Q: A body weighs \$63 \$N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

Weight of the body,\$W=63\ N\$Acceleration due to gravity at height \$h\$ from the Earths surface is given by the relation:\$g'=\dfrac{g}{(1+\dfrac{h}{R_e})^2}\$Substituting \$h=R_e/2\$,\$g'=4g/9\$Weight of body of mass m at height h is given as:\$W'=mg'\$ \$=4/9\ mg\$ \$=4/9\times 63=28\ N \$

Q: The depth \$'d'\$ at which the value of acceleration due to gravity becomes \$\dfrac {1}{n}\$ times the value at the earth's surface is \$(R =\$ radius of earth)

Acceleration due to gravity at a depth \$d\$ under the earth surface \$g' = g_s\bigg(1-\dfrac{d}{R}\bigg)\$Given : \$g' = \dfrac{g_s}{n}\$\$\therefore\$ \$\dfrac{g_s}{n} = g_s\bigg(1-\dfrac{d}{R}\bigg)\$Or \$\dfrac{1}{n} = \bigg(1-\dfrac{d}{R}\bigg)\$\$\implies\$ \$d =R\bigg(\dfrac{n-1}{n}\bigg)\$

Q: The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

\$\cfrac { GMm }{ { \left( R+h \right) }^{ 2 } } =\cfrac { 1 }{ 16 } \cfrac { GMm }{ { R }^{ 2 } } \\ \cfrac { 1 }{ { { \left( R+h \right) }^{ 2 } } } =\cfrac { 1 }{ 16{ R }^{ 2 } } \\ or,\quad { \left( \cfrac { R }{ R+h } \right) }^{ 2 }=\cfrac { 1 }{ 16 } \\ \Rightarrow \cfrac { R }{ R+h } =\cfrac { 1 }{ 4 } .\\ \therefore R+h=4R\\ \Rightarrow h=3R\$

Q: The change in the value of \$g\$ at a height \$h\$ about the surface of the earth is the same as at a depth \$d\$ below the surface of earth. When both \$d\$ and \$h\$ are much smaller than the radius of earth, then which one of the following is correct?

Acceleration due to gravity at a height \$h\$, \$g'_h = g \left (1 - \dfrac {2h}{R}\right )\$ Acceleration due to gravity at a depth \$d\$, \$g'_d = g \left (1 - \dfrac {d}{R}\right )\$\$\therefore \ g \left (1 - \dfrac {2h}{R}\right )\$ \$ = g \left (1 - \dfrac {d}{R}\right )\$or \$\dfrac {2h}{R} = \dfrac {d}{R}\$ or \$d = 2h\$.

Q: A body weighs \$72\ N\$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

At the surface of Earth, weight \$W = 72N\$\$\therefore F = mg = 72N\$\$m = \dfrac{72}{10} = 7.2 kg\$Now, at some height, h acceleration due to gravity is given as \$g' = \dfrac{g}{(1+\dfrac{h}{R})^2}\$Here, \$h =\dfrac{R}{2}\$\$g' = \dfrac{g}{(1+\dfrac{1}{2})^2} =\dfrac{4}{9}g = 4.44m/s^2\$Now, the force experienced by the body at half of the radius of the earth is given by: \$F'=mg' = 7.2 \times 4.44 = 31.97N \approx 32N \$

Q: The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Acceteration due to gravity at height h, \$g_n=g_0 (1-\dfrac{2h}{R})\$ \$ h=1km\$Acceleration due to gravity at depth d,\$d_d = g_0\left(1-\dfrac{d}{R}\right)\$\$g_h=g_d\$\$g_0(1-\dfrac{2h}{R})=g_0(1-\dfrac{d}{R})\$\$\rightarrow d=2h\$\$= 2\times 1 km\$\$d=2km\$

Q: A body weighs \$72 N\$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Weight of body on the surface of the earth \$= mg= 72 N\$Acceleration due to gravity at height h is \$g_{h}=\dfrac{gR_{E}^{2}}{\left(R_{E}+h\right)^{2}}\$Substitute \$ h=\dfrac{R_{E}}{2}\$ in above expression:\$g_h=\dfrac{g{R_{E}^{2}}}{\left(R_{E}+\dfrac{R_{E}}{2}\right)^{2}}\$\$=\dfrac{4}{9}g\$Gravitational force on body at height h is \$F=mg_{h}\$\$=m\times\dfrac{4}{9}g=\dfrac{4}{9}\times mg\$\$=\dfrac{4}{9}\times 72N=32N\$

Question 1

Discuss the variation of g with(a) altitude (b) depth

Accepted Answer

( a )  The acceleration due to gravity on the earth is given by&#10;&#10;$g = \dfrac {GM}{R^2} i.e. GM = R^2g$  . . . . . . .( 1 )&#10;&#10;The acceleration due to gravity at height 'h' from the surface of the&#10;earth is given by&#10;&#10;$g_h = \dfrac{GM}{(R+h)^2} i.e. GM = (R+h)^2 g_h$ . . . . . . ( 2 )&#10;&#10;From ( 1 ) and ( 2 ) we have,&#10;&#10;$\dfrac{g_h}{g} = \dfrac {R^2}{(R+h)^2}$&#10;&#10;$\dfrac{g_h}{g} = (1 - \dfrac{2h}{R})$  &#10;&#10;( b )  The acceleration due to gravity on the surface of the earth&#10;is given by.&#10;&#10;$g = \dfrac{GM}{R^2}$ . . . .( 1 )&#10;&#10;let 'Q' be the density of the material of the earth.&#10;&#10; Now, mass = volume $\times$ density&#10;&#10; $M = \dfrac{4}{3} \pi R^3 \times \rho$&#10;&#10; Substituting in equation ( 1 ) we get &#10;&#10; $g = \dfrac{G}{R^2} \times \dfrac{4}{3} \pi R^3 \times \rho = \dfrac{4}{3}&#10;\pi GR \rho$&#10;&#10; $g = \dfrac{4}{3} \pi GR \rho$ . . .. .. ( 2 )&#10;&#10; $g_d = \dfrac{4}{3} \pi G(R-d) \rho$  . . . .. .( 3 )&#10;&#10; Dividing equation( 3 ) by ( 2 ) we get &#10;&#10; $\dfrac{g_d}{g} = \dfrac{R-d}{R} = (1 - \dfrac{d}{R})$&#10;&#10; $g_d = g ( 1 - \dfrac{d}{R})$

Question 2

A body weighs $63 $N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

Accepted Answer

Weight of the body,$W=63\ N$Acceleration due to gravity at height $h$ from the Earths surface is given by the relation:$g'=\dfrac{g}{(1+\dfrac{h}{R_e})^2}$Substituting $h=R_e/2$,$g'=4g/9$Weight of body of mass m at height h is given as:$W'=mg'$       $=4/9\ mg$       $=4/9\times 63=28\ N $

Question 3

The depth $'d'$ at which the value of acceleration due to gravity becomes $\dfrac {1}{n}$ times the value at the earth's surface is $(R =$ radius of earth)

Accepted Answer

Acceleration due to gravity at a depth $d$ under the earth surface   $g' = g_s\bigg(1-\dfrac{d}{R}\bigg)$Given :  $g'  = \dfrac{g_s}{n}$$\therefore$   $\dfrac{g_s}{n} = g_s\bigg(1-\dfrac{d}{R}\bigg)$Or   $\dfrac{1}{n} = \bigg(1-\dfrac{d}{R}\bigg)$$\implies$  $d =R\bigg(\dfrac{n-1}{n}\bigg)$

Question 4

The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

Accepted Answer

$\cfrac { GMm }{ { \left( R+h \right)  }^{ 2 } } =\cfrac { 1 }{ 16 } \cfrac { GMm }{ { R }^{ 2 } } \ \cfrac { 1 }{ { { \left( R+h \right)  }^{ 2 } } } =\cfrac { 1 }{ 16{ R }^{ 2 } } \ or,\quad { \left( \cfrac { R }{ R+h }  \right)  }^{ 2 }=\cfrac { 1 }{ 16 } \ \Rightarrow \cfrac { R }{ R+h } =\cfrac { 1 }{ 4 } .\ \therefore R+h=4R\ \Rightarrow h=3R$

Question 5

The change in the value of $g$ at a height $h$ about the surface of the earth is the same as at a depth $d$ below the surface of earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct?

Accepted Answer

Acceleration due to gravity at a height $h$,  $g'_h = g \left (1 - \dfrac {2h}{R}\right )$ Acceleration due to gravity at a depth $d$,  $g'_d = g \left (1 - \dfrac {d}{R}\right )$$\therefore \  g \left (1 - \dfrac {2h}{R}\right )$ $ = g \left (1 - \dfrac {d}{R}\right )$or $\dfrac {2h}{R} = \dfrac {d}{R}$ or $d = 2h$.

Question 6

A body weighs $72\ N$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

Accepted Answer

At the surface of Earth, weight $W = 72N$$\therefore F = mg = 72N$$m = \dfrac{72}{10} = 7.2 kg$Now, at some height, h acceleration due to gravity is given as $g' = \dfrac{g}{(1+\dfrac{h}{R})^2}$Here, $h =\dfrac{R}{2}$$g' = \dfrac{g}{(1+\dfrac{1}{2})^2} =\dfrac{4}{9}g = 4.44m/s^2$Now, the force experienced by the body at half of the radius of the earth is given by: $F'=mg' = 7.2 \times 4.44 = 31.97N \approx 32N $

Question 7

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

Accepted Answer

Acceleration due to gravity at a depth d from surface of earth $g' = g \left(1 - \dfrac{d}{R} \right) $ ....(1)Where g = acceleration due to gravity at earth's surface Multiplying by mass 'm' on both sides of (1)$\implies mg' = mg \left(1 - \dfrac{d}{R} \right) \, \,\, \, Here, \left(d = \dfrac{R}{2} \right)$$= 200 \left(1 - \dfrac{R}{2 R} \right) = \dfrac{200}{2} = 100 N$

Question 8

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Accepted Answer

Acceteration due to gravity at height h, $g_n=g_0 (1-\dfrac{2h}{R})$  $ h=1km$Acceleration due to gravity at depth d,$d_d = g_0\left(1-\dfrac{d}{R}\right)$$g_h=g_d$$g_0(1-\dfrac{2h}{R})=g_0(1-\dfrac{d}{R})$$\rightarrow d=2h$$= 2\times 1 km$$d=2km$

Question 9

A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Accepted Answer

Weight of body on the surface of the earth $= mg= 72 N$Acceleration &#10;due to gravity at height h is &#10;$g_{h}=\dfrac{gR_{E}^{2}}{\left(R_{E}+h\right)^{2}}$Substitute $ h=\dfrac{R_{E}}{2}$ in above expression:$g_h=\dfrac{g{R_{E}^{2}}}{\left(R_{E}+\dfrac{R_{E}}{2}\right)^{2}}$$=\dfrac{4}{9}g$Gravitational force on body at height h is $F=mg_{h}$$=m\times\dfrac{4}{9}g=\dfrac{4}{9}\times mg$$=\dfrac{4}{9}\times 72N=32N$

Question 10

Explain why the value of $g$ is zero at the centre of the earth.

Accepted Answer

At the centre of Earth, the force due to upper half of the Earth will cancel the force due to lower half. In the similar manner, force due to any portion of the Earth at the centre will be cancelled due to the portion opposite to it. Thus, the gravitational force at the centre on any body will be $0$. Since, from Newton's law, we know $F=mg$. Since, mass $m$ of an object can never be $0$. Therefore, when $F = 0$, then $g$ has to be $0$. Thus, the value of $g$ is zero at the centre of Earth.

Acceleration due to gravity and its variation with height

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REVISE WITH CONCEPTS

Variation in the Value of Acceleration Due to Gravity

QUICK SUMMARY WITH STORIES

Variation of g with Height

Variation of g with depth

Discuss the variation of g with
(a) altitude (b) depth

A body weighs $63$ N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

The depth $^{'} d^{'}$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R =$ radius of earth)

The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

The change in the value of $g$ at a height $h$ about the surface of the earth is the same as at a depth $d$ below the surface of earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct?

A body weighs $72 N$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Explain why the value of $g$ is zero at the centre of the earth.

Acceleration due to gravity and its variation with height

Language

Rate

REVISE WITH CONCEPTS

Variation in the Value of Acceleration Due to Gravity

QUICK SUMMARY WITH STORIES

Variation of g with Height

Variation of g with depth

Discuss the variation of g with (a) altitude (b) depth

A body weighs 63N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

The depth ′d′ at which the value of acceleration due to gravity becomes n1​ times the value at the earth's surface is (R= radius of earth)

The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

The change in the value of g at a height h about the surface of the earth is the same as at a depth d below the surface of earth. When both d and h are much smaller than the radius of earth, then which one of the following is correct?

A body weighs 72 N on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

A body weighs 72N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Explain why the value of g is zero at the centre of the earth.

Discuss the variation of g with
(a) altitude (b) depth

A body weighs $63$ N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

The depth $^{'} d^{'}$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R =$ radius of earth)

The change in the value of $g$ at a height $h$ about the surface of the earth is the same as at a depth $d$ below the surface of earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct?

A body weighs $72 N$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Explain why the value of $g$ is zero at the centre of the earth.