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)\$

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>>Class 11
>>Physics
>>Gravitation
>>Variation in Value of Acceleration due to Gravity
>> $Variation in the Value of A...$

Variation in the Value of Acceleration Due to Gravity

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

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

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 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 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\$

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

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\$

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

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.

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\$

Physics

definition

Variation in acceleration due to gravity

Value of acceleration due to gravity changes with height and depth from the earth's surface. It is maximum on the earth's surface. Its variation with height (or depth) is shown in the plot.

result

Approximate variation of acceleration due to gravity with height and depth

At a height

h < < R_{e}

, approximate expression for acceleration due to gravity is given by:

g^{'} = g (1 - \frac{2 h}{R _{e}})

At a depth

d

, expression for acceleration due to gravity is given by:

g^{'} = g (1 - \frac{d}{R _{e}})

definition

Effect of non-spherical shape of earth on the value of acceleration due to gravity

Earth is not perfectly spherical in shape. It is flattened at poles and bulging at the equator. Hence, value of acceleration due to gravity is more on the poles than at the equator.

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Variation in acceleration due to gravity

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Acceleration due to gravity and its variation with height

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Quick Summary With Stories

Variation of g with Height

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Variation of g with depth

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Important Questions

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

Hard

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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 ?

Medium

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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)

Medium

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

Medium

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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?

Medium

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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?

Medium

NEET

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

Medium

NEET

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

Medium

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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?

Medium

View solution

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:

Medium

NEET

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definition

Variation in acceleration due to gravity

result

Approximate variation of acceleration due to gravity with height and depth

definition

Effect of non-spherical shape of earth on 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 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?

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

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

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?

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:

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 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.

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?