Ã¢ÂÂ R
2
= 0
$$ \Rightarrow $$ h = $${{ - R \pm \sqrt {{R^2} + 4{R^2}} } \over 2}$$
$$ \Rightarrow $$ h = $${{ - R + \sqrt 5 R} \over 2}$$

Answer

$${R \over 2}$$

Answer

$${{\sqrt 5 R - R} \over 2}$$

Answer

$${{\sqrt 3 R - R} \over 2}$$

Answer

$${{\sqrt 5 } \over 2}R - R$$

Gravitation

The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

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